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How do i read this circuit?

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Syafiq

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Hi

This is a circuit of an Operational Transcondcutance Amplifier. This is part of my project, but I am not so well-versed in reading schematics. Can anyone explain this OTA schematic in a simple manner? Thanks!
 

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Hi

This is a circuit of an Operational Transcondcutance Amplifier. This is part of my project, but I am not so well-versed in reading schematics. Can anyone explain this OTA schematic in a simple manner? Thanks!
as per my little knowledge, i assume that on left side are the two current sources
one each on Vdd and Vss
At the middle it is a differential amplifier at the middle with having current sources on each drain. the common part at centre bottom is a adjustable bias using a FET.
it could perhaps be useful as a modulating input either.

finally at the right it could be called a complementary push pull , with common drains outputting the amplified signal.
 
This is a long tailed pair. Here goes:
Put a bias voltage on VBIAS. This causes a current to flow in the drain of M5. This current is split between M1/M3 and M2/M4. If the input voltage Vin+ exceeds Vin-, then all the current flows down the M2/M4 combo. M4 and M6 form a current mirror. If they have the same turn on voltage the current i2 should be equal to current i6. Current i2 creates a certain gate-source voltage (=VDD - Y) and if M4 and M6 are matched, since this same gate-source voltage is applied to M4 and M6, equal currents flow. Likewise with M3 and M8. Therefore by changing Vin+ - Vin-, you can change the amplitude of i2, hence the amplitude of i6. Likewise with i8 and i7, so Iout is equal to i6 - i7.
 
This is a long tailed pair. Here goes:
Put a bias voltage on VBIAS. This causes a current to flow in the drain of M5. This current is split between M1/M3 and M2/M4. If the input voltage Vin+ exceeds Vin-, then all the current flows down the M2/M4 combo. M4 and M6 form a current mirror. If they have the same turn on voltage the current i2 should be equal to current i6. Current i2 creates a certain gate-source voltage (=VDD - Y) and if M4 and M6 are matched, since this same gate-source voltage is applied to M4 and M6, equal currents flow. Likewise with M3 and M8. Therefore by changing Vin+ - Vin-, you can change the amplitude of i2, hence the amplitude of i6. Likewise with i8 and i7, so Iout is equal to i6 - i7.
Thanks!
 
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