How accurately does Vbe determine collector current?

dannyf

Joined Sep 13, 2015
2,197
"3) Ie = alpha*Ie is the best predictor of Ic. Comments/questions welcome"

That's almost like saying that your body weight in pound is the most accurate predictor of your body weight in kg.
 

cabraham

Joined Oct 29, 2011
82
Yet Eber-Molls (eq. 2 -- assumedly without parsing your text) precisely accounts for temperature. Alpha does not.
Sure it does. The alpha variation with temp is so minute it can be neglected with an error of 1 or 2%. But Ies varies greatly with temp, and one can account for it in the equation. But as temp varies, Ic can vary by a huge factor, much more so than alpha, or even beta. The OP question is "which parameter most accurately describes Ic, Ib or Vbe?"

My reply was a direct answer. Ies varies from part to part, and hugely with temp. Alpha also varies from part to part and with temp. But the variation is so miniscule with alpha, it is safe to say that Ie is a much better method of computing Ic. Even eq 1), with beta being as sloppy as it is, is more accurate than Ebers-Moll. But, do not misunderstand me. I never design networks relying on eq 1). A factor of 6 to 1 is still too sloppy for most of my work. If your point is that I should avoid beta-dependent design practices, I assure you I am well aware of that.

Eq 1) Ic = beta*Ib is indeed a very sloppy way to do it. However it is used. For op amps and discrete power amps using the ubiquitous 3-stage topology by Dr. Lin, the 2nd stage where voltage gain occurs, known as the "VAS", eq 1) is employed. The 1st stage is an emitter coupled pair, also called differential pair or in Europe a long tail pair. The output of the diff stage is a current source, the collector of a bjt.

This current is inputted to a bjt base with its emitter at ground (or the rail, or both rails for a complementary pair VAS). So the VAS bjt receives a base current as its input. What is Ic for this 2nd stage? It is merely beta*Ib. Of course, this beta value is not very reliable, and normally such practice is foolish.

But op amps and audio power amps employ global negative feedback. This mitigates open loop gain variation to result in fixed closed loop gain. If you ever examine the data sheet of an op amp, the open loop gain, "Aol", is not fixed but varies by a factor of 5, 8, or 10. This is due to beta variation in 2nd stage.

I hope we can agree that relying on Ib and beta to determine Ic is BAD design practice. Hopefully you say amen to that. But relying on Vbe, Vt, n, and Ies is much worse. These are not predictable either. Sure one can look up the equation for temp function of Ies, n, Vt, etc. But I can examine data sheets and extract temp function for beta and alpha. Some semiconductor text books elaborate on this.

Anyway, I think we have universal consensus that eq 1) is not the right way to implement bjt networks. Eq 2) is horrible, one can never force a Vbe value and think they have a stable Ic value. Also forcing Vbe with a low impedance voltage source likely kills the bjt. Only eq 3) predicts Ic to a high degree of accuracy. Designing around Ie and alpha is optimum.

Comments/questions always welcome. Thank you for your interest. Best regards.

Claude
 

WBahn

Joined Mar 31, 2012
26,074
The circuit above is not a good practice, the bjt likely dies.
Yes and no. If you just pick a voltage source off the shelf then, yes, it will likely kill the transistor. If it is a very well controlled source, then you readily use it to map out the response. But you wouldn't want to use it open-loop because of the thermal runaway that is being invited.

Using Ic = alpha*Ie is fine, when you can do it. But that's just because with any significant beta, the emitter current is within a percent or so of the collector current.

But there are plenty of circuits, particularly IC circuits, for which you can't determine Ie without treating the transistor as a voltage-controlled current source.

The classic current mirror is one of them -- and it is used on millions of integrated circuits just as shown.

upload_2016-1-28_17-38-13.png

Is it a "good" current source? Not particularly -- but it is "good enough" for millions of integrated circuits.

Try to explain how this works using only Ic = alpha * Ie and, in particular, without relying on Vbe as playing a critical role.

