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I see that there are multiple ways to bias a simple Common Source NMOS transistor but I want to understand about biasing using current source.

I put up this circuit in SPICE:

enter image description here

The above is simple Common Source Amplifier biased with help of current source without a constant gate voltage. My first question is, "is this really biased with current source? or is it biased with -1.4V below the current source?"

I called this "biasing using current source" because changing the current value changes the operating point and Source potential is not fixed (while operation).

Well, now arises an interesting observation. I see that this circuit amplifies the voltage variations at Gate (which is what i wanted). However, the source potential varies with gate. (acts as source follower). But to amplify, shouldn't the source potential be fixed? Only then Gate-Source potential would vary and produce the output right? So, I wanted to know how current variation happens across the resistor to amplify the signal? (shouldn't the current just stay constant?)

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    \$\begingroup\$ Neither, since there is no ground connection on the output side. \$\endgroup\$ – Brian Drummond Oct 2 '17 at 13:02
  • \$\begingroup\$ When you place a current source (I1) in series with a voltage source (V1) the alarm bells in your head should start ringing. Because one of the two will be obsolete. You can also add a ground so things make sense again. \$\endgroup\$ – Bimpelrekkie Oct 2 '17 at 13:11
  • \$\begingroup\$ "redundant" rather than "obsolete"? \$\endgroup\$ – Leon Heller Oct 2 '17 at 13:26
  • \$\begingroup\$ Thanks. I believe its "redundant". So, I removed the voltage source but the behavior remains the same. \$\endgroup\$ – sundar Oct 2 '17 at 13:34
  • \$\begingroup\$ but the behavior remains the same Well obviously, since you just did "something" without understanding what you're doing. Your circuit cannot work since the right part of the circuit has no ground reference which is what Brian commented already. \$\endgroup\$ – Bimpelrekkie Oct 2 '17 at 14:28
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How are you measuring gain? You have to run AC analysis to measure gain. Also, connect the other end of the current source to ground.
By the way, this circuit cannot have any gain at low frequencies because the current source acts as an AC open. Also, this is a biasing circuit and is not intended as a gain block.

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  • \$\begingroup\$ Thanks for the Answer. So, this indeed is a biasing circuit. \$\endgroup\$ – sundar Oct 2 '17 at 20:44
  • \$\begingroup\$ My overall objective was to understand that if a differential amplifier is biased by a current source, why not a simple common source NMOS? The circuit became little wierd but biasing is accomplished by the value of the current in current source. \$\endgroup\$ – sundar Oct 2 '17 at 20:46
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    \$\begingroup\$ In differential amplifier, we are able to get gain (differential) because tail current node acts as a virtual ground. We do not get any degeneration for differential signals. As a first order estimate, the gain of the MOS circuit is given by ratio of impedance seen at the drain to that at source. Since we get open circuit for the current source load at the source, gain becomes zero. \$\endgroup\$ – sarthak Oct 2 '17 at 20:52

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