Applications of Operational Amplifiers: Current to voltage converter

Current to voltage converter:

  The circuit shown in fig. 1, is a current to voltage converter.

Fig. 1

Due to virtual ground the current through R is zero and the input current flows through R_f. Therefore,

v_out =-R_f * i_in

The lower limit on current measure with this circuit is set by the bias current of the inverting input.

Example –1:

For the current to current converter shown in fig. 2, prove that

Fig. 2

Solution:

The current through R₁ can be obtained from the current divider circuit.

Since, the input impedance of OPAMP is very large, the input current of OPAMP is negligible.

Thus, 

Example - 2

(a). Verify that the circuit shown in fig. 3 has input impedance.
                     
(b). If Z is a capacitor, show that the system behaves as an inductor.
(c). Find the value of C in order to obtain a 1H inductance if R₁ = R₂= 1K.

Fig. 3

Solution:

Let the output of OPAMP (1) be v and the output of OPAMP (2) be v_o. Since the differential input voltage of the OPAMP is negligible, therefore, the voltage at the inverting terminal of OPAMP (1)will be v_i.

For the OPAMP (2),

or, 

(b). Let the input voltage be sinusoidal of frequency (ω / 2π)

If Z is a capacitor, then Z = 1 / ωC

Let L = C R₁ R₂
Therefore,and the system behaves as an inductor.

(c). Given R₁ = R₂ = 1 K, L = 1 H

Example - 3

Show that the circuit of fig. 4 is a current divider with i_o = i_i / ( 1 + R₂ + R₁) regardless of the load.

Fig. 4

Solution:

since non-inverting and inverting terminals are virtually grounded.

Therefore, V₁ = V_L.

Now applying KCL to node (1)

Now current is 

Example - 4

(a). For the circuit shown in fig. 5 prove that

.

(b). Verify that if R₃ / R₄ = R₁ / R₂, the circuit is an instrumentation amplifier with gain A = 1 + R₂ / R₁.

Solution: Here

Fig. 5

(b).

So in this condition circuits as an instrument amplifier with gain .

Example - 3

Obtain an expression of the type i_O = V_i / R - V_O / R_O for the circuit shown in fig. 6. Hence verify that if R₄ / R₃ = R₂ / R₁ the circuit is a V-I converter with R_O =∞ and R = R₁ R₅ / R₂ .

Fig. 6

Solution:

Here

i_O = current through the resistor.

where 

So when 

then,