I'm not totally satisfied... I'd like to know whether the other contributors on the thread agree with me that comparing the input transistor to a diode is pointless and confusing, and that the comparison should be removed from the tutorial.

Your explanation sounds pretty simple and obvious but I don't think it is. In my simulation circuit, there is a fixed voltage applied, so it's not simply that a fixed current is now divided between the collector and the base, so the base current must drop.

The current is not limited by the voltage source, and it would be quite possible for the total (emitter) current to increase, with 100 mA collector current and 1 mA base current, if that was how the transistor behaved.

Clearly though, current in the collector-emitter path affects the base-emitter diode's behaviour - for a given, fixed forward voltage, the base-emitter diode draws much less current if there's current flowing from collector to emitter. That is not obvious or simple, although (I assume, from t_n_k's comments) the Ebers-Moll model would predict it.

BTW the collector voltage doesn't have to exceed the base voltage for the effect to occur. If it did, Vce(sat) would be higher than Vbe. I did an LTSpice simulation with a fixed Vbe and a ramping Vce. Ib starts dropping immediately when Vce rises above 0V and has pretty much levelled out by the time Vce has reached +0.2V. That's with a 2N3904.

 

Your explanation sounds pretty simple and obvious but I don't think it is. In my simulation circuit, there is a fixed voltage applied, so it's not simply that a fixed current is now divided between the collector and the base, so the base current must drop.

BTW the collector voltage doesn't have to exceed the base voltage for the effect to occur. .

KrisBlueNZ, I have tried a simple qualitative explanation - and I think, there was nothing wrong.
You shouldn`t forget that there are TWO pn junctions.
You have started the simulation at Vce=0, which means that both pn junctions are "open": Large base current of app 1.3mA.
EDIT: That means: The current Ib into the base node is much larger than the current through the B-E junction.
Thus, it is NOT the "normal" base current you have in mind.

Thus, it is clear that the base current drops for rising Vce voltages (but still below Vbe) because the current through upper pn junction drops with rising Vce.
Therefore, this kind of Ib reduction is NOT primarily caused by the effect I have mentioned in my former post. Thus, we have two effects that cause the current Ib to reduce with rising Vce voltage.

 

Sorry if my reply seemed rude. I appreciate you taking the time to write your explanation.

I think my problem is that I think of transistors from a behavioural point of view. I don't understand the physics.

So when you say that electrons can be attracted away from the base by the collector, which is why the base current drops, I don't know how to visualise it. Why would the collector attract electrons away from the base when the base is at a higher voltage than the collector? Because in my simulation, the base current had dropped from 1 mA (collector disconnected) to 36 µA when Vce was only 0.2V. (Vbe was ~0.7V.)

I guess if I understood the physics, that might make sense. But the way I see it, the diode characteristic of the base-emitter junction is being greatly affected by the collector current. I apply a fixed voltage across the junction, to get a certain current flow, and then when current is allowed to flow into the collector, the base-emitter diode no longer draws much current with the same voltage applied.

I can accept that description even though I can't explain it because I don't understand the physics. I'm just getting deeper into the details of BJT behaviour.

You have started the simulation at Vce=0, which means that both pn junctions are "open": Large base current of app 1.3mA.
EDIT: That means: The current Ib into the base node is much larger than the current through the B-E junction.
Thus, it is NOT the "normal" base current you have in mind.

Thus, it is clear that the base current drops for rising Vce voltages (but still below Vbe) because the current through upper pn junction drops with rising Vce.

Yes you're quite right. In my simulation, with Vce forced to 0V, almost all of the base current flows out the collector, not the emitter!

But in my first simulation, I adjusted the base voltage to cause 1 mA to flow in the base-emitter diode. When I set up that voltage, the collector was floating. I expected that the same base-emitter voltage would cause the same base-emitter current, regardless of the collector current.

I don't understand why, or how, electrons would go to the collector, at 0.2V, instead of the base, at 0.7V. But that's my problem. I just see it as a change in the characteristics of the base-emitter diode depending on the collector-emitter current.

All of this is only incidental to the issue I'm trying to discuss in this thread.

Thanks for your input.

 

I don't understand why, or how, electrons would go to the collector, at 0.2V, instead of the base, at 0.7V. But that's my problem. I just see it as a change in the characteristics of the base-emitter diode depending on the collector-emitter current.

No - the majority of carriers travel NOT yet to the collector under these conditions (Vce=0.2 V). I have tried to explain that in this case it is the base-collector junction which has caused the base current decrease and NOT the first effect I have mentioned (carriers to be attracted by the collector).

