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?"Yet Eber-Molls (eq. 2 -- assumedly without parsing your text) precisely accounts for temperature. Alpha does not.
I think we have universal consensus
Please show me a current mirror that doesn't.But relying on Vbe, Vt, n, and Ies is much worse.
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.The circuit above is not a good practice, the bjt likely dies.
Then why do millions of IC's do EXACTLY that?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.
And how do you design around Ie if the emitter is tied to the supply?Designing around Ie and alpha is optimum.
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.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.
If you were current limited to that degree, this it is highly unlikely that you damaged them.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.
I think that is what the debate is about. Certainly there is a relationship between base voltage and collector current.The word choice of "accurately" is poor.
And a very accurate one at that.Certainly there is a relationship between base voltage and collector current.
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.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.
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.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.
I just did.Try to explain it using Ic = beta * Ib.
I just did.You can't.
Best regards.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.
Indeed! Error factor of only a million. Not bad!And a very accurate one at that.
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.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.
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.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.
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?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.
by Duane Benson
by Jake Hertz
by Duane Benson
by Jake Hertz