You can't.

Try to explain it using Ic = beta * Ib.

You can't.

Why not? Simple -- because the left-hand side of the circuit doesn't ensure that some value of Ie or Ib is transferred to the right-hand side, it ensures that the value of Vbe is transferred to the right-hand side.
 

WBahn

Joined Mar 31, 2012
26,074
I hope we can agree that relying on Ib and beta to determine Ic is BAD design practice. Hopefully you say amen to that. But relying on Vbe, Vt, n, and Ies is much worse. These are not predictable either.
Then why do millions of IC's do EXACTLY that?

Designing around Ie and alpha is optimum.
And how do you design around Ie if the emitter is tied to the supply?

When you do have an element, such as a resistor, in the emitter path and use that to set Ie, the mechanism being used is to employ negative feedback to set Vbe.
 

joeyd999

Joined Jun 6, 2011
4,425
Then why do millions of IC's do EXACTLY that?



And how do you design around Ie if the emitter is tied to the supply?

When you do have an element, such as a resistor, in the emitter path and use that to set Ie, the mechanism being used is to employ negative feedback to set Vbe.
Further, I'd like to seem him design or analyze a band-gap reference without Eber-Molls. And these things are *far* more stable than 1-2% over a broad operating temperature range.
 

Thread Starter

dsharp02

Joined Aug 8, 2015
17
I want to say thank you to everyone that has replied. At the very least I think I understand now that using Vbe to characterize Ic is only valid for a single instance of transistor, and not generally across all transistors of that type.

As for my voltage sources, I was using a lab PSU with current-limited to 1ma on the base and 50ma on the collector, so hopefully I didn't damage any of my transistors.
 

WBahn

Joined Mar 31, 2012
26,074
I want to say thank you to everyone that has replied. At the very least I think I understand now that using Vbe to characterize Ic is only valid for a single instance of transistor, and not generally across all transistors of that type.

As for my voltage sources, I was using a lab PSU with current-limited to 1ma on the base and 50ma on the collector, so hopefully I didn't damage any of my transistors.
If you were current limited to that degree, this it is highly unlikely that you damaged them.

If you are going to try to run Vbe directly from a supply, keep in mind that the current response is exponential. A difference of only about 60 mV will result in a factor of ten change in collector current.
 

cabraham

Joined Oct 29, 2011
82
MOD NOTE: Edited to remove responses from the quoted text -- notice how the need to use colored text disappears.

Yes and no. If you just pick a voltage source off the shelf then, yes, it will likely kill the transistor. If it is a very well controlled source, then you readily use it to map out the response. But you wouldn't want to use it open-loop because of the thermal runaway that is being invited.
What you just described is a constant current source. Adjusting voltage so as to avoid thermal runaway is not a constant voltage source. Varying V so as to insure steady I is precisely "current control". An ideal voltage source has zero series resistance, insuring destruction of the bjt. Current driven is how they must be operated.

Using Ic = alpha*Ie is fine, when you can do it. But that's just because with any significant beta, the emitter current is within a percent or so of the collector current.

But there are plenty of circuits, particularly IC circuits, for which you can't determine Ie without treating the transistor as a voltage-controlled current source.

The classic current mirror is one of them -- and it is used on millions of integrated circuits just as shown.

View attachment 99599

Is it a "good" current source? Not particularly -- but it is "good enough" for millions of integrated circuits.

Try to explain how this works using only Ic = alpha * Ie and, in particular, without relying on Vbe as playing a critical role.

