Question

Consider the two-sided limiter circuit below. Let the load resistor RLoad = 1 kQ. The gain must be 0.8 V/V in the region where the diodes are not limiting. Use a constant voltage drop model for the diodes with VD = 0.7 V. 3. out D2 (1+2 pts) Find the value for voltage source V1 that causes D1 to start limiting the output voltage at -5 V. a. D3 D1 Vin R2 b. (1+2 pts) Find the value for voltage source V3 that causes D3 to start V3 V1 limiting the output voltage at +6 V. (2+2 pts) Find the value for resistor R1 to set the gain to 0.8 V/V in the region where the diodes are not limiting. c. d. (2+2 pts) Find the value of R2 that causes the gain when only D2 is conducting to be 0.2 V/V. Voltage source V2 is 4.5 V.

Answer 1

Ans a) Diode D1 will conduct only when forward biasing voltage is more than 0.7 V with anode positive and cathode negative.

It is given Vout = -5 volt

Therefore V1 = - 5 + 0.7 =  - 4.3 volt

Ans b) Similarly the output voltage can be limited by D3 with Vout = + 6 volt when

V3 = +6 - 0.7 = 5.3 volt

Ans c) when no diode is limiting in that case

Vout = Vin * RL / (RL + R1)

Also RL = 1K ohm and gain: Vout / Vin = 0.8

Therefore , 0.8 = 1K / (1K + R1)

Hence R1 = 0.25K ohm

Ans d) Only diode D2 is conducting with

V2 = 4.5 volt

voltage gain = 0.2 V/V

Applying nodal analysis at output node

( Vin - Vout ) / R1 = Vout / RL + ( Vout - 0.7 - 4.5) / R2

or, Vout (1 / R1 + 1 / RL + 1 / R2) = Vin / R1 + 5.2 / R2

Therefore , gain = 1 / ( R1 * (1 / R1 + 1 / RL + 1 / R2) )

or, 0.2 = 1 / (1 + R1 / RL + R1 / R2)

or, (1 + R1 / RL + R1 / R2) = 10 / 2

or, (1 + 0.25K / 1K + 0.25K / R2) = 5

or, R2 = 0.0667K ohm