operational transconductance amplifier (OTA)

budhunter

Joined Feb 29, 2012
3
Hello PG -

Here is my take on your questions with an IC design perspective:

Q1: The output of the OTA is a current source, as opposed to regular two-stage op-amp voltage mode. Naturally you always want your current sources to be high impedance. The internal design of the OTA uses current mirrors, and if they do not match currents very well then your DC output voltage will have a systematic +/- offset. Also, high output impedance will provide higher gain and bandwidth for the amplifier, especially since the OTA only has one high Z node and thus usually significantly less amount of gain compare to traditional two-stage opamp that has two high Z nodes.

Q2: You are correct, removing the resistor will not change the functionality of the amplifier, i.e. the gain Vo/Vi is still equal to -gm*RL.

Q3: Buffers are used to isolate the network that is on the input of the voltage follower from the output, while maintaining the desired DC level (Vin = Vout in case of voltage follower buffer). For example, the input to the buffer may be a DC biasing circuit. The output or load may be a circuit that requires the DC bias but is switching at high frequency. Without the buffer this output circuit would likely couple a lot of AC noise on the DC bias input, causing it to be noisy as well.

Q4: This really depends on context and how Iout is really considered. Knowing the internal circuitry of the OTA, it can either source or sink current to or from the load, so I would think Vout is the amplitude and not peak-peak.

Q5: I agree this seems off. I think Iout = K*Ibias seems more appropriate, but this could be a context issue.

Hope this helps,

Bud
 

Thread Starter

PG1995

Joined Apr 15, 2011
832
Thank you very much, Bud, and welcome to the forums. I do have some follow-on queries to make and will ask them soon. Thanks.

Best wishes
PG
 
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