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Looking at the information shown here there is nothing showing a circuit that has a jfet in it. See http://www2.eng.cam.ac.uk/~dmh/ptialcd/csink/csink.htm for an example. I am willing to draw up such a circuit. Chaney44145 ( talk) 18:50, 28 December 2007 (UTC)
Against: Galvanostat article is more technical, this keeps more general. Tswsl1989 12:06, 8 May 2007 (UTC)
Doesn't belong in article. Not a current source. Same argument as Talk:Voltage source#Potential divider "source"?? - Omegatron 14:53, August 8, 2005 (UTC)
Most people seem to want to limit the discussion of current sources to those of which they have heard and are not prepared to look at the wider view of the concept.
If people want to limit this page's content to current source circuits based on semiconductors then that is fine by me. But then it should be called 'current sources made from semiconsuctors' or such like and we put the other sorts of current sources in a different article. Exactly the same argument I apply to te voltage source page where people are trying to limit content to what they have personal experience of. Surely Wikipedia is not intended to be narrow minded like that is it?? Comments please from other interested parties Light current 16:22, 8 August 2005 (UTC)
Unfortunatley ther seems to be a little confusion over the terms describing current sources in this article. Some of the terms that have already been used and applied (too) loosely are:
Ideal
Non_ideal
Independent
Non-independent (or controlled)(CCS)
Practical
Theoretical
Voltage controlled (VCCS)
Current controlled (CCCS)
(Real world)??
In order to avoid gross confusion and endless arguments it is important for editors to understand the proper meaning of these terms before attempting to use them in this article. Any good book on electrical or electronic engingeering should define these terms adequately. Light current 16:47, 8 August 2005 (UTC)
Just a thought to try to resolve this issue. If we we put all the types of current (voltage) sources that 'O' and others like under heading called Active Device Sources and put the others under another heading (in the same article )would that be acceptable to everyone. Please discuss.Answers in no more than 50 words pls Dave Bowman Light current 20:03, 8 August 2005 (UTC)
How about this for terminology, this may seem obvious but shouldn't the simple transistor constant current source be called a constant current sink? It is not sourcing current, but rather sinking current from the load...Just a thought.
I agree with this last comment - all the circuits described are current sinks rather than sources. Mike. 80.195.172.5 ( talk) 10:05, 7 March 2012 (UTC)
I have some questions on the bootstrap insertion.
1) It says the bootstrap cct is used to bias the class A stage. But this is the Class A stage so how is the transistor biased by the this cct?
2) How is the collector current of Q determined by R1 or R2 exactly?
3) Could you explain your last paragraph in more detail
THanks Light current 21:36, 9 August 2005 (UTC)
When I studied electronics at university (many years ago now), we were taught that bootstrapping is a form of positive feeback used to increase the input impudance of a stage and improve the dynamic range of large signal circuits. Your circuit does not have any feedback so can you say how your circuit achieves these two aims? Or has the definition of bootstrapping in the high end audio world been changed without my knowledge? Light current 21:49, 9 August 2005 (UTC)
Added references section, the page pointed out by omegatron is the source for the boot-strap section here. Now, should this section be moved back to the article or is it best omitted...? Rohitbd 08:55, August 10, 2005 (UTC)
Although not as good as the active constant current sources depicted above, the boot-strap current source is commonly found in many audio power amplifiers to bias the class-A voltage gain stages in them. These offer a cheap alternative to active sources where the output impedance of the CCS is of secondary concern. As can be seen in the image on the left, a boot-strap CCS is formed by resistors R1, R2 and capacitor CB. Transistor Q is wired as a class-A voltage gain stage. The circuit works as follows:
At no signal (i.e., DC conditions), CB acts as an open circuit and the collector current is determined simply by (R1 + R2).
Assume that R1 and R2 are chosen such that the DC bias current (collector current) is much larger than the load connected to the collector. When a signal is applied to the transistor's base, the collector voltage will change. This change in voltage is coupled back to R2 by CB in such a way that the voltage across R2 remains relatively constant. Constant voltage across R2 means constant current through R2 which is also the collector current, and hence R1, R2 and CB form a CCS. This CCS is a passive current source, and the transistor Q forms the load for it.
Calculation of R1 & R2: the collector voltage (VC) is assumed to be at half the total supply (Vsupply) - in this case since the emitter is shown floating, it could be either connected to an equal −Vcc or ground. In either case we assume VC equals half the total supply.
The reason for assuming that VC is half the total supply is that this particlar configuration appears in a VFB design - similar to that of an op-amp in discrete form with a differential pair input stage and DC coupling internally. When negative feedback is applied around the ckt, the o/p (the collector of the transistor, in this case) always sits mid-way of the total supply - i.e., if we have a split ±VCC, then the o/p will be at 0V. If a single supply (VCC) is used then the o/p will be at VCC/2.
If the next stage requires say, Y mA of current, then this transistor must be biased at atleast 1.2 to 2 times this for class-A operation. We take twice that (2Y mA) for calculation, call it X mA. Now, KΩ. Usually R1 = R2. Therefore KΩ. Since VC = Vsupply / 2, therefore, KΩ. The boot-strap capacitor is calculated simply by assuming a lowest frequency (f) of operation (say 1 Hz) and applying the formula Farads, where RP = R1 || R2. Since most power amps have very little gain below about 10 Hz by design, it means that at low frequencies (as 1 or 2 Hz), R1 and R2 by themselves are sufficient to ensure adequate bias. For R1 = R2 = 2.2KΩ, and f = 1Hz, CB turns out to be about 114.7µF and most amps use the standard 100µF with no degradation in performance. Rohitbd 09:05, August 10, 2005 (UTC)
Moved from my talk page Rohitbd 08:04, August 11, 2005 (UTC)
I assume you wrote the 'bootstrap' piece in current sources.