I repeat again: Under these conditions the current into the base consists of TWO parts: Ibe and Ibc (because the upper junction is still conducting).
Thus, if Vce rises, the junction between base and collector draws less current (Ibc) and therefore the current into the base node also decreases until (for further rising Vce) the base current Ib goes through the BE junction only (with a reduced value). Only now we have Ib=Ibe. However, at the same time the rising collector voltage makes what I have mentioned before: Nearly all of the electrons are now attracted by the collector voltage.

You really should try to understand that there are two overlapping effects for rising Vce (starting at 0 volts): The current Ibc reduces (down to zero) and the current Ic increases (until Ibe has reached its nominal value Ic/current gain). OK?

 

Yes, I understand that Ibc reduces as Vce increases. I'm only talking about the base-emitter junction. When I adjusted Vbe to give Ib = 1 mA, the collector was disconnected, not tied to the emitter; I was only considering the base-emitter diode. When Ice is allowed to flow, Ibe drops by a factor of ~30 even though Vbe has not changed.

 

I was only considering the base-emitter diode. When Ice is allowed to flow, Ibe drops by a factor of ~30 even though Vbe has not changed.

Yes - and that is not the whole truth. You cannot undestand these effects by considering the B-E diode only. Ibe drops by this factor because the majority of the carriers emitted by the emitter do NOT anymore constitute the current Ibe.

 

OK. Thanks for your advice

 

Hi KrisBlueNZ,

Thanks for the feedback.

There are a number of people on the AAC forum who abhor anything to do with Spice simulation and consider it next to useless.

Perhaps to satisfy myself and the skeptics, I decided to do an actual careful measurement of the two conditions for an actual transistor.

For no particular reason I chose an NPN BD139 as the candidate. Test circuit schematic attached.

To obtain 1 mA in the base-emitter junction with the collector floating required a base-emitter voltage of 671 mV. When I connected the collector to the base the current rose to 6.1 mA with the same base-emitter voltage of 671 mV. So again this confirms the predicted outcome.

As one might expect, the current flow is rather sensitive to the transistor temperature - with a regulated voltage drive. A short touch with the fingers is enough to disrupt the current stability. Hopefully the self-heating effect (if any) didn't unduly alter the conditions between the two tests.

Thanks for starting an interesting thread.

 

t_n_k, thanks for that!

I'm never sure how much to trust simulations either. But it seems in this case that you have proven them right - or at least, not proven them wrong!

I guess you didn't measure the base and collector currents independently when both were connected together? If you had, I guess you would have seen that the base current had dropped a long way.

671 mV Vce is plenty for the transistor to operate as a current amplifier, and with a total current of 6.1 mA, if the base current was still 1 mA that would imply a current gain of 5.1. I know the BD139 is not a high-gain transistor, but I'm sure it's not that low!

Thanks to you and the other contributors for MAKING this thread interesting! I have learned quite a bit. Nothing that has affected my objection to comparing the input transistor of a current mirror to a diode, though, which is what this thread was originally supposed to be about!

 

671 mV Vce is plenty for the transistor to operate as a current amplifier, and with a total current of 6.1 mA, if the base current was still 1 mA that would imply a current gain of 5.1. I know the BD139 is not a high-gain transistor, but I'm sure it's not that low!

Thanks to you and the other contributors for MAKING this thread interesting! I have learned quite a bit. Nothing that has affected my objection to comparing the input transistor of a current mirror to a diode, though, which is what this thread was originally supposed to be about!

Good on you for sticking to your ideas.

I did measure the Beta value as70.

Cheers!

 

with a total current of 6.1 mA, if the base current was still 1 mA that would imply a current gain of 5.1. I know the BD139 is not a high-gain transistor, but I'm sure it's not that low!

KrisBlueNZ,
just to avoid misinterpretations:

Both currents - Ib=1mA and Ic=6.1mA - were measured NOT at the same time and NOT under the same condidtions. Therefore, it is not correct to take the ratio and to find the "current gain".
It was discussed earlier that - and WHY - the Ib drastically drops in case the collector is not floating anymore (in the described case: 671mV)

 

Both currents - Ib=1mA and Ic=6.1mA - were measured NOT at the same time and NOT under the same condidtions. Therefore, it is not correct to take the ratio and to find the "current gain".

Right. My point was that a current gain of 5.1 is not reasonable, so I assumed that the base current must have been a lot less than 1 mA in the second test.

t_n_k confirmed this; current gain was 70 with Ib+Ic = 6.1 mA, so Ib would have been about 86 µA, about 12 times lower than it was for the same forward voltage with the collector floating.

