Discussion in 'Homework Help' started by amangupta1219, Oct 28, 2012.
please tell the output voltage mathematically or graphically
But where is the circuit ?
sorry i uploaded it again
Homework HELP requires that you post your initial idea of what the answer should be. This forum doesn't simply do you homework for you, but will provide a lot of help and guidance to get you to the right answers.
As a start, what would the output voltage be with 0 volt input?
What would be the effect of a positive and then negative voltage input on the diodes?
What would those changes do the the current through the resistors?
If the current changes, what happens to the voltage across the resistors?
After a time, you will learn to ask questions like this yourself when analyzing any circuit.
My approach :
as long as the Vi is < 9.3 V , the diode D1 will be ON because in that case VA will be < 10 V and current can flow through 10k resistor. And if this is true then D2 will also be 0N and Vo = Vi.
Now second interval i.e Vi > 9.3 V, same reasoning for D3 i.e D3 will be ON as long as Vi > -9.3 V & D2 is ON & D4 is also ON. But in this case if we track from input to the output through D3 & D4 , then i get Vo = Vi and this is making D1 ON which i assumed to be OFF. Summarizing my results :
0<Vi<9.3 all diodes are ON & Vo = Vi
Vi>9.3 all diodes are ON & Vo=Vi..... but this is not possible because this makes VA > 10 which shows current can't flow from +10V battery. So in short i know i am doing but the question is where ????
are you sure that for Vi>9.3V all diodes are on?
i think slow but safe approach with diode circuits is to:
1. assume all diodes are off
2. calculate Vf for each diode (forward voltage)
3. if Vf for one or more diodes is greater than Von (such as 0.7V), turn on one diode (the one with greatest Vf)
4. repeat from step 2 (until result is repeated and no change occurs, this happens when all diodes that are still assumed to be off, have Vf<Von)
you would have to do this for variety of input conditions (intervals) until all cases are covered.
i don't want to reveal too much so i'll consider case you already know,
supposedly we work on interval
-9.3V<= Vi <= +9.3V
and Vi happens to be ... +9V...for example.
1. all diodes are off, hence no current through resistors, therefore Vd=Vi=9V, Va=10V,Vb=0V, Vc=-10V
3. All of them are greater than Von (0.7V for example) but we only turn on D3 because it has largest forward drop
2. Vd=Vi=9V, Va=10V,Vb=0V, Vc=Vd-Von=8.3V
3. now we turn on D2 and D3 and repeat again...
but for this part we already know the result, I'm just showing working. yess in this case all of diodes will end up conducting and Vo=Vi.
then you assume input that is outside of this interval, such as Vi>9.3V and repeat all steps. you will see that not all diodes turn on. then you can use KVL to find solution. don't forget to also need to cover case for Vi<-9.3V
thanks panic mode......... now i got how to solve diode circuits.
but for vi = 9V , when D2 & D3 are ON then current from +10V battery will flow through D2 only and using KVL i get Vo = 4.65 V & D1 , D4 will be OFF. This answer is matching........ Vo = Vi upto Vi=4.65 V and after that Vo= 4.65 V. Vo is symmetric so same explanation for Vi < 0.....
but one more question when i vary Vi from 4 to 5 V , if +10V battery can force the current through both D1 & D2 for Vi<4.65V , why it can't force the current through D2 & D4 for Vi > 4.65 V as it can provide adequate potential at Va i.e Va = Vi+0.7 ??????????
i think you are confusing Vi and Vo (they are different):
for -9.3V<=Vi<=+9.3V all diodes conduct.
when this happens, diodes D1 and D3 force voltages in points A and C.
as a result of this, you get voltage at point B that matches that of point D.
when Vi>9.3V, you will get two diodes to conduct (D2 and D3) and the other two are off because in this case we only force voltage of point C (not point A). what happens now is that point A and B are no longer linked to specific Vi value - they only depend on supply voltage and ratio of resistors. as you have found, you get Vo that is result of voltage divider (2x10k) ensuring output is half of (10V-Von)=0.5*9.3V=4.65V (same thing just backwards happens for Vi<-9.3V).
Reason that only D2 (and D3) conduct but not D1 and D4 is high potential at C so D1 and D4 are no longer on. if the input is Vi=9.5V for example, Vc=9.5V-0.7V=8.8V. Vb = 4.65V.
Vd4=Vb-Vc=4.65 - 8.8 = -4.15V note that this is negative voltage (D4 is reverse biased).
At the same time
Vd1=Va-Vd=5.35-9.5V= -4.15V so D1 is also reverse biased.
Note: I have calculated Vf for diodes ("f" for forward voltage). This means that voltage difference is Vanode - Vcathode. Diode is polarized device so order is very important!
when Vi is less than -9.3V, voltage at point C is no longer dependent on Vi. D1 however forces point A to low potential
Va=Vi+0.7V (remember that Vi is large negative value), therefore D2 and D3 are off. D4 is forward biased because if there was no current through it, Vb would be 0V which is greater than Vc ( = -10V). but since that difference is greater than Von, D4 conducts and we have again voltage divider...
can you tell me what will be the output voltage Vo for Vi = 5V ?
According to me, Vi=5V so all diodes can be ON & in that case VA = Vi+0.7 = 5.7V and Vo = 5.7-0.7 = 5V, this will make Vc = 4.3 V again possible. So is this the correct output for Vi=5V ?
yes, that is correct
but this is wrong according to the textbook.
Vo = Vi for -4.65 < Vi < 4.65
& Vo = 4.65V for Vi > 4.65 V
& Vo = -4.65V for Vi < -4.65V
yes - it looks like i made a too quick of an assumption in the post #6 with "but for this part we already know the result". an easy mistake at 4AM, sorry...
but you need to realize that this is not my homework, otherwise i would make sure that it is correct (i would have gone through every step myself). your comment tells me that you didn't work the problem either. i have an excuse but what is yours?
you may need to engage a bit more. there are several ways to verify things: analytically, in simulation or real circuit (breadboard for example).