THe trouble is that, as O pointed out, half the diagram is missing. I dont think it can be copied 'en bloc'. because of copyright. It needs redrawing fully before the text surrounding it can be discussed. Are you any good at drawing ccts??
It does seem to act as a current source drive and so maybe should be included. But maybe it should be included in an audio amplifier page. THis is where I think these are most used.:-) Light current 15:04, 10 August 2005 (UTC)
Ic = BIb Where B is the gain of the transistor. This is the only way the collector current can be modulated. The collector has a very high resistance so its current is not affected by the voltage on it (much). Please try to get hold of some books on transistors. I think you will find them quite interesting and informative. :-) Light current 00:48, 11 August 2005 (UTC)
BTW Ic = BIb + Iceo.( where B is the gain of the transistor and Iceo is the leakage current). This is the basic transistor equation developed by Ebers + Moll. Collectors are not biased as you thought, it is the base emmitter junction that has to be biased to control the collector currrent in accordance with the above equation. THe collector terminal is a high impedance terminal that does not respond to applied voltages(much). THat is why the collector of a transistor is sometimes used as a current source (because it is a high impedance point) Do you understand it now? Light current 13:13, 11 August 2005 (UTC)
If you dont know how a transistor operates, who are you to tell me about current sources?? Light current 01:37, 11 August 2005 (UTC)
THe effective emitter load is R1||R2||R3. Let us call this R' For an input Vat the base, the voltage across R4 is V-KV wher k is the the voltage gain of the transistor and is approx 1. From the poin of view of input signal (ac) the input of the stage appears as follows: a simple emitter follower with a resitor R4 connected to the base. THe input to the stage is still directly at the base as normal, but the other end of R4 now as avoltage on it (from the emitteer) of KV (Yes?). So, the current thro this resistor R4 is: I =V(1-K)/R4. So the total input resistance as seen by the input signal is now the original input resistance in parallel with Reff. Where Reff= V/i= R4/(1-K) NOW, as K approaches 1, THe new input resistance approaches Rin which is roughly (1+hfe)R'. So the input resistance of the stage has been increased by the bootstrap and the effects of the bias chain are now negligible. THe important point is that an extra source of energy has to be used in a bootstrap circuit and this is provided by the current supplied to the capacitor from the transistor emitter. This is why your circuit does not work-- there is no extra source of energy invoked. Hope this resolves your difficulty. Light current 14:57, 11 August 2005 (UTC)
Moved from my talk page Rohitbd 15:30, August 11, 2005 (UTC)
I have replied on Talk:current source. I hope you will find the argument convincing. BTW Keep Calm -I'm only interested in the accuracy of WP :-) Light current 15:22, 11 August 2005 (UTC)
Regardless of whether this is right or not, you really shouldn't be posting a circuit that you designed. We have a policy of no original research. You can definitely write about the one included on the Westhost site, though. - Omegatron 23:50, August 11, 2005 (UTC)
Please read this article: http://www.eecs.utoledo.edu/~rking/elecdesign/EDmanualB.pdf
The bootstrapping technique is described just before the middle of the article. Although this schematic also has an emitter-follower, according to it, R2A + R2B (R1 + R2 as per my description & schematic) determines the Q-point current of the transistor - A point that I made, and was opposed by Light Current. Also the article has no mention of the emitter follower being required (or otherwise), so I guess my assumption for the sake of describing it is neither right nor wrong. The effect of the emitter follower is simply that the voltage at the output end of the bootstrap capacitor is just less than that at the collector - in my case it would simply be equal. In short, the resulting constant current is given by (K.Vx + Vb)/R2B as per the article. In the PDF article, K < 1 (due to emitter-follower Vbe drop), Vx is the collector voltage and Vb is the voltage at the emitter of the emitter-follower which is 0.65V less than Vx. In my case, looking at the schematic, K = 1 and Vb = 0 since no emitter-follower is there, so as per the equation it still works. It becomes Vx/R2 in my schematic.
I hope this clarifies my point of view a bit. Rohitbd 15:12, 12 August 2005 (UTC)
is incorrect. R2a and R2b do NOT determine the Q point current of Q2. What they do do is determine the VOLTAGE at the collector of Q2 (in conjuction with the bias network (which is probably an amplified diode voltage source)). The rest of the page does appear to describe bootstrap operation correctly although I have not checked every line. This article apears to have been written by a student and therfore can not be given much credence. Any one agree?? Light current 15:27, 12 August 2005 (UTC)
Exactly!! Light current 16:11, 15 August 2005 (UTC)
Moved from my talk page Rohitbd 07:35, August 12, 2005 (UTC)
As it says on each and every editing page:
If you do not want your writing to be edited mercilessly and redistributed at will, do not submit it
Now please dont be a sore loser. We all have to learn. Its just that you have chosen the hard way to do it. Light current 01:13, 12 August 2005 (UTC)
This discussion has been going on for very long now. A couple of points...
