can someone help solving this out

Bernard

Joined Aug 7, 2008
5,360
It is easy by inspection but doing it the old fashion way calculate the current in each parallel branch then calculate V drop in each R. Noting that V-A & V-C are = then no current flow
thru diode ??
 

MrAl

Joined Jun 17, 2014
6,972
Hello there,

This is partly a topology problem where first you have to REMOVE the diode completely (open circuit there).
Once you remove the diode you have two voltage dividers, so then calculate VA and VC.
Now if VA>VC and ONLY if VA>VC (not VA=VC) then the diode is conducting and so replace it with a short circuit and compute VA and VC again (VA will equal VC now) and calculate the current flow through the diode which is now a short circuit..
Now if however VA<=VC then there is no current flow so your VA and VC are correct as is.

So the procedure is:
1. Remove diode and calculate VA and VC.
2. If VA>VC then replace diode with short circuit, otherwise keep it open circuited.
3a. If short circuit, recalculate VA and VC and those are your results (and then VC=VA and calculate the current in the short).
3b. If open circuit, VA and VC are already calculated so those are your results and VA is less than VC.

So in one case you end up with two separate voltage dividers, in the other case you end up with a bridge circuit with a shorted central section which then reduces to a SINGLE voltage divider and VA=VC then, and you can calculate the current though the short.

Note the word "conducting" means that the diode has non zero current through it. You can not determine this by only knowing the voltage across the diode when the diode is ideal (ideal here is when the diode conducts the voltage is zero).
 
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Papabravo

Joined Feb 24, 2006
12,770
You are merely repeating what I said but in different words.
Not true -- you said there is no voltage drop. When any current flows through an ideal diode there is the forward bias drop. The current may be limited by external components, but the diode itself does not impose any current limit.
 

MrAl

Joined Jun 17, 2014
6,972
Not true -- you said there is no voltage drop. When any current flows through an ideal diode there is the forward bias drop. The current may be limited by external components, but the diode itself does not impose any current limit.
Hi,

Not sure if you realize what you are saying here. An ideal diode that is said to have no voltage drop actually has no voltage drop in the forward conduction mode, period. This is when the diode acts as an ideal "switch" where when it is 'off' it has infinite resistance and when it is 'on' it has zero resistance. This appears to be very common in academia and we see problems like this all the time.
Of course if the external circuit can not produce a voltage difference when the diode is taken out of the circuit then there is still no voltage drop even though there is no current, but if the external circuit produces a negative voltage difference then the diode is open circuit and so has a reverse voltage and so no current flows.

There is also the possibility of making the ideal diode very slightly non ideal, so that the voltage drop is maybe 1e-12 volts. That can help find a solution too while still maintaining an approximate zero voltage drop in the forward mode.

The way i did it myself was to assume that the diode was slightly non ideal having a small resistance, then let that resistance go to zero later to get the final solution or have it go to infinity if the diode is reverse biased.
The voltage difference comes out to be:
Vdiff=(E*(R2*R3-R1*R4))/((R1+R2)*(R3+R4))
and if that is positive then the diode is made a short and the analysis done again, but if negative then that is the true voltage difference with the diode being open. If Vdiff is zero then the voltage difference is naturally zero and the diode does not conduct because there is no current though it (zero volts but no current).

Another interesting way to look at this circuit is to place a capacitor across R4 and do a time response analysis. The cap has to charge up so we see the diode conduct for a while, then stop conducting once VA=VB.
 
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fernandopv

Joined Oct 20, 2018
10
in case
a) Id == 0
then
Va == Vc
else
Va < Vc , because any Id > 0 will cause a further drop in Va and grow in Vc, which is incompatible with Id > 0.

Consequently Id must be 0 and Va == Vc == ER1/(R1+R2)
An ideal diode is compatible with any cero or greater than cero current. Vd == 0 is a singular point in its idealized I-V characteristic curve.
 

MrAl

Joined Jun 17, 2014
6,972
in case
a) Id == 0
then
Va == Vc
else
Va < Vc , because any Id > 0 will cause a further drop in Va and grow in Vc, which is incompatible with Id > 0.

Consequently Id must be 0 and Va == Vc == ER1/(R1+R2)
An ideal diode is compatible with any cero or greater than cero current. Vd == 0 is a singular point in its idealized I-V characteristic curve.
Hello,

Interesting reply, but i think a better way to put it is:

If Id==0 then
Va<=Vc
else
Vc=Va

This simply states that if there is no current flow then Va must be less than or equal to Vc, buf if there is current flow then Vc must be forced to be equal to Va.

Diodes like this are switches that change state based on other circuit conditions and consequently the analysis has to include a solution for exactly when the diode starts to conduct and when the diode stops conducting, and there could be several start and stop locations as time progresses in time dependent circuits.
This is one of the most complicated things that has to be done in circuits like rectifier circuits where the input is constantly changing.
 

fernandopv

Joined Oct 20, 2018
10
I just not considered the case cause bridge resistors were given values that ideally it was not the case to consider va < vc.
But Yes, I agree. That's way is better in general.
 
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