It was discussed earlier that - and WHY - the Ib drastically drops in case the collector is not floating anymore (in the described case: 671mV)

Yes.

 

I'm not totally satisfied... I'd like to know whether the other contributors on the thread agree with me that comparing the input transistor to a diode is pointless and confusing, and that the comparison should be removed from the tutorial.

Okay, let me take a shot at explaining a basic current mirror from the opposite side of the circuit. As we've been doing all along, we will ignore temperature and Early effects so that the collector current is strictly an exponential function of the base-emitter voltage.

So let's start with the following circuit:

As we adjust Vbe to different voltages, we get different collector currents, correct? The tricky part is that it only takes about a 60mV change in Vbe to produce a factor of ten change in Ic. Not very practical.

What we need is a device that, when we put the desired collector current through it, produces the proper Vbe voltage needed to get the transistor to produce that same Ic.

Since the base-emitter junction of a transistor looks like a diode and since the collector current is simply β times the base current, the relationship between the collector current and the base-emitter voltage is also that of a diode. So, in theory, we can use the following circuit:

The voltage at the base of the transistor will be clamped to something between, say, 0.5V and 0.8V. Thus the voltage across R2 will be relatively constant (between 14.2V and 14.5V). Thus the current in R2 will be dominated by the value of R2. The current will flow almost entirely through the diode. Some will flow into the base, but likely less than 1%. This will be determined by the β of the transistor, which we are assuming is at least 100 and more likely 200 to 300. If you happen to be using power transistors that typically have β values in the range of 20 to 40, this assumption would need to be revisited.

The problem with this circuit is that we need to find a diode that closely matches the Ic vs Vbe characteristic of Q1. There are two parameters that dominate the charactersitic: the scale current and the quality factor. Finding an off-the-shelf diode that closely matches our transistor in both of these parameters, particularly the quality factor, is going to be difficult.

But what if we could easily construct a diode that has the necessary parameters? It turns out we can, by simply taking a matching transistor and configuring it so that it behaves like a diode. This is, not surprisingly, called a diode-connected transistor.

A diode connected transistor has an I-V characteristic that is identical to the Ie vs Vbe characteristic of the transistor. It only closely approximates the Ic vs Vbe characteristic we are interested it. It is off by a factor of α, but for a transistor with a β>100, α>0.99, so the error is small and, in most cases, negligible.

The final circuit thus becomes:

By connecting our diode to the base of the transistor, a small amount of the input current is diverted from the diode to feed the transistor base. This redirected current no longer goes through the diode and hence no longer participates in generating the needed Vbe across the diode. This results in a further slight mismatch between the input current and the output current.

The end result is that the relationship between the input and output currents of the basic current mirror is

Iout = α² Iin

For transistors having β=100, there is a 2% mismatch. For transistors having β=300, the mismatch is less than 0.7%. There are several other error sources that will overshadow either of these.


Hopefully, this explanation shows that talking about the diode-connected transistor AS a diode is not only not misleading, but quite useful. Hopefully it also shows that using the alpha parameter is also not unreasonable, even though I did not walk through the derivation of that last equation step by step.

 

WBahn, thank you very much for that detailed explanation. I really appreciate the effort you have put into it. I will get back to you when I've studied it carefully and thoroughly.

Thank you very much!

 

OK, that's a very clear explanation the current mirror. It ties in exactly with my previous understanding, and I still believe that describing the input transistor as a diode is misleading.

What we need is a device that, when we put the desired collector current through it, produces the proper Vbe voltage needed to get the transistor to produce that same Ic.

Nicely put.

Since the base-emitter junction of a transistor looks like a diode and since the collector current is simply β times the base current, the relationship between the collector current and the base-emitter voltage is also that of a diode.

Perhaps, but it's not the same as the relationship between base current and base-emitter voltage in the output transistor.

The input and output transistors are matched. The output transistor's base-emitter junction is described as a diode, and is connected in parallel with the input transistor, which is also described as a diode, and is a matched component. I believe that this clearly implies that they will have the same characteristics, and will draw the same amount of current, given the same forward voltage. But they won't.

Hopefully, this explanation shows that talking about the diode-connected transistor AS a diode is not only not misleading, but quite useful.

I'm afraid not :-/

Hopefully it also shows that using the alpha parameter is also not unreasonable, even though I did not walk through the derivation of that last equation step by step.

