operational transconductance amplifier (OTA)

Discussion in 'Homework Help' started by PG1995, Apr 28, 2012.

  1. PG1995

    Thread Starter Well-Known Member

    Apr 15, 2011

    Could you please help me with the queries enclosed in the attachment? For high resolution copy of the attachment, please use this link: http://img210.imageshack.us/img210/7664/floydota15.jpg.

    Please note that there is a minor error in the attachment at one place, it should read: "This formula is referenced in Q5".

    Last edited: Apr 29, 2012
  2. budhunter

    New Member

    Feb 29, 2012
    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,

    PG1995 likes this.
  3. PG1995

    Thread Starter Well-Known Member

    Apr 15, 2011
    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