# Current mirror explanation (misleading diode equivalent)

Discussion in 'Feedback and Suggestions' started by KrisBlueNZ, Dec 18, 2013.

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1. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
Hi all.

I believe the tutorial section on current mirrors at
has a problem. I believe that representing the "input" transistor as a diode is misleading. All of the text between the first diagram (showing a transistor with a fixed base bias voltage) and the third diagram (showing the NPN and PNP versions of the current mirror) is wrong in places, and misleading generally.

The circuit explanation fails to explain that the collector current in the "input" transistor is essential to the operation, and understanding, of a current mirror. Almost all of the input current flows into the collector of the input transistor; the two base currents are much lower (due to the current gain of the transistor).

Therefore representing the input transistor as a diode, or describing it as operating like a diode, or even being "diode connected", is misleading, and prevents proper understanding of the current mirror's operation.

The current mirror is properly explained in other places on the web. Here are some descriptions I found:

https://en.wikipedia.org/wiki/Current_mirror - The Wikipedia article. It is not as clear as it could be.
users.ece.gatech.edu/~mleach/ece3050/notes/bjt/bjtmirr.pdf‎ - W. Marshall Leach's description. Accurate but not highly descriptive.

I did not find a really clear explanation, but I did not do a long search.

I also found some other descriptions that make reference to the base-emitter junction acting as a diode. I have to assume that the people who wrote these descriptions do not properly understand the operation of the current mirror.

I posted some more detailed material in a comment thread on a YouTube video. This thread was what directed me to the AAC article in the first place. You can find the thread at http://www.youtube.com/all_comments?v=U6qZPx4uD0g if you search on the page for my username, KrisBlueNZ.

I would be happy to rewrite that section of the tutorial if you want.

Last edited by a moderator: Dec 18, 2013
2. ### WBahn Moderator

Mar 31, 2012
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I agree that the E-book description is lacking. I don't believe it is really possible -- though I could well be wrong -- to adequately describe the operation of a current mirror while ignoring -- or mentioning only in passing -- the effect of the base-emitter voltage.

The key to understanding the operation is to recognize that the configuration forces the base-emitter voltage of both transistors to be the same. Because the collector current is a function of the base-emitter voltage (in the base Ebers-Moll model), this means that the collector currents of the transistors will be the same (assuming matched transistors).

If you want to rewrite the tutorial, keep in mind that you will have to do it using charge-carrier flow (i.e., electron flow) instead of charge flow (i.e., conventional current), as that is how the owner of the E-book wants it.

3. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
I agree that the base-emitter voltage is important to the description, and I'm not suggesting that that detail should be removed.
Thanks for the explanation that I already knew At least it confirms that I'm not alone in my understanding. When I tried to explain it in the comments on the YouTube video, I got arguments from people claiming that the input transistor behaves like a diode, which is also what the tutorial says.
Yes I noticed that. It's a bit unfortunate in my opinion, but never mind

I don't particularly WANT to update it. Perhaps you (or someone else) could contact the "owner" and see what he wants to do.

4. ### WBahn Moderator

Mar 31, 2012
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Oh, I wasn't suggesting that you didn't understand the importance of the base-emitter voltage. That was more for completeness in establishing my viewpoint for others that are reading this.

The input transistor IS acting like a diode. If you put it into a box with just the two terminals (collector and emitter) exposed but, inside the box, with the collector connected to the base, you wouldn't be able to tell the difference between that box and another box that truly contained just a diode, at least not as long as you constrained your measurements to the forward-biased region. So it is not unreasonable at all to refer to it as a diode-connected transistor or to describe it as behaving like a transistor. You simply have to step up from the fixed-forward-voltage-drop model of a diode (or transistor) and use the exponential-forward-voltage-drop model. Even here, it is sufficient to simply acknowledge this and point out that there is a one-to-one monotonic relationship between diode voltage and diode current (keeping in mind that we haven't thrown in temperature or other effects yet) and that this relationship is the same as the relationship between base-emitter voltage and collector current (keeping in mind that we haven't thrown in Early voltage effects yet, either) of the image transistor.

If you disagree that it behaves like a diode, then please do explain why you believe that to be an inaccurate description, preferably in significant mathematical detail, so that we can discuss it and try to come to a consensus. Even if we don't, we will all walk away at least aware of a different perspective that we might not have been aware of before.

