Analyzing input/output characteristics of a diode circuit

circuit analysisdiodes

In "Fundamentals of Microelectronics" textbook. The question number 27 of chapter 3 is:

The part b of the question is:

And the solution of b from the solution manual of this textbook is:

Can anyone explain simply how did he got the solution ?

Best Answer

Assume the input voltage starts at 0 and then slowly increases. For all positive input voltages, D2 will be off and can be ignored. Until D1 turns on, the circuit is a simple voltage divider so the output is R2/(R1+R2) times the input voltage. When the voltage across D1 rises to its turn on voltage, it will turn on and short R2. Thus from that point on the output voltage equals the input voltage (a slope of 1) with an offset equal to the D1 turn on voltage. At what input voltage does D1 reach its turn on voltage? When R1/(R1+R2) times the input voltage reaches the D1 turn on voltage. That is given by (R1+R2)/R2 times the D1 turn on voltage (R1 and R2 act as a voltage divider but now the voltage of interest is across R1). This explains the positive half of the graph. The negative half is done in a similar fashion. In this case, D1 will never turn on and can be ignored. For small negative voltages, the slope will be the same as for small positive voltages since both diodes are off (R1/(R1+R2). When the negative voltage rises enough to turn on D2, the output will remain at the D2 turn on voltage since D2 is now acting as a battery across the output terminals. At what input voltage does D2 turn on? When the output voltage reaches its turn on voltage, i.e. when R2/(R1+R2) times the input voltage reaches the D2 turn on voltage. That voltage is given by -(R1+R2)/R2 times the D2 turn on voltage.