Re: Operational Amplifier

Hello,
Thanks for ur reply! I am trying to remove the offset in my sinusodial
or some other signal passing through the current source. I put a
capacitor between the load and the non- inverting input of the OPA131.(
or capacitor in series with the load) What it does that it attenuates
the first few cycles of the sinusodial waveform. The problem is that
when there is no input to the INA133, It outputs DC current, now the
capacitor stops the DC current to pass through the Load, so the current
source gets independent of the load and became open circuit and the
OPA131 gets floating and ouputs the rail voltage like plus or minus 15
volts. The capacitor gets charged and it takes sometime to discharge
and during that discharging it attenuates the signal. ( if any signal
is applied to the INA133)...
If a put a capacitor at the input if the INA133 then probably it will
solve the offset problem coming form the DAC but my application needs
zero current when there is no input present at the INA133. And as u
mentioned that there is an offset of INA133 (300uvolts) prsent which
will allow 6nA current (50Kohm) to the load and my application needs
no flow of current when there is no input to INA133. Would you please
advice me some other current source or some modifications to the
exsisting current source that can solve this problem.
Please advice!
John.
Mal-2 wrote:
> john wrote:
>
> > John Popelish wrote:
>
> (snip)
> > > With a +- 2.5 volt peak input, the INA will deliver +-2.5 volts plus
> > > whatever voltage is provided to the reference input (where the
> > > feedback opamp is shown connected. This forces the +- 2.5 volts to
> > > drop across R (50k, in this case, so +- 50 uA will pass through R at
> > > the peaks of the waveform. Of course, the INA will actually put out
> > > +-6.25 volts to the series combination of the 50k R value and the 75k
> > > load resistance. This is because, with +- 50 uA passing through the
> > > 75k load, it will drop +-3.75 volts, that will be copied by the opamp
> > > and sent back to the INA, to be added to the input voltage to produce
> > > the actual output voltage.
> > Hello,
> > Thank you very much for ur reply.. So, when it means that when there is
> > -2.5 volts across both resistors, there must be a current flowing
> > through them.
>
> Yes.
>
> > What would be the direction of current?
>
> If the voltage is positive with respect ot ground, then there is a
> positive (conventional) current through the resistors to ground. If
> you want to think of current as electrons moving, then that positive
> voltage would be pulling electrons from ground, up through both
> resistors.
>
> > The problem is
> > that I do not want any current passing through my load when there is no
> > sinusoidal wavform present at the input.
>
> As long as there is zero volts between the two INA inputs during this
> case, there will be very close to zero volts across the load. The INA
> isn't perfect in this respect, but it is pretty good. Look at the DC
> offset specifications).
>
> > so, the current will be
> > flowing to the load or leaving the load.
>
> So, what voltage across the INA inputs do you expect when there is no
> sine wave? If you want to high pass the signal to eliminate any DC
> component, that is the place to do it, not at the output of the INA.

.

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