Help me analyze this basic circuit

Can someone help me get the voltages using circuit analysis technqiues?

I need the voltage across each diode and the resistor ..... *blush*

since V1=V2 both D1 and D2 will be forward biased.
The voltage drop across D1 and D2 with be the forward voltage drop across the diode. Typically .6 for silicon and .2 to .3 for a germanium diode.
With forward drop of say .6V that means the voltage across the resistor is 3V -.6 =2.4V. Since 2.4Volts is higher than V3 of 2v D3 will be reverse biased, ie off.
So D1=D2=.6V
R1 will have 2.4V
D3 will have -.4V across it.

since V1=V2 both D1 and D2 will be forward biased.

Click to expand...

how?

Yes,why did you assume that D1 and D2 are on not D3 ?

cmos babe said:

Yes,why did you assume that D1 and D2 are on not D3 ?

Click to expand...

If D3 were on then voltage across the resistor would be arnd 1.4V. Then D1 and D2 would be both forward biased and must therefore turn on. This will imply that the voltage across the resistor should be arnd 3-0.6=2.4V.So you have a contradicton here.
Suppose however D1 and D2 only were forward biased then the voltage across the resistor should be arnd 3-0.6=2.4V. and D3 would be reversed biased and wud stay off.

Okay thanks,is this the method I have to follow in analyzing diode circuits like this?

Second,what would the voltages be if i change the V2 to 1V.

cmos babe said:

Okay thanks,is this the method I have to follow in analyzing diode circuits like this?

Second,what would the voltages be if i change the V2 to 1V.

Click to expand...

Voltage across the resistor wud still be 2.4V. Once again the same analysis can be carried out as before.
As to the method, yes I guess you have to look for a soln that is consistent with the properties of all 3 diodes.

You can suppose all the diodes are off ,and the diode that voltage across which is most high will be on and others will be off. then repeat the method again. I think it will be very simple with this method.

Okay here's another circuit...D1 in simulation became off after adding the resistor...Why ?

First you should make is this
V1-Vd1=Vr(1)
V2-Vd2=Vr(2)
V3-Vd3=Vr(3)
Next you have the follows Equation:
Vd1=Vd2=Vd of the Circuit and add (1) and (2)
V1+V2-2Vd=2Vr, then
Vd=(V1+V2-2Vr)/2 (4) and add (2) and (3) then
V2+V3-Vd-Vd3=2Vr (5) and V2-Vd=V3-Vd3 (6)
of (6) I have, Vd3=V3-V2+Vd (7) this equation in (5), then I have, V2+V3-Vd-V3+V2-Vd=2Vr, then
Vd=(2V2-2Vr)/2=V2-Vr (8) and of (8) and (4) I have
V2-Vr=(V1+V2-2Vr)/2

Here is something more intuitive, I believe. Assume that D2 is off as before. Then from the other loop and KVL you have that
3v=2Vr+VD1
Vr=(3-0.6)/2=1.2v

Vr is the voltage drop across each one of the resistors, assuming that only D1 is conducting. This result shows that across R1 you'll have 1.2v and D2 will by necessity be forward biased and consequently will pull the upper side of R1 to 0.6v below V2 i.e. 2.4v. However in this case D1 will go off or in fact will be on the edge (ideally) between conducting and off, because if it is not conducting the drop across R2 will be 0 and the VD1=0.6, but if it starts conducting, there will be some non zero VR2 which will eat some part from the o.6v necessary for D1 to conduct. Note that the voltage across R1 stays unchanged, because it is fixed by D2 and V2.

To the last circuit. Both diodes are forward biased. D1 current is abt 142µA, D2 current is 2.2mA. Voltage drop on R1 is 2.35V and drop on R2 is .142V. Diode D1 is not off but voltage drop on it is only .51V.

Borber said:

To the last circuit. Both diodes are forward biased. D1 current is abt 142µA, D2 current is 2.2mA. Voltage drop on R1 is 2.35V and drop on R2 is .142V. Diode D1 is not off but voltage drop on it is only .51V.

Click to expand...

Okay can you tell us how you got these values please

Ohm's Law
V2-VD2=VR1; 3V-.6V=2.4V
VR1/R1=IR1; 2.4V/1K=2.4mA

One thing to note when a diode is connected to a voltage source it is called a clamp. A clamp will prevent the voltage from going more than .6V above or below the Voltage source, Depending on polatity of the diode.

In your circuit D2 clamps R1 to 2.4V D1 cannot conduct as there is not enough potential across it to allow current flow. it would require more than .6V across R2 and D1 to conduct. This is due to the voltage drop across series resistor R2. THis is also why D1 is off with R2 in circuit.


I disagree that there is .51V across D1 and the Voltage on R1 is 2.35V
Now while there is .6V across D1 and R2, there is no current as D1 is not conducting. A diode drops .6V when conducting, if it is off the circuit is basically open ie NO current.

my 2 cents.

Please simulate circuits and you will understand how the things are in reality. Simplified approach can be missleading.

I agree that sometimes the simplified approach could be misleading, but that's not the problem here. The guy can't understand how to analyze the circuit and getting results from the simulation is not going to help him very much. In fact he got simulation results for the second version and couldn't explain them, that's why he asked. I think that adopting a simplified approach - >0.6v on, otherwise off will help better in understanding. Later on one can assume more realistic approach where the diode conducts somewhat even at 0.5v.
But jumping straight to simulation without understanding is WRONG.

By simulation allmost real results can be acheived. Analyzing results can lead to better understanding. One of the important things is to familiarize yourself with characteristics of involved elements. Simplified analisys will give you simbolic results.

Yes, it will give you symbolic results PLUS intuition. Well, not always, evidently. But if you work by yourself through the steps you'll get it. Afterthat when you see that the simulation shows something more or unexpected, then you are in better position to trace back where the difference comes from. Simulation alone can give you 5v and what? It is no worth if you don't know where this is coming from. However, you are right, one should be familiar with the characteristics of the devices.

I think for such simple circuit, it will be better to build one by his own and test directly the current and the voltage.
Then he will have better understanding of why a diode will drop approx. 0.6V, why a series resistor will have a voltage drop, and how diodes/battery circuits works.

cmos babe said:

Okay here's another circuit...D1 in simulation became off after adding the resistor...Why ?

Click to expand...

I'll explain using an arm waving technique. Assume D2 is conducting. This implies a 0.6V drop across it. Which means net X (point where D1, D2 and R1 connect), is 2.4V. The current through R1 is simply 2.4mA, by ohm's law.

That 2.4mA has to come from somewhere. And there are 2 voltage sources, V1 and V2. At this point, you don't know exactly how much of the 2.4mA current each V1 and V2 actually supplies. But at least you know that both supplies combined together will supply 2.4mA.

2 scenarios.

1st scenario:
Now assume D2, for some reason, doesn't need a lot of current to conduct. So most of the current flows through D1, which also flows through R2. So the voltage drop across R2 is 2.4V. The net on the left side of D1 will be 0.6V. Which automatically makes D2 reverse biased. So it's off. Which also means this scenario is impossible.

2nd scenario:
The other extreme: D2 conducts most of the current. Some current must still try to flow through D1, but in order to do so, will cause a small voltage drop across R2. This forces the net on the left side of D1 to be a little less than 3V, which also makes the drop across R2 less than 0.6V. Which also means that this is impossible.

There you go. Hope you pass your homework with flying colours.....