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voltage to current converter

Discussion in 'Electronic Design' started by john, Nov 16, 2007.

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  1. john

    john Guest


    I am trying to simulate a DC block circuit in PS PICE. I am using the
    following practical circuit

    I replaced the voltage to current section with the current source
    symbol given in PS pice. Please find the circuit diagram in the
    following link

    I have following questions

    1. How can I see the capacitor charging and discharging in PSPICE? I
    found the capacitor always charged whenever I run the simulations.

    2. What is the difference between voltage to current converter and
    constant current source? The INA circuit that I am using could be a
    constant current source ( plz. refer link1) and can be replaced with
    the current source given in PSPICE. I tried to use the PSpice models
    of INA and opa131 but did not work .

    3. I am trying to get rid of the DC leakage current produced by the
    INA 133 circuit using capacitors. I have discussed this issue in my
    previous post but need more advice! There problem is

    a. At the power up the DAC outputs DC -2.5 volts. The inverting input
    of the INA133 is grounded and the 470uF capacitor is reduing the -2.5
    volts to -0.5 volts accross the 384ohm resistor. There is a volage
    difference between the inputs which reflects at the output. I solved
    this problem partly by using the waveforms that ends at zero. So,
    after sending the waveform out of DAC, the DAC holds the zero and no
    more leakage. But at power up its an issue.

    b. The leakage current is constant and I came up with the circuit
    I mentioned in the link2 to get rid of it. Any comments!!

  2. None basically.
    I would like to understand what exactly you mean by 'DC leakage'.
    In a previous post I showed you need no capacitor,

    So then for a 0 to +ma output current you need -2.5 V on the other input
    of the INA.

    No output cap either, but your load impedance must be low enough
    so the requested current can flow.
    If yo uhave a supply of plus and minus 10V, and a 5 k series resistor,
    then whatever yo udo,m the output current cannot be more thne +or- 10 / 5000.
    If your load is say 1K, then the output current can never be more then
    +or- 10 / (5000 + 1000).
    So this limits the input voltage range, as I out is U1-U2 / 5000 and
    can never be greater then Iout max.

    Using the 1M makes no sense as for 1mA you need 1000V.

    So, practially the calc goes a bit like this:

    U1-U2 = 2.5V, R=5000, Iout = 500uA, for 10V supply Rload total should be
    10 / .0005 = 20000, you already have 5000 for R, so Rload must be < 15000 Ohm.

    If I did not make any errors, but thisis the system.

    What is your Rload?
  3. john

    john Guest


    The DC leakage current will go into the load without the capacitors.
    The -2.5 volts offset is only there at the power up not after the DAC
    outputs the waveform. After the DAC outputs the waveform, the voltage
    at the non inverting input is zero so leakage gets low but still its
    there. I think that DAC zero and the ground zero has some offset and
    need trimming but can not do trimming because of PCB issues.

  4. At power up the caps have charge zero, so the ywill not help,
    but perhaps even increase startup surges / currents in teh load.
    3rd time I ask now:
    What is your load impedance.

    OK I give up.
  5. john

    john Guest


    The laod is in the range of 10kohm to 200kohm.

  6. john

    john Guest


    The resistor in parallel with the current source will not let the
    capacitor charged upto +/- 15 volts. so when the power is turned off ,
    the capacitor will have less charge to dump in the load. What do you

  7. OK, now think for a minute.

    The output was (U1-U2) / R, where R was 5000.
    Your DAC does 2.5 V or more ful lsignal.
    the output current is then 2.5 / 5000 = .5 mA
    ..5 mA in 200.000 Ohm = 100 Volt.
    Your supply voltage is? 10V?
    Go back to start and pay 100$
  8. I was thinking about the other capacitor, it is a short at power up.
    But I was also thinking that none of this can work because see my other post.
  9. Guest

  10. john

    john Guest


    Would you please confirm two more things

    1. What is the output impedance of this voltage to current converter?
    Ideally current sources have infinite output impedance.

    2. If the voltage across the load is 1000 volts and power supply is
    10volts. What would likely to happen
    a. the current will not flow
    b. the current will flow

  11. Ideally, yes.
    The output impedance is set by the open loop gain.
    The current CANNOT flow, as you need 100V for it to flow.
    As you only have a much lower supply vltage, only a lower current
    will flow, resulting in the amplifier hanging one way against the supply.
  12. Ideally, yes.
    The output impedance is set by the open loop gain.
    The current CANNOT flow, as you need 100V for it to flow.
    As you only have a much lower supply vltage, only a lower current
    will flow, resulting in the amplifier hanging one way against the supply.
  13. JosephKK

    JosephKK Guest

    Jan Panteltje posted to
    Gosh, a fixed value source versus a controlled source but no
    difference? That does not make sense to me. Spice has controlled
  14. JosephKK

    JosephKK Guest

  15. look at this in context, he has a controlled source.
    Any constant current source can be controlled in some way,
    If you really wanted to be pedantic.

    F*ck spice.
    Reality rules.
  16. JosephKK

    JosephKK Guest

    Jan Panteltje posted to
    It seems unlike you to not follow the chain properly.
  17. Reality rules.
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