The mods see these posts, as do others that like to maintain the E-book. I don't maintain the E-book because I'm opposed to using the charge-carrier-flow model.

5. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
No! At least, not the way I interpret "behave like a diode", and certainly not like the base-emitter junction of the output transistor.

Here is my explanation by example. Assume that the transistor's base-emitter characteristics are such that with exactly 0.65V between the base and emitter, the base-emitter junction will draw 1 mA. And assume the transistor has a current gain of 100, and neglect the Early effect and any other imperfections.

If you put the transistor in a black box and connect only the base and emitter, then apply 0.65V to it, you will measure 1 mA.

If you then connect the collector to the base, and apply 0.65V to it, you will measure 101 mA.

This is why I believe the "diode-connected" description is a misnomer. Sure, it only conducts in one direction, like a diode, but it doesn't have the same V-I characteristic as a diode. If you test it with a curve tracer, you will see the difference. The black box equivalence test will fail.

I think a lack of appreciation of this important difference may be the reason that some people describe the input transistor as operating "like a diode".

For another explanation showing the important difference, see my comments on the YouTube video that I linked to in my first post.

I don't understand what you're saying here. AFAIK a diode HAS an exponential current vs. forward voltage curve. A transistor with its base and collector connected will have a much sharper knee, and a much lower incremental resistance for a given forward voltage, than a diode.
I certainly agree that the I-V curve is monotonic, and that superficially it looks similar to a diode's I-V curve, but I don't think that means that it behaves "like a diode".
I have explained the reason why I don't think it behaves "like a diode". The difference might be insignificant if it's used as a rectifier, but it's critical to understanding the operation of the current mirror, and to my mind, comparing the input transistor to a diode in any way is simply misleading and wrong.

I hope I've explained my position clearly enough for you to understand how I'm thinking. Perhaps you understand my issue already, and have an answer to my criticisms. If you can address my concerns directly, please do so!
I agree. The electron flow model is important when you're understanding semiconductor physics, but when you're working with component symbols and circuits, conventional current is more sensible.

Last edited: Dec 19, 2013
6. ### WBahn Moderator

Mar 31, 2012
22,980
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I give you a black box with two terminals sticking out of it. You plot the V-I curve and you get a curve that obeys the following relationship:

$
i = Io \left( \ e^{$$\frac{v}{\eta V_T}$$} \ - \ 1 \ \right)
$

You are then asked whether the box contains a diode or a diode-connected transistor.

You claim they are distinguishable, so how will you distinguish them?

In particular, if you are saying that the above equation only applies to one or the other, which does it apply to and what is the equation that applies to the other?

You are correct to point out that I got a bit sloppy at one point. I said that the I-V characteristic for the diode-connected transistor matches that of the Ic vs Vbe for a matched transistor. This is not precisely correct (though it has no effect on whether the diode-connected transistor has a diode characteristic). While this is a common approximation, the correct statement would be that it matches the Ie vs Vbe characteristic of a matched transistor.

7. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
I see that the V-I curve of a "diode-connected" transistor has the same shape as the V-I curve of a diode, but not the same scale. I don't have the maths skills to be able to be more specific, but I think they might be distinguishable because of the value of one of the parameters in the formula, because of the relative steepness of the curve for the transistor. I don't know if it would be possible to find (or imagine) a diode that would match the curve of a high-gain transistor. I suspect not.

After thinking about your point for a while, I need to be be more specific about my objection to the diode comparison. The problem with describing the input transistor as a diode is that its characteristics are very different from the base-emitter junction of the output transistor, which is also "a diode".

It must be clearly explained that the "diode" formed by the input transistor conducts far more current (for a given forward voltage) than the diode formed by the base-emitter junction of the output transistor.

If this is not pointed out clearly, the reader would naturally assume that since the transistors are identical, the "diode" formed by the input transistor will behave like the base-emitter junction diode of the output transistor.