Yes, I can see how the alpha parameter can be useful in calculating the theoretical error due to base current. Iout/Iin can be described as α² or 1-(2/HFE).

 

OK, that's a very clear explanation the current mirror. It ties in exactly with my previous understanding, and I still believe that describing the input transistor as a diode is misleading.

Nicely put.

Perhaps, but it's not the same as the relationship between base current and base-emitter voltage in the output transistor.

SO WHAT?

You are the only person that is fixated on this notion that that is what is required in order for a diode-connected transistor to be described as behaving like a diode!

The base-emitter diode is NOT the only diode in the world!

If the V-I charactersitic of a diode-connected transistor is the same as a diode (note the phrase "same as A diode", not "same as the base-emitter diode") over a particular region of operation, then what -- specifically -- is in any way misleading about saying that it behaves like a diode within that region of operation?

The input and output transistors are matched. The output transistor's base-emitter junction is described as a diode, and is connected in parallel with the input transistor, which is also described as a diode, and is a matched component.

So? Is everything that is described as a diode have to match everything else that is described as a diode?

Yes, the underlying base-emitter diode charactersitic of the diode-connected transistor is matched to the base-emitter diode characteristic of the reflector transistor. But that's not of any concern. In fact, given two transistors, one that matches the Ic vs Vbe characteristic but not the Ib vs Vbe characteristic and a second one that matches the Ib vs Vbe characteristic but not the Ic vs Bve charactersitic, which one would still be useable for making a mirror?

I believe that this clearly implies that they will have the same characteristics, and will draw the same amount of current, given the same forward voltage.

On what do you base this implication? The current flowing into one is the COLLECTOR current PLUS the BASE current while the current flowing into the other is just the BASE current. Why on earth do you think that these two charactersitics would our should be matched? Doesn't it make a LOT more sense that the COLLECTOR currents of the two would be matched, or at least nearly so?

But they won't.

That's right. Again. SO WHAT? We don't WANT a diode that matches the characteristics of the Ib vs Vbe characteristics of the transistor. We want a diode that matches the Ic vs. Vbe characteristics of the transistor. A diode-connected transistor constructed from a matched transistor accomplishes this. It behaves like a diode and that diode has the desired charactersistics. We don't give a damn about the Ib vs Vbe charactersitic. In fact, the whole point is to remove that parameter from consideration because that allows us to ignore β in terms being sensitive to its exact value. As long as the two transistors are well matched (i.e., have the same β) we don't care if it is 100 or 300.

I'm afraid not :-/

Then I don't think I can help you much more. My last attempt would be this.

Let's say that no one had ever come up with the diode-connected transistor and, instead, the second circuit I should was how we made current mirrors. Namely, each company that made transistors also frabricated diodes specifically intended for this purpose. Would you say that calling them a diode is misleading? Would you complain because these diodes had characteristics that matched the Ic vs Vbe characteristic of the transistor? Would you rather use a diode that matched the Ib vs Vbe charactersitic of the transistor? If so, what circuit would you use? How well would it perform?

What if then, many years later, someone realized that you could take a transistor and create a two-terminal device that had that same charactersitic as the diodes that everyone had been using for years and that, in fact, resulted in much more accurate current mirrors. Would you complain about them calling this transistor configuration that matches the characteristics of the diodes that everyone has been using for years a diode-connected treansistor? What would your basis be for claiming that such a name is misleading? Or that is it misleading to say that this two-terminal circuit that is simply replacing the diode in a circuit is now behaving like the diode that used to be there?

The onus really is on you to state precisely what the error is and what is misleading and how it should be described so as to not be misleading. Continuing so simply say that you feel it is misleading won't cut it any more.

 

Needless to say that I completely agree with WBahn and his contributions in this thread.
Perhaps one additional remark concerning the mentioned error due to the base currents:
Current mirrors are not used just for fun but for a particular application.
And one of these applications is the replacement of ohmic collector resistors in a differential amplifier by such a current mirror (active load).
Now: By proper matching/selection of the SECOND stage which is connected to this diff. amplifier (to the path without the BJT-diode) it is possible to compensate this systematic error introduced by unbalanced base currents.

 

You are the only person that is fixated on this notion that that is what is required in order for a diode-connected transistor to be described as behaving like a diode!

Apparently!

If the V-I charactersitic of a diode-connected transistor is the same as a diode (note the phrase "same as A diode", not "same as the base-emitter diode") over a particular region of operation, then what -- specifically -- is in any way misleading about saying that it behaves like a diode within that region of operation?

