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- From: bwhite@cobra.camb.inmet.com (Bill White)
- Newsgroups: sci.electronics
- Subject: Grounded Emitter Amplifier Query
- Message-ID: <1992Dec29.214231.1135@inmet.camb.inmet.com>
- Date: 29 Dec 92 21:42:31 GMT
- Sender: news@inmet.camb.inmet.com
- Organization: Intermetrics Inc, Cambridge MA
- Lines: 96
- Nntp-Posting-Host: cobra
-
- 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.
-