How can the amplification of a bipolar transistor be influenced by Ic / Vt?

[king.oslo]

Hello there,



To me this looks like transconductance of a bipolar transistor is how Ic changes as an inverse function of temperature, as thermal voltage is directly proportional to temperature:



My tutor said the transconductance gm "says a lot about the gain of a transistor". How can this be? To me it looks like transconductance says how collector current of the transistor changes a function of temperature.

Thank you for your time.

Kind regards,
Marius

[jpb]

Transconductance is the amount that the output current changes for a change in the input voltage.

For a FET it is the ratio of delta ID to delta VG (I know most about FETs).

For a bipolar device it should be the ratio of collector current to base-emmiter voltage. You seem to have swapped base-emitter voltage with the junction voltage VT.
CORRECTION - it is not the ratio but the differential as it is small signal gain.

The transconductance determines the gain of the device depending on input and output impedance. Normally the input impedance is very large compared to the output so that
the input current is very small compared with the transconductance output current.

[king.oslo]

That's what I thought. But both Wikipedia and my tutor disagrees: http://en.m.wikipedia.org/wiki/Transconductance



I think this says fuck all about gain

What is going on?

Thanks M

[vlf3]

I wonder where your tutor is getting his course material from,  cynical me.

[jpb]

That's what I thought. But both Wikipedia and my tutor disagrees: http://en.m.wikipedia.org/wiki/Transconductance

My definition above agrees with the Wikipedia link that you give.

Remember that the collector current is given approximately by the diode equation
I0*(exp(VBE/mVT)-1)

to get the small signal transconductance we need to differentiate (I didn't make that clear in my first post but have corrected it now)

dIC/dVBE = (I0/mVT)*exp(VBE/mVT) which is approximately IC/mVT as the -1 term is generally much smaller than the exponential.

So yes your tutor is correct approximately and ignoring the ideality factor m and the fact that the base current should come off this as well.

The VT term comes in as it gives the slope of the curve which determines the transconductance.


Perhaps the best way of looking at this is not to think about IC influencing the transconductance, instead IC gives the corresponding VBE value (like reading a point on a curve by using the Y co-ordinate instead of the X co-ordinate) then VT comes into it because it affects the whole shape of the curve. The transconductance is just the slope at that point.

If it was simply the curve y = x^2 and I asked you what the slope was at y=4 you could read across and see that this was x=2 and the slope was 2x = 4. The value of y doesn't determine the slope in a logical way it just determines the point on the curve and each point has a different slope.

[king.oslo]

That's what I thought. But both Wikipedia and my tutor disagrees: http://en.m.wikipedia.org/wiki/Transconductance

My definition above agrees with the Wikipedia link that you give.

Remember that the collector current is given approximately by the diode equation
I0*(exp(VBE/mVT)-1)

to get the small signal transconductance we need to differentiate (I didn't make that clear in my first post but have corrected it now)

dIC/dVBE = (I0/mVT)*exp(VBE/mVT) which is approximately IC/mVT as the -1 term is generally much smaller than the exponential.

So yes your tutor is correct approximately and ignoring the ideality factor m and the fact that the base current should come off this as well.

The VT term comes in as it gives the slope of the curve which determines the transconductance.


Perhaps the best way of looking at this is not to think about IC influencing the transconductance, instead IC gives the corresponding VBE value (like reading a point on a curve by using the Y co-ordinate instead of the X co-ordinate) then VT comes into it because it affects the whole shape of the curve. The transconductance is just the slope at that point.

If it was simply the curve y = x^2 and I asked you what the slope was at y=4 you could read across and see that this was x=2 and the slope was 2x = 4. The value of y doesn't determine the slope in a logical way it just determines the point on the curve and each point has a different slope.

Thanks,

What is the reverse bias saturation current, or I0, from your equations?

Kind regards,
Marius

[jpb]

What is the reverse bias saturation current, or I0, from your equations?

Kind regards,
Marius

I'm not a great expert on this (my experience has been with FETs and HEMTs), as the diode approximation is just that - geometrically the base region is very thin and the collector and emitter regions are much larger. Under reverse bias conditions both the emitter-base and collector-base "diodes" are reverse biased. There will be some leakage but how it flows will depend on geometry, doping etc., I don't think it would bear any simple relation to the I0 in the rather crude diode approximation used to estimate the transconductance.

[king.oslo]

What is I0 an abbreviation for then? Thanks M

[jpb]

What is I0 an abbreviation for then? Thanks M

I just picked I for current and the 0 is supposed to be subscript 0 to show it is a constant in the equation. I was typing things in a bit of a hurry so didn't bother with subscripts etc.
i.e. it should be

I₀(e^V_BE/mV_T-1)

but that is a pain to type.