Help with Opamp/PNP Transistor Current Source

Discussion in 'The Projects Forum' started by sohcahtoa, Aug 28, 2008.

  1. sohcahtoa

    Thread Starter Member

    Nov 7, 2006
    15
    0
    I have been working with a voltage controlled current source circuit mainly made out of a 741 opamp and a 2N3906 PNP transistor. The current source is needed to create control current for an operational transconductance amplifier that in turn will be part of a voltage controlled oscillator. The output of the current source should sweep between 0 to 0.5mA in a linear fashion.

    I got the current source idea from a lecture on OTAs found here: http://users.ece.gatech.edu/~lanterma/ems/

    Here's a schematic of the current source, with the voltage values I've recorded. VCC should be set at 8.15V, forgot about that, sorry:
    [​IMG]

    The operation, in theory, should work as follows:

    The opamp V+ terminal connects to ground. This ensures that the opamp V- terminal (in the ideal case) is also set to ground. The current flowing between the VCC input (at 8.15V) and the V- terminal should be:

    8.15V / 100K = 81.5uA

    Since V- is virtual ground (with infinite impedance into the opamp), the value at the end of the emitter should be -VCC as the current flowing into and out of the V- node should be equal, or 81.5uA.

    When calculating the emitter current, one then looks at the 100K and 22K resistors like they are in parallel because of the virtual ground at V- and actual ground (I'm not so certain about this). Therefore, the emitter current should be:

    -Vi / (100K || 22K) = (-Vi)(0.055) mA

    Which leaves you at approximately a linear sweep from 0 to 0.45mA with 8.15V being your top value. I'm using 9V batteries here if you haven't guessed already. The source output is the collector.

    The problem I'm having is that I'm getting -5mA out of the collector not 0.5mA. I've checked this a number of times. -5.37V / 1K is certainly -5.37mA.

    Things I've done:

    1. Used a 10K pot to offset null the opamp. Is this ever really possible? I've gotten it to where blowing on the pot will change which rail the opamp hits, but I never get 0V.

    2. Checked the PNP terminals a number of times to ensure the emitter and collector are the proper terminals.

    3. Rechecked the pins on the opamp to ensure the proper connections.


    I guess the real question is, does this circuit make sense? Should I be assuming a real opamp can create that virtual ground? Currently it won't budge from around 1.3V or so.

    Thank you very much for any help you can provide. Sorry for the long post.

    EDIT:

    I forgot to add that the opamp Vout is -6.27V.
     
    Last edited: Aug 28, 2008
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I think you would come out better if you used the configuration illustrated here.

    A glaring error in your circuit is that the 1k resistor needs to be referenced to a load referenced to the positive power supply.

    The 741 opamp is not an OTA.

    hgmjr
     
  3. sohcahtoa

    Thread Starter Member

    Nov 7, 2006
    15
    0
    Thanks hgmjr. I have seen that configuration as well. I'll likely give it a try after playing around with this one for awhile.

    I left the OTA out of the schematic. The control current leaving the collector of the PNP is supposed to drive an OTA input that needs a control current. The 741 is working together with the PNP to make a stable, or supposedly stable, current source. I was just grounding the resistor coming out of the collector to test the circuit.

    Perhaps it's just a bad design overall?
     
    Last edited: Aug 28, 2008
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I have some experience with the topology that I sent you a link to and it works fairly well. What works even better is using an n-channel mosfet instead of a BJT.

    hgmjr
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Here's another way you might do it. See the attached.
     
  6. sohcahtoa

    Thread Starter Member

    Nov 7, 2006
    15
    0
    Thanks to both of you for your suggestions.

    How would one go about connecting one of these to the control current terminal of an OTA? The OTA terminal really can't act as a load that connects to VCC and then returns the current to the transistor, as far as I know. In other words, the current that the OTA sinks isn't returning. I'm thinking that's why the OpAmp/PNP combination was used?

    For reference, I 'm thinking of using an LM13700.

    http://www.national.com/mpf/LM/LM13700.html

    Thanks again.
     
    Last edited: Aug 28, 2008
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    It's late, and I'm out of it.

    What's an OTA? In the military, an OTH is an Other Than Honorable discharge.

    I'm not really up to reading thru everything on that page and listening to all of the lectures at the moment, however I will in the future.

    But if you want to make some progress, explain your acronyms - better yet, what's your target circuit?
     
  8. sohcahtoa

    Thread Starter Member

    Nov 7, 2006
    15
    0
    OTA stands for operational transconductance amplifier. It's a differential voltage controlled current source, but has an additional terminal for a control current that controls the transconductance of the circuit. Basically it can act like a current controlled potentiometer. I don't think they're all that common anymore. I'm using it as a building block for a voltage controlled oscillator.

    The control chain essentially looks like:

    (Volt-to-Amp Current Source)-->(OTA)-->(Voltage Controlled Oscillator)

    The voltage controlled oscillator is really current controlled by the OTA.

    I guess the main issue is that there aren't two terminals that let you plug in the OTA to act like a load (Rl).
     
  9. BobaMosfet

    Active Member

    Jul 1, 2009
    109
    11
    Hi!

    Your equation result is wrong:

    8.15V / 100K = 81.5uA

    The answer is 81.5mA (0.0815), or 8.15uA.

    Perhaps this will help.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    This thread is a year old, but 8.15V / 100K still equals 81.5uA.

    And WTH does this mean?
    81.5ma≠8.15uA
     
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