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Draw the circuit diagram and explain the operation of the V -I converter, State its application areas.
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Solution:

In a voltage-to-current converter, the output load current is proportional to the input voltage According to the connection of load there are two types of V-to-I converters.

In floating type v to I converter $R_L$ is not connected to the ground whereas, in grounded type, one end of RL is connected to the ground.

voltage to current converter with floating Load. Fig Shows a voltage-to-current converter in which load resistor $R_L$ is floating.

As the input current of the opamp is zero,

\begin{aligned} & I_L \approx I_i=\frac{V_i}{R_1} \\ \therefore \quad I_L \propto V_i \end{aligned}

Thus the load current is always proportional to the input voltage and the circuit works as voltage to current converter.

If the load is a capacitor, it will charge or discharge at a constant rate. Hence converter circuits are used to generate the sawtooth or triangular waveform its proportionality constant is generally $1 / R_{\text {, }}$ hence this circuit is also called a transconductance amplifier.

It is abs called a voltage-controlled arrest source.

The expression $I_L=V_i / R_1$ holds regardless of the type of load. It can be linear (eg; a resistive) or nonlinear C e.g, a light-emitting diode) or it can have time-dependent characteristics [eg, a capacitor].

No matter what the load is the op-amp will draw the current Ii whose magnitude depends only on $V_i$ and R.

voltage to current converter with Grounded load:

when one end of the load is grounded it is no longer possible to place the load within the feedback loop of the op-amp.

Fig (b) Shows a voltage-to-current converter in which one end of load resistor $R_L$ is grounded. It is also known as 'The Howland current converter from the name of its inventor.

The analyze is of the circuit is accomplished by first determining the voltage $v_1$ at the noninverting input terminal and then establishing the relationship between $v_1$ and the load current.

Applying $k C L$ at node $V_1$ we get,

$I_1+I_2=I_L$

\begin{aligned} &\frac{V_i-V_1}{R}+\frac{V_0-V_1}{R}=I_L \\ &V_i+V_0-2 V_1=I_L R \\ &\frac{V_i+V_0-I_1 R}{2}=V_1 \end{aligned}\\

The gain of the op-amp in noninverting mode is given as $A=1+\frac{R f}{R_1}$. For this circuit, it is $1+\frac{R}{R}=2$. Hence output voltage can be written as.

\begin{aligned} V_0 &=2 V_1=V_i+V_0-I_L R \\ 0 &=V_i-I_L R \\ V_i &=I_L R \\ I_L &=\frac{V_i}{R} \end{aligned}\\

From the above equation, we can say that the load current depends on the input voltage $V_i$ and resistor R.