Hi,
In this pic I'm only sure about D1 which is ON and for D2 and D3 I'm not really sure about them! So, please help me out how can I determine whether if they are on or off and how?
thanks in advance.

 

D1 is on
D2 + D3 is off

Oops .....Recalculating

Nope D1 on, D2 + D3 off.

 

I get D2=off, D1 & D3 =on

But the more I look at it the less certain I am.

 

Iterestingly enough my sim tells me that not all of them are ON. Sadly I don´t have the mental capability right now to figure out how to reliably come up with the result.

Edit: it seems there could be a bit of uncertainity coming from the assumed Vf of the diodes, which might turn D3 on as well. My sim seems to assume Vf of 0.56V at this sort of current with 1N4148 and states that D3 is off.

 

D1 & D3 on............D2 off?

 

Hi,
In this pic ...

What assumption are you making about the forward voltage drop of a conducting diode?
0V or 0.6V or something else?

 

The symbol is not for a schottky so I assumed a silicon diode with .7 Volt drop.

 

Ignore D2 and D3. Assume all diodes are perfect. Assume all diode's Vf are 0.6 V.

Loop current is 14.4 V / 30.2 K = 476.8 uA.

The node voltages are:

8.2K/12K = +5.49V. This is greater than 4.4 V, so D2 is off.

12K/10K = -0.232 V. This greater than -0.6 V, so D3 is off.

ak

 

AK, I used the same logic. Agreed (I am sure Mike is plugging this into spice as I type.)

 

D1 is on
D2 + D3 is off

Oops .....Recalculating

Nope D1 on, D2 + D3 off.

Well you consider D1, D2 on and D3 off! Sorry it's wrong answer!

 

I get D2=off, D1 & D3 =on

But the more I look at it the less certain I am.

you got them right! But how? I spent an hour trying to find how but i couldn't. plz show me how you found that, thanks

 

Iterestingly enough my sim tells me that not all of them are ON. Sadly I don´t have the mental capability right now to figure out how to reliably come up with the result.

Edit: it seems there could be a bit of uncertainity coming from the assumed Vf of the diodes, which might turn D3 on as well. My sim seems to assume Vf of 0.56V at this sort of current with 1N4148 and states that D3 is off.

What about D2?

 

D1 & D3 on............D2 off?

Well, you got it, but how?!

 

What assumption are you making about the forward voltage drop of a conducting diode?
0V or 0.6V or something else?

0V

 

Ignore D2 and D3. Assume all diodes are perfect. Assume all diode's Vf are 0.6 V.

Loop current is 14.4 V / 30.2 K = 476.8 uA.

The node voltages are:

8.2K/12K = +5.49V. This is greater than 4.4 V, so D2 is off.

12K/10K = -0.232 V. This greater than -0.6 V, so D3 is off.

ak

Sorry but the answer isn't correct!

 

Try making the circuit to better understand..

 

Well isn't that special. You present an electrical puzzle and you leave out critical data. Then after several opinions are entered, you release the critical data at the same time you proclaim the winners and losers. Now where is that "Ignore" button I have been hearing about.

 

I just a simulation on my phone and it say D1 on , D2 &D3 off..

 

Well isn't that special. You present an electrical puzzle and you leave out critical data. Then after several opinions are entered, you release the critical data at the same time you proclaim the winners and losers. Now where is that "Ignore" button I have been hearing about.

This is NOT electrical puzzle this is example I found on google and I have been trying to solve it! And I shared it here so everyone can give me an idea how s/he got it. It's not my problem what do you think and how do you think! I asked question and I didn't write who is winner or loser YOU wrote it. And have fun with ignore button!

 

Try making the circuit to better understand..

So, then Mesh analysis? I'll end up with 4 Loops? But how can I determine which one is on and which one is off?!