In my explanation of the circuit, I described it from the point of view of designing and not analysing the circuit - although I termed it as "analysis" - for want of a better word (sorry about that). Also, I do not feel that the ckt requires in-depth analysis or design steps, it can be described simply in a few words. If needed, the schematic can be updated to include an emitter-follower, but IMHO it makes no difference to the operating principle. Rohitbd 08:56, August 15, 2005 (UTC)
Alfred, my view is assuming CCS loads and global -ve feedback circuits. All along I had a CCS instead of a collector load in mind while discussing the bootstrap principle. I don't know, maybe repetitive work with VFB amps has perhaps put me into the habit of always using a CCS to bias the individual stages of the amp. Nevertheless, as defence of my argument that the collector resistor can be used to bias the transistor, let me give an example, I just designed and simulated this CE amplifier as per my method — These are the known parameters and design goals:
Step 1: Assumptions
Step 2: Calculate Rc & Re
Step 3: Calculate base voltage divider
After calculations, the actual voltage gain into a 4K load is about 9.6 as simulated (instead of the intended 10), practically we can expect about 9. In terms of decibels, this is a difference of about 0.35db (1db practically), and the voltage gain is stable across following transistors: BC547, BC337, 2N2222 and BD139. This can be improved by using a CCS instead of Rc and using global negative feedback. Rohitbd 18:03, August 15, 2005 (UTC)
<insert>
</insert>
In the last section of this article, there is the statement that an ideal voltage source cannot be connected to an ideal short circuit. Well, here is a thought experiment to consider. Connect a resistor to a voltage source. The voltage across the resistor is the voltage of the source and the current is just the ratio of the source voltage and the resistance of the resistor. Now, let the resistance tend to zero... At all times, the voltage across the resistor is the voltage across the source and the current increases without bound. In the limit, we have that the voltage across the resistor of zero ohms is still the source voltage. How can this be? It is because the current through the short circuit has passed to infinity and recall that infinity multiplied by zero is an indeterminate form. In order to determine the value, a limit process must be used just as we did above. A similar argument can be made for an ideal current source without an external circuit connected (infinite voltage multiplied by zero conductance). Enjoy! Alfred Centauri 02:09, 16 August 2005 (UTC)
Ok, now that I have received enough brick-bats, would anybody care to present an analyis of the bootstrap CCS? I tried and was opposed vehemently (not to mention personal attacks)...so I guess it is only fair of me to ask my detractors to give an analysis themselves. Of course, there are no obligations - but for a change, why don't you try and then include it in the article? Rohitbd 13:19, August 16, 2005 (UTC)
(see also Current and Voltage Source Suggestion)
Hi everybody! Two months ago, I placed a link on constant current source page pointing to my story about this subject. In this way, I tried to resume the discussion about constant current source and the dual voltage source. Browsing through these and talk pages I found a lot of brilliant thoughts. Only, the pages look quite cluttered; there isn't good structure and hierarchy in their arrangements. So, I suggest rearranging the materials according to the principles below:
1. At every page, expound the subject step-by-step by moving from simple to complex (imperfect to perfect, passive to active, transistor- to op-amp versions etc.)
2. At every step, first reveal the basic circuit idea; then show the concrete circuit solution.
In order to illustrate my suggestion, I have made a web page containing two versions of the contents - a short and a long one. Here is the short version:
1. Theoretical current source.
1.1. Ideal current source: definition, examples of natural current sources.
1.2. Comparison between current and voltage sources.
1.3. Real current source: imperfections.
2. Practical current sources.
2.1. Ohmic resistor current source. Imperfections.
2.2. Dynamic resistor current sources.
2.3. Current sources with compensating voltage.
2.4. Current sources with compensating current.
2.5. Current sources using negative feedback.
1. Theoretical voltage source.
1.1. Ideal voltage source: definition, examples of natural voltage sources.
1.2. Comparison between voltage and current sources.
1.3. Real voltage source: imperfections.
2. Practical voltage sources.
2.1. Voltage divider source. Imperfections.
2.2. Dynamic resistor voltage sources.
2.3. Voltage sources with compensating negative resistance.
2.4. Voltage sources using negative feedback.
I realize that I have proposed major changes in the structure of these pages. So, I would like first to coordinate them with you, especially with those who are created these pages; then, I will try to modify gradually (in small steps) the pages. But yet, don't remember that I am a Wiki newbie; so,
don't bite me desperately:)
Circuit-fantasist 16:24, 29 June 2006 (UTC)
I have reordered a bit the galvanostat voice and have written a brief comment on its talk page Talk:Galvanostat: please give it a look :). Daniele.tampieri 14:54, 24 December 2006 (UTC)
Rsduhamel, you have to connect a resistor in series with the zener diode on the circuit diagram of an op-amp current source. Circuit-fantasist 08:49, 26 January 2007 (UTC)
Circuit-fantasist, I presume that you are referring to Figure 7, where the Zener diode is connected between the op-amp's non-inverting (+) input and ground. No resistor is needed in this portion of the circuit. Current limiting for the Zener diode is provided by resistor between the op-amp's non-inverting input and the power supply. In fact, including a resistor in the same branch as the Zener diode will prevent the voltage across the sense resistor from tracking the voltage across the Zener diode. That would be prevent the circuit from working properly as a constant current source. The circuit is designed so that if the voltage across the sense resistor should increase (indicating an increased current flow), the op-amp will reduce the output voltage, thereby causing a reduction in the current through the load and sense resistor. If the voltage across the sense resistor should decrease (indicating a reduced current flow), the opposite happens. For the constant current in the load to be stable, the voltage at the op-amp's non-inverting input needs to be stable and the resistance of the sense resistor needs to be stable for the expected currents. Thus, no additional resistor is needed. OhioFred ( talk) 22:53, 17 September 2014 (UTC)
Circuit-fantasist, perhaps you meant to remark on the fact that the op-amp inputs are biased near the Zener diode voltage and the op-amp output is biased near the power supply voltage. See the comment below (in section Mistake in Figure 7: Typical op-amp current source) regarding the placement of the load resistance. The solution recommended there would reduce the bias differences. (Don't forget that the op-amp positive and negative supply voltages may differ from the circuit voltages. For best circuit performance, the op-amp should operate half-way between its positive and negative supply voltages.) OhioFred ( talk) 02:35, 18 September 2014 (UTC)
Alfred Centauri, I highly appreciate your last edits. Now, the section about the so-called " resistor current source" is where it has to be - at the beginning of the page.
The ESP link is a 'leftover' from the original text I edited. Had I bothered to read it, I would have removed it myself. Good catch.
