From: bwhite@cobra.camb.inmet.com (Bill White)
Newsgroups: sci.electronics
Subject: Grounded Emitter Amplifier Query
Date: 29 Dec 92 21:42:31 GMT
Organization: Intermetrics Inc, Cambridge MA

Hello.  I am reading Horowitz and Hill's book on Electronics, but 
I am having trouble understanding the grounded emitter circuit.  In
particular, I am having trouble understanding why the grounded emitter
is not thermally stable.  I believe it, I just don't understand why
it is true.  

I have the first edition if H&H.  If the second is more
clear on transistors I will go buy it, but it is very expensive.
I read somewhere that the only real difference is in the digital
parts, which I am not particularly interested in. I also wrote up
this note from memory, so it is very possible that I got the
formulas below wrong.  If so, please forgive me.

To review the bidding, the grounded emitter circuit looks something
like this:

	-----+-------------+---------------------- V_{CC}
             |             |
             |             |
             \             \
             /             / R_{C}
             \ R_1         \
             /             /
             |             |
             |             |
             |             +--------------------- V_{OUT}
             |             | (collector)
             |        +---------+
             |        |         |
      V_{B}->+--------+   npn   | (I've never seen a good
             | (base) |         |  ascii transistor.)
             |        +---------+
             |             | (emitter)
	     \             |
             /             |
	     \ R_2         |
             /             |
             |             |
             |             |
             |             |
           -----         -----
	    ---           ---
             -             -

	NOTE: I write V_{CC} for "V with subscript CC".

The axioms are:
    a The Ebers-Moll model of transistor behavior states that under
      certain conditions conditions which establish a well-conditioned
      environment the following two equations hold.

	I_{C} = I_{S}[exp(V_{BE}/V_{T})+1]	(*)
	I_{B} = I_{C}/h_{FE}			(**)

      Where V_{T} depends on temperature, and is kT/q =
      25mV@20degreesC, and h_{FE} is a constant.  Now, I_{C} >> I_{S},
      so the "+1" term in (*) may be ignored.
    b V_{BE} changes -2.1mV/degreeC with temperature.  That is,
      V_{BE} decreases as the temperature increases.
    c I_{C} increases 10X for each 60mV increase in V_{BE}.  (I
      may have this wrong.  It may be 60mV increase in V_{B}, which
      changes things tremendously.)

The problem is to show that if R_1 and R_2 are set to hold V_{C}
equal to (1/2)V_{CC} that and 8degreeC change in temperature will
cause the transistor to saturate.  It seems to me that as temperature
increases, V_{BE} gets smaller and V_{T} (= kT/q) gets bigger.  This
means V_{BE}/V_{T} must get smaller.  The exponential function is
always increasing, so I_{C} = exp[V_{BE}/V_{T}] must get smaller
when T gets bigger.  If I_{C} gets smaller, this takes the transistor
*away* from saturation, not toward it.

I thought of the following answer.  The voltage divider formed by R_1
and R_2 holds V_{B} at one diode drop above ground.  If this is
sufficiently stiff, V_{BE} cannot decrease, even if the transistor
wants it to.  Now, an increase of 8degreesC means that the transistor
wants to decrease V_{BE} by 16mV, but the voltage divider won't let it
happen.  Therefore, the 16mV is actually a net increase.  From axiom a
we get:

	I_C  = I_S[exp(V_{BE}/V_{T})]	   (axiom a)
	I_C' = I_S[exp(V_{BE}-16mV)/V_{T}] 
					   (Why doesn't V_{T} change?)
	     = I_S[exp(V_{BE}/V_{T})]/[exp(16mV/V_{T})]
	     = I_C/[exp(16mV/V_{T})]

If V_{T} is 25mV, the multiplier is 
	1/[exp(16/25)] = 1/[exp(0.64)]
		       = 0.53.

This makes me think that I_{C} is going to decrease as well.
Furthermore, the transistor could act as a small resistance
between the base and emitter to make the input impedance as
small as it wants.  That is, there isn't any rule which tells
us that the voltage divider can be stiff enough to hold V_{BE}
to be one diode drop.
Newsgroups: sci.electronics