# Ebers-Moll: Ic is nonconstant, but Vbe is approximately constant.

1. Nov 13, 2013

### Mark9

Hello all,

The Ebers-Moll equation as I understand it is: Ic = Is*exp((Vbe/Vt) -1). Question1: Assume fixed temperature; if Vbe is approximately 0.6 volts, as is typically used in transistor current source analysis, then how does Ic change at all? A core issue here is that I was under the impression that since Vbe is basically due to diode forward drop, that Vbe should be approximately constant. I'm looking to understand the NPN transistor as transconductance device(Vbe controls Ic), but the way I'm reading things, the equation just gives me a constant Ic(for fixed Temp). Question2: If the answer is that Vbe does indeed vary, then please clarify how this is possible. I thought that diodes maintained constant(approximately) forward voltage drop over a wide range of currents.

Thank you in advance to any takers!
Mark

2. Nov 13, 2013

### analogdesign

Question 1: Vbe is approximately 0.5 - 0.7 volts depending on temperature and various other factors, but it does change. If you change the current through it, Vbe changes to accommodate that. Note the Ic is exponentially related to Ic, so only a small change in Vbe is needed to bring about a large change in Ic.

Also, don't confuse Large Signal analysis with Small Signal analysis. The large signal analysis would be to look at the DC value of Vbe to get the quiescent or DC value of Ic. Then you would super impose a small sine wave (for example) on Vbe which is translated into a sinusoidal variation in Ic. This is transconductance and its a small signal property.

Question 2: The answer is in your question! The diode only maintains "approximate" forward voltage drop, and a little variation in voltage can translate into a big variation in Ic because of the exponential relation.

3. Nov 13, 2013

### Mark9

Thank you analogdesign! If you don't mind, I would like to try to check my understanding by restating in different words something you've said, and then ask a quick follow up question. Please correct me if I'm mistaken. Then the reason why I can use a value in the range 0.5-0.7 as an estimate of Vbe in,say, designing a transistor current source, is because my driving base voltage(Vb relative to ground) will be orders of magnitude larger than Vbe. So the small differences(0.5,0.6,0.7 etc..) won't significantly impact any such calculations. BUT, when considering Ic, even small deviations count, so the ebers-moll exponential relationship is necessary.
FOLLOW UP QUESTION: Since(at room temp) Vt = 23.5mV, this means that(according to ebers-moll) Ic>0 whenever Vbe>0.0235. But I've read that to ensure that the emitter is conducting, (i.e., Ic >0) I must make Vbe>0.6. But the difference 0.6 - 0.0235 is quite large and NOT negligible from the perspective of the ebers-moll equation. So why the large discrepancy, and do I really need Vbe > 0.6? Is there some reason why it is necessary in practice to make Vbe so much bigger than Vt=23.5mV??

4. Nov 13, 2013

### analogdesign

Hi Mark,

Almost, but not quite. 0.1 V is a big change. I was thinking like a sinewave on the order of mV riding on the 0.6 or 0.7. Large signal BJT calculations are almost trivial because just remember Ic about equals Ie and Ic = Beta*Ib.

The BJT is a very complex device from a physics standpoint. The transition from OFF to ON is not super sharp, so there *is* current flowing even when Vbe is small. Consider the "reverse saturation" current. It is the "Is" in the Ebers Moll eqn and it is very small. In fact it is typically less than picoamp (it depends on the details of the transistor's construction).

So, if VBE = 0.6 V, vt = 23 mV, and Is= 1pA then Ic = is*exp(vbe/vt - 1) = 78 mA *big*

Now if VBE = 0.1 (the BJT is "off") then Ic = 28 pA.

So the current is 9 order-of-magnitude smaller when VBE = 0.1. For most practical (but not all!) the device can be considered to be OFF.

Make more sense now?