On another note, please do keep in mind that this is an encylopedia, not a text book. That is, Wikipedia is a reference. To keep the articles concise, there must be some assumption of familiarity with the subject material.
I've looked at some of your contributions (voltage-to-current converter and others) and, while I'm impressed with the skills you display for presenting this material for learning and the depth of the information you have put into these articles, it is my opinion that Wikipedia is not the proper venue for this type of presentation. Have you considered making contributions to Wikibooks? The EE sections there could certainly use your help.
Best regards, Alfred Centauri 00:53, 10 April 2007 (UTC)
Solar cells are generally regarded as a current source in parallel with a diode and a shunt resistance with an extra resistor connected in series. My background is in Physics as opposed to electronics but are there any objections to my adding this to the page? —The preceding unsigned comment was added by 155.198.186.155 ( talk) 17:30, 16 May 2007 (UTC).
The simplest way to drive an LED from line-voltage AC is with a large series resistor, which functions as a current source. Very inefficient, due to power loss in the resistor.
The article implicitly only covers DC current sources. AC is a related but more complex subject. But some simple cases are important. We could also drop line voltage to a low voltage to drive an LED with an AC capacitor. We can calculate power in the capacitor by V x A, but the result is not real watts - the answer is ideally all "imaginary" VA? In the real world, how do we calculate power dissipation in real capacitors used thus: how hot they would get, what would be safe design, etc? For example, if two LEDs are anti-parallel, and you want the total average current through the pair to be 20 mA, from a series capacitor, at 120 VAC, what capacitance is needed, what capacitor types would be appropriate, and would power issues in the capacitor be a concern?- 69.87.204.197 13:20, 28 May 2007 (UTC)
91.75.70.167 ( talk) 04:08, 3 February 2008 (UTC) A batterie's function is to maintain a constant potential difference about itself/ in the ckt. Does a current producing source, work on the same basis? ie. at every instant of time, its job is to provide a constant supply of current? If yes, then how will we balance the produced current?
"The current then flowing is the IDSS of the FET."
Please list typical values for IDSS. What is the physical reason for the limitations on this value? It seems from CLD datasheets that this is only a solution for currents below about 15 mA. See [6], for instance.
If more than one CLD can be placed in parallel on a PCB, why can't they manufacture multiple CLDs in parallel on the chip? Wouldn't that be exactly the same as paralleling a bunch of transistors on the same chip to make a power transistor? — Omegatron 18:47, 21 March 2008 (UTC)
Where is the inductor on the LM317 circuit in the section "Inductor type current source"? Section title typo? I think it should be "Linear type current source". — Preceding unsigned comment added by 213.63.100.45 ( talk • contribs) 02:03, 11 December 2010
A section title says "Current sources with series negative feedback" and then the first part "Simple transistor current source" doesn't even mention the word feedback. Later there is a negative series feedback circuit presented; however it is not mentioned that the first bjt circuit already in itself has a negative series feedback.
If temperature doesn't change, then Re emitter resistor is not necessary (in theory only!) as the Ic collector current is set by the Vbe base-emitter voltage. But in this case we can't make the assumption that Vbe=0.65V or so, because a very little change in Vbe results in big changes in Ic. (See http://diranieh.com/Electrenicas/Figures/BJT_EbersMollGraph.jpg for example, don't know why the bjt article doesn't have these curves). And of course that's one reason we use the RE resistor, as we can't guarantee a constant base potential (though the zener diode is a good try). The Re resistor kind of forces the current onto the transistor, as small changes in Vbe will not change the emitter resistor's current much. But still it is possible to change the output current not only by Re but by changing the base voltage (if using a potmeter instead of a Zener).
Now for the feedback: the voltage across emitter resistor Re is proportional to the output current. If the temperature rises, then Ic collector current increases - but then the Ve emitter potential will also increase. As the Vb base potential is fixed by the diode or resistor divider, and Vbe=Vb-Ve so Vbe will decrease, which acts towards decresing the collector current. This is a series negative feedback. If you simulate the circuit without the emitter resistor you get a much worse temperature dependency. Of course Re is only good so far, but can be useful in many cases.
As for earlier comments: in the output curve, the collector current does have a very little dependancy on Vce, which can be a problem for a very precise circuit and this is also compensated by the feedbacks. (Also the finite value of h22e is also due to this). Hoemaco ( talk) 11:35, 21 April 2011 (UTC)
I think the load resistance in this circuit diagram should be located above the transistor. The reason for this is that if most of the voltage drops at the load (as intended), the voltage at the emitter of the transistor can easily become too big for it to conduct (since the base voltage needs to be at least 0.7V above the emitter voltage). For an PNP transistor the depicted way would be okay, but this is an NPN transistor. A detailed discussion of this kind of circuit is given at http://www.mikrocontroller.net/articles/Konstantstromquelle#Konstantstromquelle_mit_Operationsverst.C3.A4rker_und_Transistor (sorry, in German).
I already produced a corrected version of the image but can not replace the file because it's imported from wikimedia. — Preceding unsigned comment added by 5.158.164.237 ( talk) 01:09, 12 May 2013 (UTC)
This paragraph is misleading. A Van de Graaff generator is essentially a current source and must be regulated to maintain a constant voltage. The high voltage capacitor is charged by a belt on which charge is placed at low voltage and mechanically moved to a region of high voltage. When used a particle accelerator, the load current is determined by an ion source, which generates a constant current and does not resemble a resistor.
