# Expression for Current in a Small Signal Differential Amplifier

Discussion in 'Homework Help' started by HunterDX77M, Sep 29, 2013.

1. ### HunterDX77M Thread Starter Active Member

Sep 28, 2011
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2
I have the diagram attached for a small signal model of a differential amplifier that my professor gave me. I am having a little trouble understanding where he got the expression for the current ib2.

According to his notes, it says that when source e2 is 0 V (that is, shorted to ground), the expression for the current is

$i_{b2} = \frac{-e_1}{2 \times h_{ie}}$

I've been staring at this for a while now trying to understand his logic and am stumped. How did he get this expression? What loop do you have to go through? Thanks in advance for any help.

• ###### Small Signal Diff Amp.PNG
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Last edited: Sep 29, 2013
2. ### WBahn Moderator

Mar 31, 2012
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It would help to have the schematic of the circuit for which this is the small signal model.

3. ### LvW Active Member

Jun 13, 2013
674
100
I suppose, it is obvious that the corresponding circuit is the classical two-transistor differential pair with an ohmic resistor Ro in the common emitter path.

*However, the mentioned expression R=Ro(1+beta) applies only if one of the signal sources is set to zero volts (in the case under discussion e2=0).
* Now - it can be assumed that the value of R is much larger than hie since the value of hie typically is in the lower kOhm range.
* In this case, it is obvious that ib2=-e1/2*hie.
* This result confirms the known fact that the signal input resistance of a common-collector-common base combination has an input resistance of 2*hie (identical transistor parameters assumed).

Last edited: Oct 1, 2013
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4. ### HunterDX77M Thread Starter Active Member

Sep 28, 2011
99
2
As requested, here is the original circuit. My question wasn't so much an analysis question as it was an algebra question, which is why I didn't include it originally and went straight to the small signal model.

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5. ### HunterDX77M Thread Starter Active Member

Sep 28, 2011
99
2
Thank you for your response; it makes a lot of sense. My professor wasn't clear that he took R >> hie, which is why I wasn't sure where he got that expression from. It is now obvious that since you effectively have 2hie there, it is basically a voltage divider cutting e1 in half.

6. ### LvW Active Member

Jun 13, 2013
674
100
Hunter, because you have mentioned the expression R=Ro(1+beta) I have assumed an ohmic emitter resistance. Now - with respect to your circuit diagram - the value of Ro seems to be the dynamic internal resistance of the BJT current source in the common emitter path.
Thus, the mentioned assumption (R>>hie) is fullfilled even better.