Question 8.3: One-Sided Limiter Find the output voltage of the circuit sho...

The output voltage can be calculated by applying KVL. If the voltage across the diode is known, the output voltage will correspond to:

v_o(t)=V+E     (8.7)

On the other hand, if the current flowing through the diode is known, the output voltage will correspond to:

v_o(t)=v_{i}(t)-I \times R     (8.8)

The knowledge of v or I depends on whether the diode is ON or OFF, respectively. If the diode is ON, the voltage across the diode v = 0.7, and based on Equation (8.7):

v_o(t)=0.7+E     (8.9)

If the diode is OFF, the current flowing through the diode I = 0, and based on Equation (8.8):

v_o(t)=v_{i}(t)     (8.10)

If the voltage difference between v_{i}(t) and E exceeds 0.7 V, the diode will turn on. In other words, to turn the diode on, the following must be true:

v_{i}(t)>E+0.7     (8.11)

Accordingly, the output voltage corresponds to Equation (8.9). On the contrary, the diode will be turned off if:

v_{i}(t)<E+0.7     (8.12)

In this case, the output voltage will correspond to Equation (8.10).

As a summary, the output voltage is given by:

v_{o}(t)=\left\{\begin{array}{cc}E+0.7 & v_{i}(t)>E+0.7 \\v_{i} & v_{i}(t)<E+0.7\end{array}\right.     (8.13)

Plots of the transfer characteristic, input voltage, and output voltage for the limiter circuit of Figure 8.17 are shown in Figure 8.18. As shown in the figure, it is evident that the output voltage doesn’t exceed E + 0.7. The upper limit of the output voltage can be controlled by adjusting the value of E.