I think we need to include the "interlinked circles" symbol for a current source. See link. This is very commonly used in UK engineering practice and in text books/app notes/papers, and is what was used throughout my university course. I much prefer this symbol because it can be quickly and unambiguously interpreted as a current source. Amj4321 ( talk) 22:31, 22 November 2015 (UTC)
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Looking at the information shown here there is nothing showing a circuit that has a jfet in it. See http://www2.eng.cam.ac.uk/~dmh/ptialcd/csink/csink.htm for an example. I am willing to draw up such a circuit. Chaney44145 ( talk) 18:50, 28 December 2007 (UTC)
Against: Galvanostat article is more technical, this keeps more general. Tswsl1989 12:06, 8 May 2007 (UTC)
Doesn't belong in article. Not a current source. Same argument as Talk:Voltage source#Potential divider "source"?? - Omegatron 14:53, August 8, 2005 (UTC)
Most people seem to want to limit the discussion of current sources to those of which they have heard and are not prepared to look at the wider view of the concept.
If people want to limit this page's content to current source circuits based on semiconductors then that is fine by me. But then it should be called 'current sources made from semiconsuctors' or such like and we put the other sorts of current sources in a different article. Exactly the same argument I apply to te voltage source page where people are trying to limit content to what they have personal experience of. Surely Wikipedia is not intended to be narrow minded like that is it?? Comments please from other interested parties Light current 16:22, 8 August 2005 (UTC)
Unfortunatley ther seems to be a little confusion over the terms describing current sources in this article. Some of the terms that have already been used and applied (too) loosely are:
Ideal
Non_ideal
Independent
Non-independent (or controlled)(CCS)
Practical
Theoretical
Voltage controlled (VCCS)
Current controlled (CCCS)
(Real world)??
In order to avoid gross confusion and endless arguments it is important for editors to understand the proper meaning of these terms before attempting to use them in this article. Any good book on electrical or electronic engingeering should define these terms adequately. Light current 16:47, 8 August 2005 (UTC)
Just a thought to try to resolve this issue. If we we put all the types of current (voltage) sources that 'O' and others like under heading called Active Device Sources and put the others under another heading (in the same article )would that be acceptable to everyone. Please discuss.Answers in no more than 50 words pls Dave Bowman Light current 20:03, 8 August 2005 (UTC)
How about this for terminology, this may seem obvious but shouldn't the simple transistor constant current source be called a constant current sink? It is not sourcing current, but rather sinking current from the load...Just a thought.
I agree with this last comment - all the circuits described are current sinks rather than sources. Mike. 80.195.172.5 ( talk) 10:05, 7 March 2012 (UTC)
I have some questions on the bootstrap insertion.
1) It says the bootstrap cct is used to bias the class A stage. But this is the Class A stage so how is the transistor biased by the this cct?
2) How is the collector current of Q determined by R1 or R2 exactly?
3) Could you explain your last paragraph in more detail
THanks Light current 21:36, 9 August 2005 (UTC)
When I studied electronics at university (many years ago now), we were taught that bootstrapping is a form of positive feeback used to increase the input impudance of a stage and improve the dynamic range of large signal circuits. Your circuit does not have any feedback so can you say how your circuit achieves these two aims? Or has the definition of bootstrapping in the high end audio world been changed without my knowledge? Light current 21:49, 9 August 2005 (UTC)
Added references section, the page pointed out by omegatron is the source for the boot-strap section here. Now, should this section be moved back to the article or is it best omitted...? Rohitbd 08:55, August 10, 2005 (UTC)
Although not as good as the active constant current sources depicted above, the boot-strap current source is commonly found in many audio power amplifiers to bias the class-A voltage gain stages in them. These offer a cheap alternative to active sources where the output impedance of the CCS is of secondary concern. As can be seen in the image on the left, a boot-strap CCS is formed by resistors R1, R2 and capacitor CB. Transistor Q is wired as a class-A voltage gain stage. The circuit works as follows:
At no signal (i.e., DC conditions), CB acts as an open circuit and the collector current is determined simply by (R1 + R2).
Assume that R1 and R2 are chosen such that the DC bias current (collector current) is much larger than the load connected to the collector. When a signal is applied to the transistor's base, the collector voltage will change. This change in voltage is coupled back to R2 by CB in such a way that the voltage across R2 remains relatively constant. Constant voltage across R2 means constant current through R2 which is also the collector current, and hence R1, R2 and CB form a CCS. This CCS is a passive current source, and the transistor Q forms the load for it.
Calculation of R1 & R2: the collector voltage (VC) is assumed to be at half the total supply (Vsupply) - in this case since the emitter is shown floating, it could be either connected to an equal −Vcc or ground. In either case we assume VC equals half the total supply.
The reason for assuming that VC is half the total supply is that this particlar configuration appears in a VFB design - similar to that of an op-amp in discrete form with a differential pair input stage and DC coupling internally. When negative feedback is applied around the ckt, the o/p (the collector of the transistor, in this case) always sits mid-way of the total supply - i.e., if we have a split ±VCC, then the o/p will be at 0V. If a single supply (VCC) is used then the o/p will be at VCC/2.
If the next stage requires say, Y mA of current, then this transistor must be biased at atleast 1.2 to 2 times this for class-A operation. We take twice that (2Y mA) for calculation, call it X mA. Now, KΩ. Usually R1 = R2. Therefore KΩ. Since VC = Vsupply / 2, therefore, KΩ. The boot-strap capacitor is calculated simply by assuming a lowest frequency (f) of operation (say 1 Hz) and applying the formula Farads, where RP = R1 || R2. Since most power amps have very little gain below about 10 Hz by design, it means that at low frequencies (as 1 or 2 Hz), R1 and R2 by themselves are sufficient to ensure adequate bias. For R1 = R2 = 2.2KΩ, and f = 1Hz, CB turns out to be about 114.7µF and most amps use the standard 100µF with no degradation in performance. Rohitbd 09:05, August 10, 2005 (UTC)
Moved from my talk page Rohitbd 08:04, August 11, 2005 (UTC)
I assume you wrote the 'bootstrap' piece in current sources.
