# Differential BJT

I'm trying to understand the differential mode for a BJT. For the second picture, when we apply +1V to the base of Q1 and 0V to the base of Q2. Why is the emitter voltage .3V instead of -.7V? Similarity, for the third picture, why is the emitter voltage -.7V instead of 1.7V?
Thanks for the help!

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## Answers and Replies

NascentOxygen
Staff Emeritus
I'm trying to understand the differential mode for a BJT. For the second picture, when we apply +1V to the base of Q1 and 0V to the base of Q2. Why is the emitter voltage .3V instead of -.7V?
Because VBE of the ON transistor must be +0.7V.
Similarity, for the third picture, why is the emitter voltage -.7V instead of 1.7V?
For the same reason. If VBE is any value other than +0.7V it follows that an NPN transistor will not be functioning.

Because Q1 pulls the emitters to +0.3V while Q2 can't sink the voltage. It results from the non-linear behaviour of bipolar transistors, whose emitter current increases brutally if the base-emitter voltage exceeds the threshold.

By the way, 0.7V is not a universal value. A low-power bipolar has rather 0.6V or 0.65V, while many-amps bipolar can have 1.5V, most of it being wasted in the stray resistances to access the emitter and the base.

NascentOxygen
Staff Emeritus
I'm trying to understand the differential mode for a BJT. For the second picture, when we apply +1V to the base of Q1 and 0V to the base of Q2. Why is the emitter voltage .3V instead of -.7V?
Note the constant current source, I, in the emitter leads. So as one transistor conducts better and takes more emitter current, it deprives the other transistor of an equal amount of current. Whichever transistor has the greater VBE* has the greater IB therefore its emitter (and collector) will take the greater share of the fixed bias current, I.

* being a PN junction it obeys the V-I exponential curve

Back to your first question. If the emitter voltage is 0.3V, then Q1 has VBE=0.7V and Q2 has VBE=–0.3V, whereas if the emitter voltage were –0.7V, then Q1 would have VBE=+1.7V and Q2 would have VBE=0.7V

VBE=+1.7V is impractical here, but let's imagine Q1 was to be given a lot of base current, then it would take all of the available emitter current, I, in turn depriving Q2 of emitter current and turning Q2 off, meaning VBE of Q2 would fall below 0.7V.

Note the constant current source, I, in the emitter leads. So as one transistor conducts better and takes more emitter current, it deprives the other transistor of an equal amount of current. Whichever transistor has the greater VBE* has the greater IB therefore its emitter (and collector) will take the greater share of the fixed bias current, I.

* being a PN junction it obeys the V-I exponential curve

Back to your first question. If the emitter voltage is 0.3V, then Q1 has VBE=0.7V and Q2 has VBE=–0.3V, whereas if the emitter voltage were –0.7V, then Q1 would have VBE=+1.7V and Q2 would have VBE=0.7V

VBE=+1.7V is impractical here, but let's imagine Q1 was to be given a lot of base current, then it would take all of the available emitter current, I, in turn depriving Q2 of emitter current and turning Q2 off, meaning VBE of Q2 would fall below 0.7V.

So when analyzing the circuit, one of the transistors BE voltage has to be .7 and the other has to be less than .7? If Vbe is greater than .7 then it would be wrong? Also when doing these diff bjt, is it always one transistor on and one off? or can it have both transistors on?

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NascentOxygen
Staff Emeritus