But a capacitive load doesn't require the same amount of driving, i.e. we desing op-amps to drive the pads on our chips, they can drive an off chip capacitance - they would be very slow though as they can't supply that much current. A resistive load of less than 50K would kill them - we have to use off-chip buffers.
But a capacitive load doesn't require the same amount of driving, i.e. we desing op-amps to drive the pads on our chips, they can drive an off chip capacitance - they would be very slow though as they can't supply that much current. A resistive load of less than 50K would kill them - we have to use off-chip buffers.
Guys think what will happen to your gain if you have a resistor directily at the output. DO an open loop analysis.
You will loose complete control on the loop.
A amplifier is a OTA followed by a buffer. And OTA takes care of the open looop gain in a buffer.
But a capacitive load doesn't require the same amount of driving, i.e. we desing op-amps to drive the pads on our chips, they can drive an off chip capacitance - they would be very slow though as they can't supply that much current. A resistive load of less than 50K would kill them - we have to use off-chip buffers.
Generally when you're designing on-chip circuitry area is a major concern, an OTA will drive gate capacitances etc. very readily thus there is no need for more complex amplifiers.
The main problem is the open loop gain. The open loop gain is some thing like this Av=gm * Ro, where gm is transconductance of the amplifier and ro their output resistance. If you want a high gain Ro must be high, so you can not load the output with a resistor since this would reduce Ro and consequently the gain. That is why OTA only drive capacitors.
To drive resistors you will need a driver with low output resistance.
As to the resistor load, what bastos4321 said is totally right. and for the capacitive load, I think its function is to compensate the frequency response to stablize the opamp.
one missing thing to bastos's explanation is that OTA has (small) limited current drive capability so it is also harder to supply current for low resistance loads.
when large currents are needed than buffers get into picture.
OTA drive a small resistor that is said by bastos4321 . bastos4321 should want to test the output current. If current of output is not large enough, OTA can't drive a small resistor.
The output of the OTA is a current source at to with current Io and current sink at bottom with the same current Io, if we connect a resistor at the output than it will disturb the DC conditions as Resistor draws the currrent to establish the voltage(ie the sink current now will not be Io). If we keep a cap load than this cap does not draw any DC current in steady satate and the DC conditions are not disturbed
to drive a capacitor, if we have a constant current source. that is good. ota acts constant current source., i.e, it's o/p i s current instead of voltage.
The way you compensate an OTA to make it stable is by placing a cap on the output to ground. Your pole is set as gmC, hence why an OTA driving a cap can be used a gmC filter. The larger the cap on the output, the more stable the amp becomes. the gain is given by gm*x*Rout where x would be your scaling factor of your current mirrors. If you drive a resistor with this amp, it is going to kill your gain.