THe trouble is that, as O pointed out, half the diagram is missing. I dont think it can be copied 'en bloc'. because of copyright. It needs redrawing fully before the text surrounding it can be discussed. Are you any good at drawing ccts??
It does seem to act as a current source drive and so maybe should be included. But maybe it should be included in an audio amplifier page. THis is where I think these are most used.:-) Light current 15:04, 10 August 2005 (UTC)
Ic = BIb Where B is the gain of the transistor. This is the only way the collector current can be modulated. The collector has a very high resistance so its current is not affected by the voltage on it (much). Please try to get hold of some books on transistors. I think you will find them quite interesting and informative. :-) Light current 00:48, 11 August 2005 (UTC)
BTW Ic = BIb + Iceo.( where B is the gain of the transistor and Iceo is the leakage current). This is the basic transistor equation developed by Ebers + Moll. Collectors are not biased as you thought, it is the base emmitter junction that has to be biased to control the collector currrent in accordance with the above equation. THe collector terminal is a high impedance terminal that does not respond to applied voltages(much). THat is why the collector of a transistor is sometimes used as a current source (because it is a high impedance point) Do you understand it now? Light current 13:13, 11 August 2005 (UTC)
If you dont know how a transistor operates, who are you to tell me about current sources?? Light current 01:37, 11 August 2005 (UTC)
THe effective emitter load is R1||R2||R3. Let us call this R' For an input Vat the base, the voltage across R4 is V-KV wher k is the the voltage gain of the transistor and is approx 1. From the poin of view of input signal (ac) the input of the stage appears as follows: a simple emitter follower with a resitor R4 connected to the base. THe input to the stage is still directly at the base as normal, but the other end of R4 now as avoltage on it (from the emitteer) of KV (Yes?). So, the current thro this resistor R4 is: I =V(1-K)/R4. So the total input resistance as seen by the input signal is now the original input resistance in parallel with Reff. Where Reff= V/i= R4/(1-K) NOW, as K approaches 1, THe new input resistance approaches Rin which is roughly (1+hfe)R'. So the input resistance of the stage has been increased by the bootstrap and the effects of the bias chain are now negligible. THe important point is that an extra source of energy has to be used in a bootstrap circuit and this is provided by the current supplied to the capacitor from the transistor emitter. This is why your circuit does not work-- there is no extra source of energy invoked. Hope this resolves your difficulty. Light current 14:57, 11 August 2005 (UTC)
Moved from my talk page Rohitbd 15:30, August 11, 2005 (UTC)
I have replied on Talk:current source. I hope you will find the argument convincing. BTW Keep Calm -I'm only interested in the accuracy of WP :-) Light current 15:22, 11 August 2005 (UTC)
Regardless of whether this is right or not, you really shouldn't be posting a circuit that you designed. We have a policy of no original research. You can definitely write about the one included on the Westhost site, though. - Omegatron 23:50, August 11, 2005 (UTC)
Please read this article: http://www.eecs.utoledo.edu/~rking/elecdesign/EDmanualB.pdf
The bootstrapping technique is described just before the middle of the article. Although this schematic also has an emitter-follower, according to it, R2A + R2B (R1 + R2 as per my description & schematic) determines the Q-point current of the transistor - A point that I made, and was opposed by Light Current. Also the article has no mention of the emitter follower being required (or otherwise), so I guess my assumption for the sake of describing it is neither right nor wrong. The effect of the emitter follower is simply that the voltage at the output end of the bootstrap capacitor is just less than that at the collector - in my case it would simply be equal. In short, the resulting constant current is given by (K.Vx + Vb)/R2B as per the article. In the PDF article, K < 1 (due to emitter-follower Vbe drop), Vx is the collector voltage and Vb is the voltage at the emitter of the emitter-follower which is 0.65V less than Vx. In my case, looking at the schematic, K = 1 and Vb = 0 since no emitter-follower is there, so as per the equation it still works. It becomes Vx/R2 in my schematic.
I hope this clarifies my point of view a bit. Rohitbd 15:12, 12 August 2005 (UTC)
is incorrect. R2a and R2b do NOT determine the Q point current of Q2. What they do do is determine the VOLTAGE at the collector of Q2 (in conjuction with the bias network (which is probably an amplified diode voltage source)). The rest of the page does appear to describe bootstrap operation correctly although I have not checked every line. This article apears to have been written by a student and therfore can not be given much credence. Any one agree?? Light current 15:27, 12 August 2005 (UTC)
Exactly!! Light current 16:11, 15 August 2005 (UTC)
Moved from my talk page Rohitbd 07:35, August 12, 2005 (UTC)
As it says on each and every editing page:
If you do not want your writing to be edited mercilessly and redistributed at will, do not submit it
Now please dont be a sore loser. We all have to learn. Its just that you have chosen the hard way to do it. Light current 01:13, 12 August 2005 (UTC)
This discussion has been going on for very long now. A couple of points...
