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We have current to voltage converter opamp which has below schematic. Below is the formula to find TP1 Voltage :

Vout (TP1) = I1 x R2 or R1 (depend on the switch2 and not calculating the C1)

the question that I have is when the switch1 is open and nothing is connected to opamp V+ should Tp1 Voltage be zero doesn't matter what is the gain resistor(SW2), and if I am correct why I am not getting this in a real scenario.

I think it is because of the noise or rail power but not sure.I am using TP1 (after some other amplifier) for my 4-20mA output calculation.

schematic

simulate this circuit – Schematic created using CircuitLab

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When Vin+ = 0V then Vin- must also = 0V

Vout = I*R always if not saturated.

When SW1=closed I source drives the output = Vout/Rf so that Vin-input stays at 0V.

For this to work well with high R values, Op Amp Input bias current Iib = ? must be low. FET types are very low.

Tp1 voltage = Iib * R e.g.

for Iib=10nA, R= 100k

  • 10nA * 100k= 1mV output.
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  • \$\begingroup\$ So why when I build this circuit and test Tp1 I am not getting 0V \$\endgroup\$
    – Shahreza
    Commented Jan 8, 2019 at 15:34
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    \$\begingroup\$ does input bias current * R= Vout? \$\endgroup\$ Commented Jan 8, 2019 at 16:55
  • \$\begingroup\$ Yes, could because of feedback capacitor \$\endgroup\$
    – Shahreza
    Commented Jan 8, 2019 at 17:41
  • \$\begingroup\$ Cap leakage R must be >> R1or R2 but this is controlled by \$I_{IB}\$ not cap \$\endgroup\$ Commented Jan 8, 2019 at 17:42
  • \$\begingroup\$ So how do u cancel the input bias current so you will get the same result no matter what gain you have. \$\endgroup\$
    – Shahreza
    Commented Jan 8, 2019 at 18:10

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