My objective is to understand the need of the tail current source in a differential amplifier. So, to understand the significance of current source, I tried simulating the behavior with and without the current source.

Without the current source, I was able to simulate and get the amplified waveforms. But with the current source I am not able to simulate and getting the error.

The circuit is below: enter image description here

It gives the following error: enter image description here

The Node n003 is exactly the drain of the right NMOS Transistor.

Its mentioned in a Circuit Design book that the addition of tail current source is to reduce the dependence of the bias current on the input common mode level.

Since I did not get exactly the above, I tried to simulate the circuit with the current source. But for some reason it does not work and unable to understand the significance of the current source in a differential pair.

Here is what i tried as an alternate approach:

I tried simulating a single transistor with a resistor at top and a Current source at bottom like below: enter image description here

So, this acts like a source follower. The voltage at the source changes with respect to the gate. I see that the current is kept constant by the MOSFET action through the variation of the source voltage with respect to Gate.

But for amplification we need Source to be fixed so that Gate potential variation is turned to current variation by MOSFET transconductance. So, to fix the Source potential (to Zero), we can have other MOSFET to the right side operating exactly opposite. So, we have two source followers operating exactly opposite and I replicated this circuit with other transistor like this: enter image description here

I got the below output: enter image description here

The top differential waveform is applied at Gate, whereas the bottom differential waveform is got at the Source node, just above the current source. So, source potential is maintained at zero. But the output stays at constant voltage (in my simulation).

EDIT - After grounding the voltages, simulation atleast works but the amplification is somewhat lost at the output. The output voltage looks little wierd:

enter image description here

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    \$\begingroup\$ In your diff amplifier the voltage sources should all return to ground, no the source pin of the MOSFETS. \$\endgroup\$ – Kevin White Oct 1 '17 at 23:14
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    \$\begingroup\$ Current has to flow in a loop, trace out where you think the 63u should flow \$\endgroup\$ – sstobbe Oct 1 '17 at 23:30
  • \$\begingroup\$ I did what you just said. Thanks for the tip. The simulation is successful but amplification is somehow lost at the output. (not matching the one without the current source). I will edit the post and add a screenshot of that. \$\endgroup\$ – sundar Oct 1 '17 at 23:31

A differential amplifier amplifies the difference between two inputs. The purpose of the 'tail' current is to split that difference between the two sides of the amp. To do this the input voltage must be referenced to ground, so that any increase in current on side causes a corresponding decrease on the other, and vice versa.

enter image description here

If V1 increases then M1 conducts more and passes more current. Since I1 is fixed that means less current must flow through M2. But how does current through M2 reduce? Voltage at the Source terminals of both M1 and M2 increases, reducing M2's Gate- Source voltage and making it conduct less. The result is equal but opposite output signals on each side.

I have set V1 and V2 to produce sine waves with opposite polarity (180° out of phase). Try it with V2 amplitude = 0V and you will see that output appears on both sides even though the input signal is only on one side.

  • \$\begingroup\$ Thanks. It works now. But to the curiosity, in a differential operation, the source voltage of the transistors should be virtually grounded or at zero potential. This does not seem to happen in this circuit. To be specific, if we apply out-of-phase sine waves at the gate of the transistors, the source potential should be maintained at zero which is not happening. Any idea why? \$\endgroup\$ – sundar Oct 2 '17 at 10:29
  • \$\begingroup\$ I am afraid you are confusing source POTENTIAL (DC voltag) with the AC VOLTAGE at the sources. Of course, the DC potential is NOT zero, but the AC voltage is because both inputs are driven out of phase (ideal differential mode). \$\endgroup\$ – LvW Oct 2 '17 at 11:43
  • \$\begingroup\$ "out of phase" means: Opposite phase (180deg shift). \$\endgroup\$ – LvW Oct 2 '17 at 14:05
  • \$\begingroup\$ The Sources have a DC bias voltage, which for an enhancement mode NMOSFET is negative relative to the Gate. Some signal voltage will be impressed on this due to non-linearity in the FETs. \$\endgroup\$ – Bruce Abbott Oct 2 '17 at 17:32

Question: My objective is to understand the need of the tail current source in a differential amplifier.

Answer: Consider, what the diff. amplifier is in principle: A two-stage amplifier consisting of a common-gate stage (NMOS 2) driven by a common-drain stage (NMOS1). This description is referenced to an input at the gate of NMOS1. However, the role of both transistors is interchanged if we consider the gate of NMOS 2 as an input. That is the working principle of the whole circuit.

Note that - by its nature - the gate-source path of NMOS2 does already act as a differential voltage amplifier - however, one input resistance (gate) is very large and the other one (source) is rather low. Therefore, we use a unity-gain follower NMOS1 (with an ohmic resistor Ro) for driving the source of NMOS2.

Hence, the most basic diff. amplifier has an ohmic resistor Ro in the common source path - and the common mode gain is small (as desired) if the Ro value is large. This results from the negativ feedback effect caused by Ro (for common mode signals). That is the reason we can replace the ohmic part Ro with a third transistor acting as a very large DYNAMIC resistance (a current source). Hence, we can have a very good common-mode suppression.


The purpose of tail current is to provide high common mode rejection ratio. The intent of the differential amplifier is not only to amplify the differential signals and but also reject (provide less gain, ideally zero) for the common mode signal.
You could have taken two common source amplifiers and take the difference of their output to have high differential gain. But the circuit has high gain for common mode signals as well.
By adding the tail current source, we degenerate the source with high impedance for the common mode signals and reduce the gain for them.


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