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Differential pairs

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Bipolar differential pair

Figure 1. Basic bipolar differential pair

We first start our foray into the world of differential circuits by considering the basic bipolar differential pair in Basic bipolar differential pair.

• Set the two input voltages to zero.

What is the emitter voltage v_E? (think of the answer before revealing!)

The base voltages are the same because they are set to be so, the emitter voltages are the same because they are attached to the same node.

v_BE is about 0.6{V}, so v_E is about −0.6{V}

This is not a problem so far, but what happens when V1 and V2 are not equal? Is this even possible? How can V1 be anything different if v_BE is supposed to always be 0.6{V}?

Wait a second: why “must” v_BE be 0.6{V} in the first place?

It is merely an approximation of the base-emitter voltage in active mode that causes collector currents somewhere around 1{mA} at around room temperature (300 K).

This connection forces us to use a more accurate approximation for the collector current, see Table: Bipolar transistor modes:

\$i_E = \dfrac{i_C}{\alpha} = \dfrac{I_S}{\alpha} \exp\left(\frac{v_{BE}}{V_T}\right)\$

KCL forces the emitter currents to sum to I_tail. A substitution for each transistor’s current gives us:

\$I_{tail} = \frac{I_{S2}}{\alpha_2} \exp\left(\frac{v_{BE2}}{V_{T2}}\right) + \frac{I_{S2}}{\alpha_2} \exp\left(\frac{v_{BE2}}{V_{T2}}\right)\$

So many currents and voltages — take a minute to list out which are actually transistor properties and not part of the circuit analysis, but just have units of {V} or {A}.

• \$I_{S1}\$ and \$I_{S2}\$ — depend on individual doping, geometry, and other manufacturing variations (and junction temperature)

• \$V_{T1}\$ and \$V_{T2}\$ — thermal voltage \$k_B T / q\$, transistors may be at different temperatures!

Since the emitters are connected together, we can write the base-emitter voltages as

• \$v_{BE1} = v_1 - v_E\$

• \$v_{BE2} = v_2 - v_E\$

Assumptions

To proceed much more and arrive at the important result of this path, we need to make two assumptions:

1. Both transistors are at the same temperature.

2. The transistors are matched, meaning:

   • \$I_{S1} = I_{S2} = I_S\$, and

   • \$\alpha_1 = \alpha_2 = \alpha\$

3. Then, purely and only for convenience, we’ll approximate \$\alpha \approx 1\$

Now, factor the common v_E term out of the exponentials.

\$I_{tail} = I_S \exp\left(\frac{-v_E}{V_T}\right) \left[\exp\left(\frac{v_1}{V_T}\right) + \exp\left(\frac{v_2}{V_T}\right) \right]\$

and rearrange for later use

\$I_S \exp\left(\frac{-v_E}{V_T}\right) = \frac{I_{tail}}{\exp\left(\frac{v_1}{V_T}\right) + \exp\left(\frac{v_2}{V_T}\right)}\$

Follow this progression for the collector current in transistor Q1 (each step is simple!):

\$\begin{align} i_1 &= I_S \exp\left(\frac{v_{BE1}}{V_T}\right) \\$ \$&= I_S \exp\left(\frac{v_1}{V_T}\right) \exp\left(\frac{-v_E}{V_T}\right) \\$ \$& \\$ \$&= I_S \exp\left(\frac{-v_E}{V_T}\right) \exp\left(\frac{v_1}{V_T}\right) \\$ \$& \\$ \$&= \dfrac{I_{tail} \exp\left(\frac{v_1}{V_T}\right)}{\exp\left(\frac{v_1}{V_T}\right) + \exp\left(\frac{v_2}{V_T}\right)} \end{align}\$

Then multipy by “1,” an algebra trick that is handy in many situations:

\$\begin{align} i_1 &= \dfrac{I_{tail} \exp\left(\frac{v_1}{V_T}\right)}{\exp\left(\frac{v_1}{V_T}\right) + \exp\left(\frac{v_2}{V_T}\right)} \cdot \dfrac{\exp\left(\frac{-v_1}{V_T}\right)}{\exp\left(\frac{-v_1}{V_T}\right)} \\$ \$& \\$ \$&= \dfrac{I_{tail}}{1 + \exp\left(\frac{-(v_1 - v_2)}{V_T}\right)} \\$
\$\end{align}\$

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Figure 2. Generating single-ended voltages from DM and CM terms

Differential pair / long-tailed pair

Half-circuit analysis

Figure 3. Common-mode half-circuit

Figure 4. Differential-mode half-circuit