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Look at the attached pic:
looking at point X (=Vo) :
you need to ask, at what voltage at point X does D1 actually conducts?
The answer is:
1v +1ma*1kohm+0.7=2.7 v
if X is at 2.7v what is the state of D2 and D3 ?
The answer is they are cut-off(not conducting).
Now you ask what is the voltage at Vi
The the answer is 2.7+1ma*3khom=5.7v
If Vi is greater than 5.7v you have :
Vo=(Vi-1-0.7)*1/4 +1+0.7=Vi/4 +1.275
at Vi=10v we get Vo=3.775v
look at the graph
to complete the task you need to look at D2 and D3 and ask at what voltage do they conduct both and only one ( point Y).
looking at point X (=Vo) :
you need to ask, at what voltage at point X does D1 actually conducts?
The answer is:
1v +1ma*1kohm+0.7=2.7 v
if X is at 2.7v what is the state of D2 and D3 ?
The answer is they are cut-off(not conducting).
Now you ask what is the voltage at Vi
The the answer is 2.7+1ma*3khom=5.7v
If Vi is greater than 5.7v you have :
Vo=(Vi-1-0.7)*1/4 +1+0.7=Vi/4 +1.275
at Vi=10v we get Vo=3.775v
look at the graph
to complete the task you need to look at D2 and D3 and ask at what voltage do they conduct both and only one ( point Y).
Yes. You are correct. The diodes are reverse-biased and they are not conducting.
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The diodes are reverse-biased and are not conducting.
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Colin your analysis is totally wrong!
"Suppose we have no input voltage on Vi.
What is the state of the circuit?
The voltage-difference between top and bottom is 3v.
we will assume the diodes are conducting and dropping 0.7v"
Wrong!
the diodes are reversed bias and thus not conducting!
Sorry I did not get to this problem sooner, but I was on vacation.
It is assumed that the diodes will turn off if their voltage is less than 0.7 volts, or their current is less than 1 ma. Starting at -10 volts, it is easily observed that diode D1 will not conduct at this voltage, and D3 and D4 are conducting. That makes the output voltage -3.4 volts. At input voltage -8.5 diode D3 is current starved and turns off. Diode D2 is on until the input voltage is -6.7 volts, and it also turns off due to current starvation. All diodes are now off so the output voltage is the same as the input voltage. At input voltage +5.7 volts diode D1 turns on at 1 ma and conducts until the +10 volt limit is reached. See the plot below.
Ratch
This problem has been hanging around for ten days now. If the OP hasn't been given the solution by the teacher by now, he should change schools. Sometimes, the full solution is a way of learning something. Others who read this thread might want to see the solution, too. Since the OP has not responded to this thread, it can be assumed that he has moved on to something else.
Ratch
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