And when you have to explain this input "diode" in that way, comparing it to a diode becomes pointless, and simply misleading. It is a lot clearer to avoid the diode idea altogether, and explain the current mirror simply as two matched common emitter amplifiers with commoned inputs, where one amplifier absorbs most of the input current and the other supplies the output current.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,479
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Well I think that I understand what you are trying to say. The ordinary real diode for sure has a different I-V characteristic (slope) because of this "n" in Shockley equation.
So in real life diode I-V characteristic don't match transistor "diode" connection.
But it does not change the fact that in theory we are able to match this two devices.
See this simulation result:

As you can see I managed to get almost perfect match.

And also don't forget that it is well established to call this BJT connection a "diode-connected transistor". And it is impassible to change this.
http://www.designinganalogchips.com/_count/designinganalogchips.pdf (page 33)

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Last edited: Dec 19, 2013
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9. ### Georacer Moderator

Nov 25, 2009
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Discuss a bit more with tshuck and if you come to a conclusion and you still want to write a revised version of the article, you are welcome to do so.

It is not guaranteed that the article will be updated very soon, though.

10. ### WBahn Moderator

Mar 31, 2012
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6,875
I could get a bunch of diodes and a bunch of transistors and you could measure them and you would find (subject to the particular part numbers I picked) that there is no discernable pattern. For instance, if you were to use the current at a particular voltgage drop the device with the smallest current might be a transistor or it might be a diode. Likewise of the device with the largest current.

The important point, though, is that when we talk about the diode-connected transistor having an I-V characteristic that is (approximately) matched to the reflector transistor, we are NOT talking about the Ib vs. Vbe characteristic. We are talking about either the Ic vs. Vbe charactersitic (which is a very close approximation) or the Ie vs. Vbe characteristic (which is an exact match). Of course, the is still ignoring other effects such as temperature and Early effect.

Even though the match is only approximate for the Ic vs. Vbe charactersitic (which is usually the one we are most interested in), the match is good to withing 1/β, which typically makes it better than 1%. Considering that even transistors right next to each other on the same die may not be matched to within 1%, that's good enough for all practical purposes. Where this effect becomes significant is when you have more than one reflector transistor. It is what limits the use of BJT current mirrors from being usable to bias more than a handful of transistors. MOSFETs, on the other hand, don't suffer from this problem; I have used single FET mirrors to bias literally millions of FET reflectors.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That is most likely an incorrect assertion. A careful interpretation of the Ebers-Moll BJT model equations will show this to be the case. No doubt the current will be higher than 1mA in the latter case, but not by that substantial amount. Probably less than an order of magnitude.

12. ### WBahn Moderator

Mar 31, 2012
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Interestingly, the quality factor, 'n', is not part of the "true" Shockley equation; it was added later to account for effects seen in transistors.

I've never been able to track down a good history of the quality factor, but from what I've been able to piece together (and this may well be quite wrong), in the early days when Shockley developed his model diodes had an n-value close enough to 1 that he didn't recognize a need for any adjustment. When they started characterizing transistors they found significant imperfections and therefore added the quality factor. By the time I was studying this stuff in the ~1990 time frame, the rule of thumb seemed to be that transistors had an n-value of ~1 and diodes had an n-value of ~2. But the last few times I have tried to track down information it seems that, for diodes or transistors, the n-values are close enough to unity that the n-value is usually left out of both the diode and the transistor equations any more.

If anyone has any better information, I'd love to hear it. One of the many things on my list is to take some measurements on some (modern) diodes and transistors and see what the quality factors actually look like.

13. ### WBahn Moderator

Mar 31, 2012
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I didn't read this closely enough the first time.

If you put a transistor in a box and only connect the base and the emitter, you will probably get a lot more than the 1mA of current. Remember, the device is now being operated in the saturated region and the familiar Ebers-Moll model doesn't apply. You would have to resort to the unapproximated Ebers-Moll equations but that requires knowing things like the reverse common-emitter current gain.

I've just spent a few minutes tinkering with the unapproximated equations and applying the constraint that Ic=0, but am keep getting useless results (such as Ib=Ie.. duh). I want, of course, to see if I can get Ib(Vbe). I'm probably just overlooking some obvious manipulation.

But stepping back from that, the notion of this diode-connected transistor behaving like a diode does NOT claim that this diode has the same characteristic as the Ib vs. Vbe characteristic. We really don't care about that at all when describing a current mirror. The base currents are internal to the mirror circuitry. What we are interested in is the relationships between input current to the mirror (the collector current of the programming transistor) and the output current (the collector current of the reflecting transistor). These relationship are very well matched (for matched transistors and before other effects are considered).