In my explanation of the circuit, I described it from the point of view of designing and not analysing the circuit - although I termed it as "analysis" - for want of a better word (sorry about that). Also, I do not feel that the ckt requires in-depth analysis or design steps, it can be described simply in a few words. If needed, the schematic can be updated to include an emitter-follower, but IMHO it makes no difference to the operating principle. Rohitbd 08:56, August 15, 2005 (UTC)
Alfred, my view is assuming CCS loads and global -ve feedback circuits. All along I had a CCS instead of a collector load in mind while discussing the bootstrap principle. I don't know, maybe repetitive work with VFB amps has perhaps put me into the habit of always using a CCS to bias the individual stages of the amp. Nevertheless, as defence of my argument that the collector resistor can be used to bias the transistor, let me give an example, I just designed and simulated this CE amplifier as per my method — These are the known parameters and design goals:
Step 1: Assumptions
Step 2: Calculate Rc & Re
Step 3: Calculate base voltage divider
After calculations, the actual voltage gain into a 4K load is about 9.6 as simulated (instead of the intended 10), practically we can expect about 9. In terms of decibels, this is a difference of about 0.35db (1db practically), and the voltage gain is stable across following transistors: BC547, BC337, 2N2222 and BD139. This can be improved by using a CCS instead of Rc and using global negative feedback. Rohitbd 18:03, August 15, 2005 (UTC)
<insert>
</insert>
In the last section of this article, there is the statement that an ideal voltage source cannot be connected to an ideal short circuit. Well, here is a thought experiment to consider. Connect a resistor to a voltage source. The voltage across the resistor is the voltage of the source and the current is just the ratio of the source voltage and the resistance of the resistor. Now, let the resistance tend to zero... At all times, the voltage across the resistor is the voltage across the source and the current increases without bound. In the limit, we have that the voltage across the resistor of zero ohms is still the source voltage. How can this be? It is because the current through the short circuit has passed to infinity and recall that infinity multiplied by zero is an indeterminate form. In order to determine the value, a limit process must be used just as we did above. A similar argument can be made for an ideal current source without an external circuit connected (infinite voltage multiplied by zero conductance). Enjoy! Alfred Centauri 02:09, 16 August 2005 (UTC)
Ok, now that I have received enough brick-bats, would anybody care to present an analyis of the bootstrap CCS? I tried and was opposed vehemently (not to mention personal attacks)...so I guess it is only fair of me to ask my detractors to give an analysis themselves. Of course, there are no obligations - but for a change, why don't you try and then include it in the article? Rohitbd 13:19, August 16, 2005 (UTC)
(see also Current and Voltage Source Suggestion)
Hi everybody! Two months ago, I placed a link on constant current source page pointing to my story about this subject. In this way, I tried to resume the discussion about constant current source and the dual voltage source. Browsing through these and talk pages I found a lot of brilliant thoughts. Only, the pages look quite cluttered; there isn't good structure and hierarchy in their arrangements. So, I suggest rearranging the materials according to the principles below:
1. At every page, expound the subject step-by-step by moving from simple to complex (imperfect to perfect, passive to active, transistor- to op-amp versions etc.)
2. At every step, first reveal the basic circuit idea; then show the concrete circuit solution.
In order to illustrate my suggestion, I have made a web page containing two versions of the contents - a short and a long one. Here is the short version:
1. Theoretical current source.
1.1. Ideal current source: definition, examples of natural current sources.
1.2. Comparison between current and voltage sources.
1.3. Real current source: imperfections.
2. Practical current sources.
2.1. Ohmic resistor current source. Imperfections.
2.2. Dynamic resistor current sources.
2.3. Current sources with compensating voltage.
2.4. Current sources with compensating current.
2.5. Current sources using negative feedback.
1. Theoretical voltage source.
1.1. Ideal voltage source: definition, examples of natural voltage sources.
1.2. Comparison between voltage and current sources.
1.3. Real voltage source: imperfections.
2. Practical voltage sources.
2.1. Voltage divider source. Imperfections.
2.2. Dynamic resistor voltage sources.
2.3. Voltage sources with compensating negative resistance.
2.4. Voltage sources using negative feedback.
I realize that I have proposed major changes in the structure of these pages. So, I would like first to coordinate them with you, especially with those who are created these pages; then, I will try to modify gradually (in small steps) the pages. But yet, don't remember that I am a Wiki newbie; so,
don't bite me desperately:)
Circuit-fantasist 16:24, 29 June 2006 (UTC)
I have reordered a bit the galvanostat voice and have written a brief comment on its talk page Talk:Galvanostat: please give it a look :). Daniele.tampieri 14:54, 24 December 2006 (UTC)
Rsduhamel, you have to connect a resistor in series with the zener diode on the circuit diagram of an op-amp current source. Circuit-fantasist 08:49, 26 January 2007 (UTC)
Circuit-fantasist, I presume that you are referring to Figure 7, where the Zener diode is connected between the op-amp's non-inverting (+) input and ground. No resistor is needed in this portion of the circuit. Current limiting for the Zener diode is provided by resistor between the op-amp's non-inverting input and the power supply. In fact, including a resistor in the same branch as the Zener diode will prevent the voltage across the sense resistor from tracking the voltage across the Zener diode. That would be prevent the circuit from working properly as a constant current source. The circuit is designed so that if the voltage across the sense resistor should increase (indicating an increased current flow), the op-amp will reduce the output voltage, thereby causing a reduction in the current through the load and sense resistor. If the voltage across the sense resistor should decrease (indicating a reduced current flow), the opposite happens. For the constant current in the load to be stable, the voltage at the op-amp's non-inverting input needs to be stable and the resistance of the sense resistor needs to be stable for the expected currents. Thus, no additional resistor is needed. OhioFred ( talk) 22:53, 17 September 2014 (UTC)
Circuit-fantasist, perhaps you meant to remark on the fact that the op-amp inputs are biased near the Zener diode voltage and the op-amp output is biased near the power supply voltage. See the comment below (in section Mistake in Figure 7: Typical op-amp current source) regarding the placement of the load resistance. The solution recommended there would reduce the bias differences. (Don't forget that the op-amp positive and negative supply voltages may differ from the circuit voltages. For best circuit performance, the op-amp should operate half-way between its positive and negative supply voltages.) OhioFred ( talk) 02:35, 18 September 2014 (UTC)
Alfred Centauri, I highly appreciate your last edits. Now, the section about the so-called " resistor current source" is where it has to be - at the beginning of the page.
The ESP link is a 'leftover' from the original text I edited. Had I bothered to read it, I would have removed it myself. Good catch.
