Operational Amplifier

The output will no longer be a current source, but a ground referenced voltage source with R in series with its output. The feedback opamp forces the output voltage (the difference voltage of the two inputs , V2 and V3 times the INA gain of 1), to all appear across R, so the current through R is dependent only on the input signal, not the resistance of the load, below R. This is because the output voltage is forced to be referenced to whatever voltage is applied to pin 1. So if the load drops more or less voltage that voltage is sent back to the reference pin by the opamp, so the output from pin 6 rises or falls by the same amount. This holds the voltage drop across R unchanged, thus R passes a current that is independent of the load resistance. All this assumes that the supply voltage allows such an output.

Hi,

Thanks for ur reply! Would you please confirm two more things that

1. if INA133 is outputing a DC voltage of minus2.5 volts. Then what would be the voltage across the resistot "R" and the load ( lets say

50kohm).

1. I want to deliver a current of 50uA to the load. for that purpose I put the resistor 50kOhm. My > john wrote:

The voltage across the pair would b -2.5. How that is divided between those two impedances depends on their ratio.

Not if you are interested in the peak values. If you want 50uA RMS through your load, then it would make most sense to calculate R based on the RMS input voltage, or convert 50 uA RMS to a peak value so you can go back to working with peak voltages and currents. If sine waves are involved, this simply involves dividing or multiplying by a factor of the square root of 2 (1.414).

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.

Yes.

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.

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, 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.

If your load has a fixed resistance, so you could feed it with a voltage source as well as with a current source, this would eliminate the need for the current sensing resistor. Then you can add the capacitor in series with the load without messing up the INA operation. In this configuration, you eliminate the feedback opamp and tie the INA reference input to ground. Of course, in this mode, you really have no use for an INA, since any opamp could produce the drive voltage.

Hi,

No my load is not fixed.. Please advice!

John snipped-for-privacy@rica.net wrote:

Then putting an inductor in parallel with your load would detour most of the DC from the INA current source around it, without involving opamp offsets, but finding real inductors with high enough inductance to block most of the AC current and low enough resistance may be impractical. How much offset voltage can you stand across your load? If you say zero, active devicesay be ruled out.

hi,

yes, its zero. Please advice!

Thanks John snipped-for-privacy@rica.net wrote:

A perfect zero? Too bad. Not a nano ampere, not a pico ampere, not a fempto ampere, not an atto ampere, but exactly zero offset current.

I think you should give up.