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Subject: Re: Simple OTA gain

Date: 08/28/03 at 7:03 PM
Posted by: Russ Kincaid
E-mail: russlk-nospam[AT55]yahoo.com (replace [AT55] with @ to reply)
Message Posted:

In Reply to: Simple OTA gain posted by Jason on 08/28/03 at 6:56 AM:

When the gate is connected to the drain, the gate-source voltage (Vgs) will adjust itself to the value required by the drain current. If the drain is not connected (floating), the Vgs will reduce to the pinchoff value or less.

The only source of current in the circuit is Ib at the source of M1 and M2. If the inputs to M1 and M2 are equal, Ib divides between them. The drain current of M5 must also be Ib/2 because it is a current mirror. And, the drain current of M8 must be Ib/2 because it is a mirror of M7.

Now, if the input to M1 is made more negative, the drain current of M1 will want to increase. That can only happen if the drain current of M2 decreases. M1 acts as a source follower thus reducing the Vgs of M2.

It is easy to see that an increase in M1 current goes thru M3, M5, M7 and M8 to the output, as you have shown. Simularly, the decrease in M2 current goes thru M4 and M6 so, the change in output current is 2X the current change of M1 or M2 (or is the sum of the current changes).


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