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I posted this exercise given as homework in the wrong section and it got deleted.Nevertheless,I asked my teacher to solve it and so did he.Thing is I still have some unclear stuff.Please ,a crucial part of the exercise is the part b,so please pay attention to it.
We have the limiting circuit with diodes.The diodes have Rf=20 Ohm, Rr=infinity.Vγ=0.
a) Build the transfering characteristics.
b) Make a full explanation for every characteristic zone ( The state of the diodes,the values of the tensions in the entrance and in the exit,etc)
c) Replace the diode D2 with the resistance R.Build the transfering characteristic in this case.Compare with the previous result.
d) Change the Rf parameter of the diode to create only one breaking point.(So to pass from the characteristic of the point A to the characteristic of point C)
http://i.imgur.com/cixtsbM.jpg
There arent any equations in this type of exercise.Its more logical.
a)So first we have zone a with a horizontal line at Y=20 V which stops at approx X=20 V.The next zone slopes up at an intermediate rate till just above 20 V.The next zone is c which slopes up at a faster rate, almost X = Y. There are 2 breakpoints, one at each end of zone b.
I get zone a,but how about zone b and c? How does one determine those? I have studied many V0 and Vi graphs but we don't do them immediately,we first have to find point b).
b) For zone a, Vin is below 20V, so D1 is not conducting and D2 is conducting. Current flows through R=5K at 4mA approx. so there is ~ 80mV drop in the Rf of D2 so the actual voltage is 20V-80mV.
I get this,the output voltage is 20V-80mV so approx 20 V,hence the horizontal line (zone a)
For zone b, once Vin reaches 20V-80mV D1 starts to conduct. There is current in both diodes because of the Rf value of 20 ohms, so for the Vin range between 20V - 80mV and 20V + 80mV both diodes are sharing current to R=5K.What about output voltage here?
Also,This means we are studying both diodes,first at ]-∞;20[ (for zone a) then at 20 V -80 mV≈ 20 V(for zone b) and as we are going to zone c, ]20V + 80 mV;+∞[.I think that we should have studied the diodes at ]-∞;0[ ,[0;20], [20;+∞[.This is because at [0;20] the diodes are actually conducting,but in ]20;+ ∞[ only D1 is conducting. For zone c, as the Vin increases beyond 20V + 80mV the current is only flowing in D1, and D2 is not conducting.in R=5k increases beyond 4mA, so the effect of Rforward increases slightly as Vin increases. The output is close to Vin-80mV, but this changes as the voltage changes because the current in the Rf and R=5K changes.I get this.
c)Now it has no zone b with intermediate slope because there is only one diode, and current is not shared with another diode. The output is almost exactly 20V in zone a, when Vi is below 20V because there is no current in the output circuit. In zone c Vout is almost the same as Vin because only a small current flows through R=5K to the 20V supply. This is because the voltage difference is not large, e.g. 1V across 5K = 200uA.The voltage drop across Rf is correspondingly small, at 4mV for Vin = 21V.
I get this.
d)
Reduce Rf in both diodes as small as possible as this makes the range of voltages shared between the diode very small, so zone b is very small.I don't get this,he practically just explained the question but when I asked him about a hint,he didnt reply.How do I do this?
Homework Statement
We have the limiting circuit with diodes.The diodes have Rf=20 Ohm, Rr=infinity.Vγ=0.
a) Build the transfering characteristics.
b) Make a full explanation for every characteristic zone ( The state of the diodes,the values of the tensions in the entrance and in the exit,etc)
c) Replace the diode D2 with the resistance R.Build the transfering characteristic in this case.Compare with the previous result.
d) Change the Rf parameter of the diode to create only one breaking point.(So to pass from the characteristic of the point A to the characteristic of point C)
http://i.imgur.com/cixtsbM.jpg
Homework Equations
There arent any equations in this type of exercise.Its more logical.
The Attempt at a Solution
a)So first we have zone a with a horizontal line at Y=20 V which stops at approx X=20 V.The next zone slopes up at an intermediate rate till just above 20 V.The next zone is c which slopes up at a faster rate, almost X = Y. There are 2 breakpoints, one at each end of zone b.
I get zone a,but how about zone b and c? How does one determine those? I have studied many V0 and Vi graphs but we don't do them immediately,we first have to find point b).
b) For zone a, Vin is below 20V, so D1 is not conducting and D2 is conducting. Current flows through R=5K at 4mA approx. so there is ~ 80mV drop in the Rf of D2 so the actual voltage is 20V-80mV.
I get this,the output voltage is 20V-80mV so approx 20 V,hence the horizontal line (zone a)
For zone b, once Vin reaches 20V-80mV D1 starts to conduct. There is current in both diodes because of the Rf value of 20 ohms, so for the Vin range between 20V - 80mV and 20V + 80mV both diodes are sharing current to R=5K.What about output voltage here?
Also,This means we are studying both diodes,first at ]-∞;20[ (for zone a) then at 20 V -80 mV≈ 20 V(for zone b) and as we are going to zone c, ]20V + 80 mV;+∞[.I think that we should have studied the diodes at ]-∞;0[ ,[0;20], [20;+∞[.This is because at [0;20] the diodes are actually conducting,but in ]20;+ ∞[ only D1 is conducting. For zone c, as the Vin increases beyond 20V + 80mV the current is only flowing in D1, and D2 is not conducting.in R=5k increases beyond 4mA, so the effect of Rforward increases slightly as Vin increases. The output is close to Vin-80mV, but this changes as the voltage changes because the current in the Rf and R=5K changes.I get this.
c)Now it has no zone b with intermediate slope because there is only one diode, and current is not shared with another diode. The output is almost exactly 20V in zone a, when Vi is below 20V because there is no current in the output circuit. In zone c Vout is almost the same as Vin because only a small current flows through R=5K to the 20V supply. This is because the voltage difference is not large, e.g. 1V across 5K = 200uA.The voltage drop across Rf is correspondingly small, at 4mV for Vin = 21V.
I get this.
d)
Reduce Rf in both diodes as small as possible as this makes the range of voltages shared between the diode very small, so zone b is very small.I don't get this,he practically just explained the question but when I asked him about a hint,he didnt reply.How do I do this?
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