Understanding the Transistor Differential Amplifier

When studying amplifiers it’s impossible to leave out the differential amplifier.  It is a very common and well known building block in electronics, and is also the basis for one of the most popular RF mixers: the Gilbert Cell mixers.  Learning about the differential amplifier won’t just give you a useful tool in your electronic repertoire, but it also gives us a good excuse to introduce what CMRR – Common Mode Rejection Ratio- is.

Differential amplifiers can be easily made with a single op-amp.  Some op-amps are even specifically designed to be used as such and don’t require external circuitry.  In this article however, we’re going to focus on the transistor differential amplifier.

Transistor Differential Amplifier

Note here that we have two possible inputs: one for each transistor’s base.  We also have two possible outputs: one for each transistor’s collector.  We can use this amplifier configuration in multiple ways:

  • using only one input: this is called single-ended input
  • using both inputs: this is called differential input
  • using one output: this is called single-ended output
  • using both outputs: this is called differential output

When using differential amplifiers, two different gains can be calculated: the differential gain, and the common-mode gain.

The differential gain is determined with a differential input signal, while the common-mode gain is determined with the same signal applied at both inputs.

Let’s start with the common-mode gain.

Common-Mode Gain

Let’s apply an identical signal to both inputs.  The two transistors are identical, and both collector resistors are identical.  Therefore, since the same signal is applied to both transistors, the same current flows through them both.  At point A, the two identical currents flowing from both transistors flow into the common emitter resistor R_1.  A common mistake would be to assume that the gain of this circuit is simply \frac{R_C}{r_e + R_1}.  The reason this is not correct is because the common emitter resistor R_1 sees twice the current that either transistor handles.  So how do we find the true gain of this circuit?  A neat little trick outlined in The Art of Electronics simplifies this process nicely.  First, replace R_1 by an equivalent resistor network consisting of two 2\times R_1 resistors in parallel, like so:

Now, we know that the transistor currents don’t flow from one transistor to the other, but instead go through the common emitter resistor(s).  Therefore, opening the circuit at point C is of no consequence.  Indeed, both legs of the differential amplifier handle the same current, and the same current will flow through each 2\times R_1 resistor:

We can now easily find the gain.  Each transistor is in a classic common emitter configuration, so gain is -\frac{R_C}{r_e + 2R_1}.  Our common-mode gain is thus:

    \[ G_{CM} = -\frac{R_C}{r_e + 2R_1} \]

The output is identical regardless of which output point A or B is used.  Taking the differential output (v_{AB}) will yield zero.  In practical circuits the differential output will always show some minute voltages due the transistors not being perfectly matched, and due to the resistors’ tolerances.

Differential Gain

To find the differential gain, instead of applying the same signal to both inputs, we apply a differential input, as follows:

Here, v_{in} = v_+ -v_-.

A little math is required to find this circuit’s gain.  Using the transistor’s equivalent circuit (see here for more details), we know that each collector’s current is:

    \[ i_C = g_m v_{BE} \]

Our output voltage is thus:

    \[ v_C = R_C \times i_C = R_C g_m v_{BE} \]

With g_m = \frac{1}{r_e} = \frac{I_C}{26}, with I_C in mA.  Now we need to express v_{BE} as a function of our input signal v_{in}  For each transistor however, v_{BE} is different:

  • Left transistor: v_{BE} = v_B - v_E = v_+ - v_E
  • Right transistor: v_{BE} = v_B - v_E = v_- - v_E

To find the gain here, we need to realize one important characteristic of this circuit.  The input is differential.  That means that for every 1mV of signal increase at transistor T_1‘s base, there is an equal and opposite 1mV of signal decrease at transistor T_2‘s base.  Since the transistors and collector resistors are identical, this means that when collector voltage drops by a certain amount at T_1, then the collector voltage at T_2 increases by the same amount.  The same can be said for their emitter voltage.  Since both emitters are connected at point A, point A is a fixed voltage that doesn’t change with the input signal.  V_E is a constant.

The last thing we need to understand is the relationship between v_+, v_-, and v_{in}.  For every mV of increase in v_+, v_- decreases by the same amount.  This causes v_{in} to change by twice that amount.  Another way to see it is that for every increase in 1mV of v_{in}, v_+ increases by 0.5mV and v_- decreases by 0.5mV.

    \[ \Delta v_C = -R_C g_m \Delta(v_B - v_E) \]

    \[ \Delta v_C = -R_C g_m \Delta(v_B) \]

Here, v_B is either v_+ or v_- depending on which transistor we choose:

  • for T_1: \Delta v_C = -R_C g_m \Delta(v_+)
  • for T_2 : \Delta v_C = -R_C g_m \Delta(v_-)

With the reasoning of the previous paragraph we can now link v_+ and v_- with v_{in}:

  • for T_1: \Delta v_C = -R_C g_m \Delta(\frac{v_{in}}{2})
  • for T_2 : \Delta v_C = +R_C g_m \Delta(\frac{v_{in}}{2})

And thus, we have:

    \[ G_{diff} = \frac{R_C g_m}{2} = \frac{R_C}{2 r_e} \]

If used as a single-ended output amplifier, this configuration, depending on which collector we sample our output voltage, will either have positive or negative gain.

However, if used as a double-ended output amplifier, as shown below, the gain changes.

In this case:

    \[ \Delta v_{out} = \Delta v_{AB} = \Delta(v_A - v_B) = -R_C g_m \Delta(\frac{v_{in}}{2}) - R_C g_m \Delta(\frac{v_{in}}{2}) \]

    \[ \Delta v_{out} = - R_C g_m \Delta(v_{in}) \]

Thus, the gain becomes!

    \[ G_{diff}_{de} = - R_C g_m = -\frac{R_C}{r_e} \]

Which is twice the gain of the single-ended output configuration.

To increase thermal stability and stabilize gain, identical emitter resistors can be used before point A.

CMRR – Common Mode Rejection Ratio

We’ve covered what Common Mode signals were: this is when both inputs are identical.  In the case of the differential amplifier, when using a common mode input, the differential output is nil.  This means that this circuit can get rid of parasitic offsets that are common to both inputs.  An differential amplifier can thus clean up a noisy signal, up to a point.

An amplifier’s capability of eliminating common mode signals is measured by its CMRR: its common mode rejection ratio.  Mathematically speaking, CMRR is equal to the ratio of differential gain over common mode gain, often expressed in dB

    \[ CMRR = \frac{G_{diff}}{G_{CM}} \]

An ideal differential amplifier would have an infinite CMRR.  For the differential amplifier studied in this article, we can calculate the theoretical CMRR:

    \[ CMRR = \frac{R_C}{2 r_e} \times \frac{r_e + 2 R_1}{R_C} \]

Since r_e << R_1, we have:

    \[ CMRR = \frac{R_C}{2 r_e} \times \frac{2 R_1}{R_C} = \frac{R_1}{r_e} \]


I hope this article made differential amplifiers clearer for you.  We’ll be seeing them again, most notably when we’ll be studying Gilbert Cell mixers.  While we only focused on the BJT differential amplifier here, a differential amplifier can be built with FETs and Op-Amps as well.

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