Specifically, that the diode represents a transistor that IS MATCHED to the output transistor (and this is emphasised in the description, for good reason), so the person reading the description could easily assume that a diode that represents this transistor WILL be matched to the output transistor's base-emitter DIODE (which is connected in parallel with it).

If the input transistor is to be represented as a diode, that diode's characteristics must be significantly different from the characteristics of the base-emitter junction of the output transistor, for the mirror to work properly. The descriptions that I have seen do not emphasise (or even point out) the important difference and the potential for confusion. This is what's misleading.

I've heard the current mirror described as "often misunderstood" and there's a case of someone misunderstanding it in the comments in the YouTube video I linked in my first post. He seemed to think that the input transistor doesn't have current gain, because its collector voltage isn't higher than its base voltage. I think the diode comparison is part of the reason for his confusion.

I'm sorry my persistence is frustrating. I'm not doing this just to p*ss you off, honestly! Instead of complaining about the analogy, I will offer my own description.

The transistors operate as matched amplifiers with current outputs (at the collectors). Their inputs (bases) are tied together. Because of their β, their collector currents are much higher than their base currents, which can be neglected.

Current into the circuit makes the transistors start to conduct (from collector to emitter). The input transistor's collector provides a kind of negative feedback by absorbing most of the circuit's input current.

If the input current increases, the base bias increases and the input transistor conducts more heavily, mostly counteracting the increase in base-emitter voltage. At all times, nearly all of the input current flows into the collector of the input transistor.

The input transistor regulates the base-emitter voltage of both transistors, keeping this base-emitter voltage equal to the voltage that causes a collector current (in the input transistor) that is equal to the circuit's input current.

Since the transistors are matched, the output transistor's collector current mirrors that of the input transistor.

That description is not as good as I hoped I could do, but it avoids the distraction of the diode comparison.

So? Is everything that is described as a diode have to match everything else that is described as a diode?

If they are in matched components, where the matching is critical to the circuit's operation, and the important difference was not clearly pointed out and explained, yes, I would assume that they were matched!

Yes, the underlying base-emitter diode charactersitic of the diode-connected transistor is matched to the base-emitter diode characteristic of the reflector transistor. But that's not of any concern. In fact, given two transistors, one that matches the Ic vs Vbe characteristic but not the Ib vs Vbe characteristic and a second one that matches the Ib vs Vbe characteristic but not the Ic vs Bve charactersitic, which one would still be useable for making a mirror?

Yes, that's true. It's actually the Vbe vs. Ic characteristic that has to match. But the current mirror is described in terms of Ib vs. Ic and I think that's probably best, since transistors are nominally current-driven.

On what do you base this implication? The current flowing into one is the COLLECTOR current PLUS the BASE current while the current flowing into the other is just the BASE current. Why on earth do you think that these two charactersitics would our should be matched? Doesn't it make a LOT more sense that the COLLECTOR currents of the two would be matched, or at least nearly so?

Yes, it would if the input transistor wasn't represented as a diode!

Then I don't think I can help you much more.
[...]
The onus really is on you to state precisely what the error is and what is misleading and how it should be described so as to not be misleading. Continuing so simply say that you feel it is misleading won't cut it any more.

I've given it my best shot. Perhaps I will have to just agree to disagree. Certainly, people much more knowledgeable than me (including you, WBahn) like the diode comparison. I do believe it's likely to cause confusion but that's not really my problem anyway

Thank you for your time. I apologise for being difficult.

 

I also don't have any problem with input transistor is being called diode-connected transistor.
The beginners may have some problem with this. But they think that the base current determines the collector current. But in reality this is not the case. The main idea behind current mirrors is that the output transistor behaves just like a VCCS (Vbe control current source). So by changing Vbe we change (set) Ic current.
Ic = Is*e^(Vbe/Vt) and we don't care about base current we assume hfe = oo.
Because the only things that matters here is Vbe vs Ic characteristic.
And the first transistor (diode) job is to convert this input current into Vin (Vbe) voltage and nothing more. And the output transistor job is to convert this Vin (Vbe) voltage into Ic current. And we can do this thanks to this equation
Ic = Is*e^(Vbe/Vt). And this is why we don't care about IB vs Vbe.
Also first transistor behaves like a diode because its only has two terminals and conduct current only in one direction. And this is the definition of a diode. And this is why we can call it a diode-connected transistor.

 

And the first transistor (diode) job is to convert this input current into Vin (Vbe) voltage and nothing more.

That's a good way to explain it. That should be in the tutorial!