In "Fundamentals of Microelectronics" textbook. The question number 25 of chapter 3 is:

This is part c of the question:

And this is the solution of part c from the solution manual:

  1. For the blue circle: Is it correct to add the two voltage sources before voltage division as it is done here ? Isn't there any restriction ?

  2. For the red circle: Shouldn't it be: Vin > VB + VD,on ? where did he get (R1+R2)/R1 from ?

  3. How did he draw this graph ? How to calculate the slope here ??

Please answer even for some of those three questions.

  • \$\begingroup\$ The voltage sources are subtracted, not added. It seems correct. \$\endgroup\$ Oct 25, 2014 at 11:09
  • \$\begingroup\$ I should have written "combined". \$\endgroup\$
    – ammar
    Oct 25, 2014 at 11:34

1 Answer 1


Answering your second question first, since I think it's the key. There are clearly two regions of operation here, diode on and diode off. You need to solve for the input voltage where you move from one region to the next, which is when the voltage across R1 is just barely enough to turn the diode on (Vd,on), whilst ignoring the diode.

The input voltage under those conditions is Vin = VB - Vd,on - I*R2 (defining current I as flowing into the source, and thus Vd,on as positive).

However, we know that the current I = Vd,on/R1, so we can say that the input voltage is

Vin = VB - I * (R1+R2) = VB - (Vd,on)*(R1+R2)/R1

The solutions for the two regions of operation can then be easily written down.

Edit: It does appear that the solution has a sign error, or (alternately) that Vd,on is defined to be negative (which may be technically correct, but is just confusing).

The output voltage expression in the blue circle is correct- it's just a voltage divider when the diode is 'off', dividing down (Vin - Vb), as is the expression below that- input voltage plus the diode drop minus Vb, but the slope is 1 since there is no series resistor.

  • \$\begingroup\$ Sorry but I don't understand what you've done here. In line 5: what do you mean by input voltage?. In line 6: are you assuming that the diode is on?. In line 7: How does I*(R1+R2) equal Vd*(R1+R2)/R1 ?. I will be pleased if you made it more detailed and obvious. Thanks \$\endgroup\$
    – ammar
    Oct 25, 2014 at 11:56
  • \$\begingroup\$ Input voltage is marked as Vin on the diagram. No, I'm assuming the diode is off, but just by a smidgeon. Third, simply substitute the expression for I from the above line into the equation. \$\endgroup\$ Oct 25, 2014 at 11:59
  • \$\begingroup\$ I think that VD,on is the voltage of the voltage source when we replace the diode of constant-voltage model with an ideal diode in series with a voltage source (usually = 0,7V). Is this right? \$\endgroup\$
    – ammar
    Oct 25, 2014 at 13:01
  • \$\begingroup\$ Yes, exactly right. \$\endgroup\$ Oct 25, 2014 at 13:07
  • \$\begingroup\$ Ok. When the diode is off ( i.e. when Vout = (R2/(R1+R2))(Vin-VB) ): Vin + VD,on > VB or Vin > VB - VD,on. Right? If yes, why is it different from the solution in solution manual? If no, Why? Thanks for answering my questions. \$\endgroup\$
    – ammar
    Oct 25, 2014 at 13:15

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