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It's a bit of an anomaly directly driving the base emitter junction with a constant voltage source - nominally 0.65V - without any limiting resistance.

For the sake of the argument (however) one could consider the idea. In the first case, with the collector not in circuit (floating) it's difficult to model. But one could consider the collector tied to ground via a very high resistance - say 100MΩ. This would ensure virtually zero (negligible) collector current. Alternatively, one could tie the collector through the same high resistance to a positive supply rail sharing the same common as the base drive. This latter configuration would allow one to assume the transistor is "saturated" with effectively zero collector current - strange as that might seem.

One could then make a few assumptions, again for the sake of the argument. Suppose Vce = 10mV for this case - a random guess.
One could then set up a solution of the un-approximated Ebers-Moll equation for the case of Ic=0 [Amp]. I tried this in Excel using goal seek, with the aim of getting Vce to 10mV whilst adjusting both VBC and Beta_R - using an assumed VBE as the starting point. One comes up with a solution for VBC and Beta_R which satisfies the requirement. From this one can use the un-approximated equation for Ib to find the saturation current Is, whilst setting Ib=1mA as proposed.

As a final step, all the derived values (including the assumed forward current gain of 100) can be used to find Ic & Ib for the case of VBC=0 when the base and collector are directly connected and driven directly by a 0.65V source..

Last edited: Dec 20, 2013
15. ### WBahn Moderator

Mar 31, 2012
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Yeah, I don't think anyone is saying this; just that, from a modeling standpoint, once you determine the Vbe resulting from a set of bias conditions (or stipulate a Vbe) you can, for the sake of static analysis, replace whatever bias circuitry is resulting in that particular Vbe with a voltage source of that same amount to make the analysis easier.

What kind of value did you come up with for βr? Was it in the 0 to 20 range, which is my understanding are common values?

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In the end it's rather a lot of hand waving to resolve an initially fuzzy assumption. One could try this on any simulator using a spice model for a typical transistor. The OP may well have a point about the e-book potentially leading people to a poor understanding of the subtleties of current mirrors. Hopefully, he can come up with something clearer & help sort out any uncertainties in the confused minds of some readers.

Last edited: Dec 20, 2013
17. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
Thanks to everyone who has responded. I have certainly revised my opinion, although I still think that it's misleading to describe the input transistor as behaving like a diode.

My objection remains as I said in post #7. Describing the input transistor as behaving "like a diode", without any further explanation, implies that it behaves like the base-emitter junction of the output transistor, which also behaves like a diode and is in a matched component.

Adding the explanation that the input transistor behaves like a diode, but not like the diode formed by the base-emitter junction of the output transistor, makes the comparison with a diode pointless.

In fact I don't think that describing the input transistor as behaving like a diode helps at all in understanding the operation of the current mirror. I think it is tangential and a distraction.

Jony130:

Thanks for your reply. I think you understand my objection to describing the input transistor as "behaving like" a diode.

I see you got a very close match between the diode and the diode-connected transistor. You had to use some customised parameters with the diode and transistor models. Do you know whether there would be any real-world components with parameters like those?

That's not exactly what I said. I know forward voltage (for a given current) is quite variable. I'm talking about the SLOPE of the V-I curve.

The diode-connected transistor will have a much steeper V-I curve than a base-emitter junction, because a voltage increment that causes a 1 mA increase in current in a diode junction will cause a 101 mA increase in current for the diode-connected transistor. (Still assuming a current gain of 100 and neglecting imperfections.)

This is a significant difference, and that's why I think it may not be within the range of variation that could happen due to process and parameter differences in component manufacture.

Jony130 was able to get a match in his simulation but I don't know whether the parameters he used are reasonable for real world components.
Yes, I see that. I did not get that impression from the tutorial.

(Also, I don't see the point of the long explanation of the α parameter (defined as Ic / Ie), and the focus on emitter currents, in the tutorial, but that's a separate issue.)

This is why I think the comparison is misleading. There are two matched transistors in the current mirror. The output transistor has a base-emitter junction that behaves like a diode. If you say that the input transistor behaves like a diode, you are implying that it behaves like the matched junction.