On another note, please do keep in mind that this is an encylopedia, not a text book. That is, Wikipedia is a reference. To keep the articles concise, there must be some assumption of familiarity with the subject material.
I've looked at some of your contributions (voltage-to-current converter and others) and, while I'm impressed with the skills you display for presenting this material for learning and the depth of the information you have put into these articles, it is my opinion that Wikipedia is not the proper venue for this type of presentation. Have you considered making contributions to Wikibooks? The EE sections there could certainly use your help.
Best regards, Alfred Centauri 00:53, 10 April 2007 (UTC)
Solar cells are generally regarded as a current source in parallel with a diode and a shunt resistance with an extra resistor connected in series. My background is in Physics as opposed to electronics but are there any objections to my adding this to the page? —The preceding unsigned comment was added by 155.198.186.155 ( talk) 17:30, 16 May 2007 (UTC).
The simplest way to drive an LED from line-voltage AC is with a large series resistor, which functions as a current source. Very inefficient, due to power loss in the resistor.
The article implicitly only covers DC current sources. AC is a related but more complex subject. But some simple cases are important. We could also drop line voltage to a low voltage to drive an LED with an AC capacitor. We can calculate power in the capacitor by V x A, but the result is not real watts - the answer is ideally all "imaginary" VA? In the real world, how do we calculate power dissipation in real capacitors used thus: how hot they would get, what would be safe design, etc? For example, if two LEDs are anti-parallel, and you want the total average current through the pair to be 20 mA, from a series capacitor, at 120 VAC, what capacitance is needed, what capacitor types would be appropriate, and would power issues in the capacitor be a concern?- 69.87.204.197 13:20, 28 May 2007 (UTC)
91.75.70.167 ( talk) 04:08, 3 February 2008 (UTC) A batterie's function is to maintain a constant potential difference about itself/ in the ckt. Does a current producing source, work on the same basis? ie. at every instant of time, its job is to provide a constant supply of current? If yes, then how will we balance the produced current?
"The current then flowing is the IDSS of the FET."
Please list typical values for IDSS. What is the physical reason for the limitations on this value? It seems from CLD datasheets that this is only a solution for currents below about 15 mA. See [6], for instance.
If more than one CLD can be placed in parallel on a PCB, why can't they manufacture multiple CLDs in parallel on the chip? Wouldn't that be exactly the same as paralleling a bunch of transistors on the same chip to make a power transistor? — Omegatron 18:47, 21 March 2008 (UTC)
Where is the inductor on the LM317 circuit in the section "Inductor type current source"? Section title typo? I think it should be "Linear type current source". — Preceding unsigned comment added by 213.63.100.45 ( talk • contribs) 02:03, 11 December 2010
A section title says "Current sources with series negative feedback" and then the first part "Simple transistor current source" doesn't even mention the word feedback. Later there is a negative series feedback circuit presented; however it is not mentioned that the first bjt circuit already in itself has a negative series feedback.
If temperature doesn't change, then Re emitter resistor is not necessary (in theory only!) as the Ic collector current is set by the Vbe base-emitter voltage. But in this case we can't make the assumption that Vbe=0.65V or so, because a very little change in Vbe results in big changes in Ic. (See http://diranieh.com/Electrenicas/Figures/BJT_EbersMollGraph.jpg for example, don't know why the bjt article doesn't have these curves). And of course that's one reason we use the RE resistor, as we can't guarantee a constant base potential (though the zener diode is a good try). The Re resistor kind of forces the current onto the transistor, as small changes in Vbe will not change the emitter resistor's current much. But still it is possible to change the output current not only by Re but by changing the base voltage (if using a potmeter instead of a Zener).
Now for the feedback: the voltage across emitter resistor Re is proportional to the output current. If the temperature rises, then Ic collector current increases - but then the Ve emitter potential will also increase. As the Vb base potential is fixed by the diode or resistor divider, and Vbe=Vb-Ve so Vbe will decrease, which acts towards decresing the collector current. This is a series negative feedback. If you simulate the circuit without the emitter resistor you get a much worse temperature dependency. Of course Re is only good so far, but can be useful in many cases.
As for earlier comments: in the output curve, the collector current does have a very little dependancy on Vce, which can be a problem for a very precise circuit and this is also compensated by the feedbacks. (Also the finite value of h22e is also due to this). Hoemaco ( talk) 11:35, 21 April 2011 (UTC)
I think the load resistance in this circuit diagram should be located above the transistor. The reason for this is that if most of the voltage drops at the load (as intended), the voltage at the emitter of the transistor can easily become too big for it to conduct (since the base voltage needs to be at least 0.7V above the emitter voltage). For an PNP transistor the depicted way would be okay, but this is an NPN transistor. A detailed discussion of this kind of circuit is given at http://www.mikrocontroller.net/articles/Konstantstromquelle#Konstantstromquelle_mit_Operationsverst.C3.A4rker_und_Transistor (sorry, in German).
I already produced a corrected version of the image but can not replace the file because it's imported from wikimedia. — Preceding unsigned comment added by 5.158.164.237 ( talk) 01:09, 12 May 2013 (UTC)
This paragraph is misleading. A Van de Graaff generator is essentially a current source and must be regulated to maintain a constant voltage. The high voltage capacitor is charged by a belt on which charge is placed at low voltage and mechanically moved to a region of high voltage. When used a particle accelerator, the load current is determined by an ion source, which generates a constant current and does not resemble a resistor.
I think we need to include the "interlinked circles" symbol for a current source. See link. This is very commonly used in UK engineering practice and in text books/app notes/papers, and is what was used throughout my university course. I much prefer this symbol because it can be quickly and unambiguously interpreted as a current source. Amj4321 ( talk) 22:31, 22 November 2015 (UTC)
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