# Differential pair large signal DC analysis

Discussion in 'Homework Help' started by Nol, Jun 7, 2012.

1. ### Nol Thread Starter New Member

Jun 7, 2012
2
0
Hello everyone,

I'm having problems with analysis of one of the circuits presented on this site in the worksheet section, under this link (also see attached screenshot). As you can see it's a BJT differential pair but I'm not interested in small-signal application but large signal DC analysis.

Normally I would start from assumption that these transistors work in linear mode, calculate DC operating point and then apply small-signal model to calculate Vout/(Vinverting - Vnon-inverting). However in this case, such assumptions are not possible (?) as this circuit does not have to operate in linear region and I'm looking for a complete transfer function, regardless of operating state.

That's why I came up with an idea to use Shockley model to characterize emitter currents Ie~Ies*exp(Vbe/Vt) and Vbe voltages. However, despite all my efforts I seem to end up with too many variables and too few equations to find a solution. Maybe it could've helped me if I could somehow assume/calculate constant current flowing through common emitter resistor, but I'm unable to tell if I can do such thing and how to calculate it.

Can someone help me and explain in a step-by-step manner how to approach analysis of such circuit? Some practical engineering tips would be most welcomed. Again, I'm not interested in small-signal analysis - my inputs will have full rail to rail swing. I would like to derive full transfer characteristic, that is Vout as a function of Vinv - Vnon-inv.

P.S.
Intuitively, I know that this circuit is going to be a comparator for most of input voltage range. Nevertheless, I still need my complete transfer function, not just linear approximation.

Thanks in advance!

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2. ### jegues Well-Known Member

Sep 13, 2010
735
43
Usually in a differential pair one uses a stable current source of some sort (i.e. a current mirror) attached to the emitter of each transistor. With this is in place it is such that for identical transistors, the current is split evenly between Q1 and Q2.

The first thing you need to do is determine which mode both Q1 and Q2 are in.

Assume a particular mode (why not start with active) and verify that this is indeed the case. (e.g. Vcb > 0)

3. ### Tesla23 Active Member

May 10, 2009
318
67
One approach:

Estimate as accurately as posible the current through the 1.5k tail resistor as a function of $V_+$ and $V_-$

$I_E(V_+, V_-)$ where $V_+$ and $V_-$ are the base voltages of Q1 and Q2 respectively

a crude estimate would be $I_E(V_+, V_-) = (max(V_+, V_-) - 0.6)/1.5k$

Replace the 1.5k emitter resistor with a current source $I_E$

Derive the transfer function - something like:

$V_{out} = 0.5*R_c I_E tanh(\frac{V_{dif}}{2V_t})$

with $V_{dif} = V_+ - V_-$

Substitute in $I_E(V_+, V_-)$ into the expression for $V_{out}$

Worry about what happens when Q2 saturates.

Last edited: Jun 8, 2012
4. ### Nol Thread Starter New Member

Jun 7, 2012
2
0
Thank you for your answers. After some additional thinking I came up with conclusions similar to Tesla23. Like you said, it's best to calculate maximum emitter current and substitute the resistor with a current source. We do not need to worry about distribution of currents between transistors to come up with a transfer function, as indicated by your formulas.

BTW, this simple example is a good way to test if one is able to capture physical sense of what is going on in this circuit. Normally people tend to see only small-signal models. This one forces to think outside of common assumptions.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
The first think that you need to do when you start this type of analysis. Is to determine the input voltage range. It's very impotent because, if difference between two input voltage is greater then 150mV than we almost for sure can tell that one of a BJT is in cut-off. So we end up with a CC or CE circuit. For example if Q1 base voltage is greater (150mV) then Q2. Then we can tell that Q1 is "ON" and Q2 is in cut-off.