If you explicitly point out that the input transistor's "diode" has significantly different characteristics from the output transistor's diode, then I don't see how the comparison has any value in explaining the operation of the current mirror.

As I said before, I think it's tangential, and a distraction to understanding the current mirror's operation.

I meant that you would adjust the applied voltage until the base-emitter junction draws 1 mA. That has to happen at SOME voltage!

But I think you're saying that the collector current, and/or whether the transistor is saturated or not, has an effect on the behaviour of the base-emitter diode. Is that right? If so, I'm not surprised, but I didn't know that.

I don't understand the Ebers-Moll model. I think the maths is beyond me.
Right.
Exactly. So what's the point of describing the input transistor as operating like a diode? It doesn't help with understanding the current mirror at all, as far as I can see. It's just a distraction and a source of confusion.

If that were true, the current mirror wouldn't work!

I also said, immediately before the lines you quoted:
The base-emitter junctions of the input and output transistors have the same voltage across them, and the same base current. Their collector currents must be the same, because they are (most of) the input current and the output current, respectively, and the output current is nearly equal to the input current. So how can the input transistor's collector current be much less than 100 mA?

I think you may be raising a similar point to WBahn - that the base-emitter characteristics are affected by the collector current. But in the case of the input transistor, the collector is not disconnected, and the transistor is not saturated.

Right. That's what I intended.
You guys have lost me here. I guess my understanding of transistors only covers the basics. I hadn't even heard of the Early effect until I got involved with current mirror explanations.

I'm grateful for everyone's input.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In post #5 I think you proposed two test conditions. The first assumed only the base and emitter were connected and the second was the case with the collector connected to base. In the first case you proposed that a current of 1mA would flow in the base-emitter junction with a Vbe=0.65V. You then propose that in the second test with the collector joined to base the current draw would be (1+β)*1mA or 101mA.
This is not correct despite your objections. The Ebers-Moll equations show the contrary and the current mirror will work perfectly well in the circumstances.
For the two test conditions you propose, the current would increase from 1mA to somewhere around 3mA. As I suggest, apart from the Ebers-Moll approach, one can readily verify the two outcomes using Spice based simulations - which I have now done and verified the result of a modest increase in current consistent with the Ebers-Moll prediction.

19. ### KrisBlueNZ Thread Starter Member

Oct 17, 2012
111
14
Yes, that's what I meant. The base-emitter voltage is set so that 1 mA flows into the base, with the collector disconnected. Then the collector is connected to the base.

I'm aware that my understanding of transistors is a bit superficial, so I tried a simulation using LTSpice IV V4.20g (2013-12-17). I don't suppose the model is 100% accurate either, but it's going to be close enough for me to learn from, right?

I connected a voltage source to the base-emitter junction of a 2N3904 transistor and adjusted the voltage to get 1.000 mA. The voltage was 0.716517V.

Then I connected the collector to the base and re-measured the total (or the emitter) current. It was 9.86 mA. According to my assumptions, that would mean the current gain was 8.86.

However, the base current had dropped from 1 mA to only 33.57 µA. This answers my question to WBahn - yes, changes in the collector current and/or voltage DO affect the base-emitter diode characteristics. I'm not surprised that there is SOME effect but I didn't think it would be so much.

I didn't know this, as I have never understood the Ebers-Moll model. It's quite interesting

So in this situation, the transistor is showing a current gain of 292.7.

So you're right. The base-emitter voltage that produces 1 mA of base current with the collector unconnected will not produce 101 mA of total current with the collector connected to the base.

This doesn't affect my arguments about the inappropriateness of describing the "diode-connected" transistor in a current mirror as behaving "like a diode", but this is a good learning experience for me

20. ### LvW Active Member

Jun 13, 2013
674
100
Hi KrisBlueNZ, it seems everybody is satisfied now.
Nevertheless, may I place one short comment?
All you guys always spoke about the Ebers-Moll model - and you have indicated some problems to understand the model.
Why not try a simple (simplified?) physical explanation in one sentence?
Here it comes:
As soon as the collector node is connected to a voltage which exceeds the base voltage most of the carriers (99% or more) emitted by the emitter are attracted by the collector and travel NOT to the base node (as before with open collector) - thus drastically reducing the base current.