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- 25th December 2004, 21:22 #1

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## active load in diffrential amplifier

how to define a Differential amplifier

how is Differential amplifier with active load different fom a normal amplifier

- 25th December 2004, 21:22

- 26th December 2004, 04:53 #2

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## differential amplifier active load gain

A differential amplifier is an amplifier that amplifies a differential input, ie. the difference between its input terminals

i.e. Vout = A(Vin1-Vin2)

A differential amplifier with an active load does not have resistors inside it to convert the current gain to voltage output, instead it uses active devices as current sources which effectively have a very high output resistance therefore give a higher voltage gain.

- 26th December 2004, 04:53

- 30th December 2004, 18:34 #3

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## differential amplifier with activeload

Every OP Amp is a difference amplifier because its output is proportional to the voltage difference between the two inputs

- 30th December 2004, 18:34

- 1st January 2005, 18:15 #4

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## active load differential amplifier

See chapter 4 in Razavi's book

The differential amp is a amp that amplify the diff mode signal and try to reject the common mode signal.

The diff amp with active current load is just a way to increase the gain of the diff amp.

- 4th January 2005, 03:02 #5
## active load

Active loads increase the gain of an amplifier, since they present a high impedance in AC. But they can draw fairly large currents in DC. A simple resistor would need to have a large value for high AC gain, but this large value would make it hard or even impossible to obtain the required DC current, because the required voltages can become very high. Let's assume you need a current of 1mA in each transistor and you need a load resistance of 1MΩ. That means you need 1000V across that resistor!

Most opamps have a single-ended output, but a differential input. So, at some point a conversion is needed from differential to single-ended.

With a single resistor, like in the first figure, the gain of the differential stage is halved, since the left transistor does not contribute anything to the output voltage, only variations in the current of Q2 are seen be the load. So Vout=A*(V1-V2)/2. The load is R1 in parallel with the input impedance of the next stage. The gain depends on the load impedance so the presence of R1 only reduces the gain.

By using a current mirror, Q5 and Q6, you get an active load for Q4 and at the same time restore the gain of the differential stage to Vout=A*(V1-V2). This happens because a variation in the current of Q3 is no longer lost, but gets mirrored via Q5 into Q6, so the excess current flows into/ out of the load, which is actually the input of the next stage. Now Q3 constributes the same amount as Q4 (the changes in the currents of Q3, Q4 are equal and opposite). So Vout=A*(V1-V2). The gain will be dictated only by the input impedance of the next stage, since there is no resistor in parallel with it.

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