You can't.
Sorry, but I sure can. Vcc source supplies current through R1 into the tied together bases of Q1 & Q2. What happens when a current in a single branch incurs 2 elements in parallel. Division of course, like 2 resistors in parallel, if they have equal values current divides into halves. The bases are 1 forward junction drop above ground. Each base current is computed as ((Vcc-Vbe)/R1)/2. If Vcc >> Vbe, then Vbe has little influence on Ib. The current in R1 must divide between the 2 b-e junctions in parallel. If the 2 b-e junctions are equal area and doping and length, current divides in half. So Ib1 = Ib2 = Ir1/2. Ic2 = beta*Ie2. The key is making sure Ie1=Ie2. With the circuit shown, the match is not good at all. Forcing the 2 b-e junctions in parallel to insure equal Vbe's is not as accurate as forcing equal Ie's.
On IC chips, the 2 junctions track with temp, and emitters have inherent resistance. This helps by degenerating the b-e junction and tends to equalize current division. Still, emitter ballast resistors are recommended. With your circuit, if the Ies values of the 2 parts vary, which they do, the bjt with lower Ies value gets less current. This lesser Ie2 value results in an Ic2 value less than Ir1.

Add a resistor in each emitter, Re1 & Re2. In the 1st case, what resistance does each b-e junction present. It is hie or r_pi if you prefer. If they match, base currents are equally divided. If not, they do not match. But what is r_pi? Is it not hfe/gm? If betas don't match, each part presents differing input resistance, and mirror is uneven.

So with the emitter resistors, the impedance seen at each base is now hie+(hfe+1)*Re. The Ib value is (Vcc-Vbe)/(R1+hie+(hfe+1)*Re). So Ic=hfe*Ib=(Vcc-Vbe)*hfe/(R1+hie+(hfe+1)*Re). Notice the "hfe" in the numerator, and the "hfe+1" in the denominator. If Re is sufficiently large, and hfe is reasonable, then the mismatched "hie" values are compensated, and currents divide more evenly. I do not work at an OEM that makes chips, I design with discrete parts. I never use your rendition of current mirror because it is not good, esp with discretes.

With emitter resistors, the 2 Vbe values can be very unequal, but a large enough pair of Re parts can assure balance. Because Re assures equal current division resulting in equal Ie values. Since alpha is very close to 1, Ic values match well.

Just for thought, how about a weighted mirror? If Re1 differs from Re2, the mirrored current is scaled, not equal to the reference current. That can only be explained with alpha and Ie, not Vbe. The reason we do not worry about beta with standard current mirrors is that 2 parts on the same substrate with equal lengths and areas and doping, will track with temp and betas should be close to equal. Thus Ie values match as does Ic values.


Try to explain it using Ic = beta * Ib.
I just did. :)

You can't.
I just did. :)

Why not? Simple -- because the left-hand side of the circuit doesn't ensure that some value of Ie or Ib is transferred to the right-hand side, it ensures that the value of Vbe is transferred to the right-hand side.
Best regards.

Claude
 
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cabraham

Joined Oct 29, 2011
82
Further, I'd like to seem him design or analyze a band-gap reference without Eber-Molls. And these things are *far* more stable than 1-2% over a broad operating temperature range.
Different creature altogether. With band gaps we bias 2 junctions and measure the difference across them. Better than Zeners, if I may add. We do not use them to get voltage gain, nor current gain. A bjt provides both. Apples and oranges. Not germane to this OP question.

Claude
 

hp1729

Joined Nov 23, 2015
2,304
Okay here is an exercise. Same transistor, saturation avoided. Three different collector loads. Base voltages the same, collector currents similar, collector voltages differ greatly.
No indication you can look at base voltage and predict collector voltage but yes for collector current. Beta is pretty constant.
It does, in deed.
I don't think you can calculate ic from vbe, but knowing the parts and the circuit in question you can go from experience. Is using base current better? Perhaps. Knowing beta calculations can be made and base current varies ore than vbe so changes can be measured easier.
 

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WBahn

Joined Mar 31, 2012
26,074
What you just described is a constant current source. Adjusting voltage so as to avoid thermal runaway is not a constant voltage source. Varying V so as to insure steady I is precisely "current control". An ideal voltage source has zero series resistance, insuring destruction of the bjt. Current driven is how they must be operated.


Sorry, but I sure can. Vcc source supplies current through R1 into the tied together bases of Q1 & Q2. What happens when a current in a single branch incurs 2 elements in parallel. Division of course, like 2 resistors in parallel, if they have equal values current divides into halves. The bases are 1 forward junction drop above ground. Each base current is computed as ((Vcc-Vbe)/R1)/2. If Vcc >> Vbe, then Vbe has little influence on Ib. The current in R1 must divide between the 2 b-e junctions in parallel. If the 2 b-e junctions are equal area and doping and length, current divides in half. So Ib1 = Ib2 = Ir1/2. Ic2 = beta*Ie2.
Note how you just invoked Ic2 = beta*Ib2 (at least I'm assuming you meant to say Ib2 and not Ie2). So you can't very well go on to say that you explained the circuit operation without using Ic = beta*Ib.
 

cabraham

Joined Oct 29, 2011
82
If you were current limited to that degree, this it is highly unlikely that you damaged them.

If you are going to try to run Vbe directly from a supply, keep in mind that the current response is exponential. A difference of only about 60 mV will result in a factor of ten change in collector current.
Actually that is just part of it, there is more, and it gets worse. If a CVS (constant voltage source) had zero noise and ripple and was fine adjustable to a microvolt, the bjt still faces destruction. If one connects a bjt directly across the CVS set for 0.650 V, what happens?

A current exists in the b-e junction, and this results in collector current Ic per Ebers-Moll. With current Ic, current Ib, voltages Vce, and Vbe, there is power dissipated by the bjt. The thermal coefficient is Rth in deg C/W. Power times Rth is junction temp rise. So when dissipating power, temp must rise. But per Ebers-Moll
2) Ic = alpha*Ies*exp((Vbe/Vt)-1), where Ies(T) is a strong funnction of temp, non-linear. A rise in temp produces a large rise in Ies, non-linear. The Vt (thermal voltage) does increase but the net is that Ies dominates.

The result is that with 0.650V Vbe held constant, Ic must increase due to Ies increasing. This makes power increase, which increases temp, which increases Ies, which increases Ic, which increases power, which increases temp ---------------. I think you know the rest, as you stated above - thermal runaway.

But instead of a CVS, let's use a CCS (constant current source). We can force Ie or Ib to the CCS value. Let's use Ie set to 10 mA. The voltage Vbe is given by Ebers-Moll

2) Vbe = Vt*ln((Ie/Ies)+1).

The bjt has Ie = 10 mA, and Vbe is given by eq 2) above, we get Ic = alpha*Ie = alpha*10mA, around 9.9 mA. Power times Rth gives temp rise, so junction temp increases. The result is an increase in Ies, and increase in Vt. But Vt increases in proportion to absolute temp (Kelvin), whereas Ies is non-linear temp relation, and dominates. Ies is in the denominator so Vbe decreases with temp increasing. As long as the power dissipated is below the limit for the part, thermal stability results.

Driving a p-n junction with a CCS is stable, whereas CVS driven is runaway. THis is why unis drum into my head and my colleagues to always drive p-n junctions such as LEDs, SCRs, triacs, and bjt with current not voltage. These are "current-driven" devices. However a CVS can be used if a sufficient value of resistance is used. A CVS plus series resistor is safe. The current cannot run away since the resistor and CVS value put a limit on current and power.

Just thought it deserved to be mentioned. BR.

Claude :)
 

WBahn

Joined Mar 31, 2012
26,074
Nowhere did I say that you should design circuits such that you are driving the base-emitter junction with an open loop voltage source. I was talking about what the TS was doing, which was using such an arrangement to measure the collector current in a test circuit which would typically mean holding the Vce at a constant voltage and measuring only the initial collector current and then letting the device cool (or ramping Vbe quickly enough so that you can sweep out the curve before the junction begins to heat). If you allow the junction to change temperature (by any significant amount) then you have poisoned your data.
 
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