Engineering Calculating Voltages in a circuit (With Diodes)

[wtfman]

Say D1 is a silicon diode( .7 volts) and D is a Germanium diode (.3 volts)

Determine the voltages at A,B,C and D with respect to ground

You can use hypothetical values

Any help would be greatly appreciated

Thanks

 

[dimpledur]

Remember KVL? ƩV=0 around a loop. Apply this fact to solve for unknown voltages.

 

[wtfman]

Remember KVL? ƩV=0 around a loop. Apply this fact to solve for unknown voltages.

But the voltage in each point should be the same right even with a resultant voltage?

What about the diodes?

 

[gneill]

This is one of those questions where the answers will depend upon the level of detail and the assumptions that you make; a simply posed question that could take pages to answer!

For example, is Vcc > Vc or is Vc > Vcc? (are the diodes forward or reverse biased by the supply voltages?). What about the region where the diodes are just beginning to turn on or off and their I vs V curves are nonlinear? What sort of model are you going to use for the diodes? Real diodes have small reverse bias leakage currents, will that be taken into account?

 

[wtfman]

This is one of those questions where the answers will depend upon the level of detail and the assumptions that you make; a simply posed question that could take pages to answer!

For example, is Vcc > Vc or is Vc > Vcc? (are the diodes forward or reverse biased by the supply voltages?). What about the region where the diodes are just beginning to turn on or off and their I vs V curves are nonlinear? What sort of model are you going to use for the diodes? Real diodes have small reverse bias leakage currents, will that be taken into account?

Well let's just say its not that complicated. Let's put Vc as 12 V and Vcc as 24. The diodes are forward basis where D1 is Silicon and D is Germanium. Thats all the details given. Just straightforward.

 

[gneill]

Well let's just say its not that complicated. Let's put Vc as 12 V and Vcc as 24. The diodes are forward basis where D1 is Silicon and D is Germanium. Thats all the details given. Just straightforward.

Oh sure, spoil ALL the fun!

 

[wtfman]

lol not for me. But still any ideas on the problem?

 

[gneill]

lol not for me. But still any ideas on the problem?

Sure. Since they're forward biased, pencil in the forward voltage drops for the diodes. What does that leave for the voltage drop across the resistor? (KVL)

 

[wtfman]

Sure. Since they're forward biased, pencil in the forward voltage drops for the diodes. What does that leave for the voltage drop across the resistor? (KVL)

ok I am going to use the total voltage for the voltage drops in each diode. So for D1, the voltage drop would be Vd1 = VR - .7. Which is Vd1 = 36-.7 = 35.3

For D, the voltage drop would be Vd = VR - .3. Which is Vd = 36-.3 = 35.7

Does that help?

 

[gneill]

ok I am going to use the total voltage for the voltage drops in each diode. So for D1, the voltage drop would be Vd1 = VR - .7. Which is Vd1 = 36-.7 = 35.3

For D, the voltage drop would be Vd = VR - .3. Which is Vd = 36-.3 = 35.7

Does that help?

No, in fact it's rather bizarre The voltage drop across the silicon diode is 0.7V. That's it, that's all. The voltage drop across the germanium diode is 0.3V. That's it, that's all. So what's left for the resistor?

 

[wtfman]

No, in fact it's rather bizarre The voltage drop across the silicon diode is 0.7V. That's it, that's all. The voltage drop across the germanium diode is 0.3V. That's it, that's all. So what's left for the resistor?

the total for all voltages?

 

[gneill]

Are you familiar with KVL? Can you write the KVL equation for the loop?

EDIT: Here's your circuit with the known voltages penciled in:

 

[wtfman]

Are you familiar with KVL? Can you write the KVL equation for the loop?

VR = Vcc -.7 -.3 + Vc ?

 

[gneill]

VR = Vcc -.7 -.3 + Vc ?

Almost. Watch your polarity on Vc; when you pass through it on your way around the loop, you're going from its + terminal to its - terminal.

 

[wtfman]

Almost. Watch your polarity on Vc; when you pass through it on your way around the loop, you're going from its + terminal to its - terminal.

So VR= Vcc - .7- .3- Vc?

 

[gneill]

So VR= Vcc - .7- .3- Vc?

Is that a guess? Why don't you explain your reasoning? Take me on a "round the loop tour" of the circuit.

 

[wtfman]

Is that a guess? Why don't you explain your reasoning? Take me on a "round the loop tour" of the circuit.

ok well the voltage for the resistor is the Vcc added to the silicon diode. It is in forward basis, so its 12-.7 and then past the resistor into the second diode, again with forward basis, and then the last terminal where it goes from the + terminal to its - terminal into the ground.

Hence reasoning: VR= Vcc- .7-.3- Vc

 

[gneill]

ok well the voltage for the resistor is the Vcc added to the silicon diode. It is in forward basis, so its 12-.7 and then past the resistor into the second diode, again with forward basis, and then the last terminal where it goes from the + terminal to its - terminal into the ground.

Hence reasoning: VR= Vcc- .7-.3- Vc

Well, that's more or less it. For KVL work it's better to deal with individual potential differences across the components rather than the potential at various nodes around the circuit with respect to ground.

So I would say: beginning the loop at point C (at the right end of the resistor) and proceeding clockwise, There is a voltage drop of 0.3V for the germanium diode, followed by a drop of 12V for Vc, followed by a rise of 24V for Vcc, and then a drop of 0.7V for the silicon diode. The sum is then: -0.3V - 12V + 24V - 0.7V = 11V. This remaining 11V will be the potential drop across the final component in our loop, the resistor, bringing us back to point C where we started. Thus the potential drop across R is 11V.

 

[wtfman]

Well, that's more or less it. For KVL work it's better to deal with individual potential differences across the components rather than the potential at various nodes around the circuit with respect to ground.

So I would say: beginning the loop at point C (at the right end of the resistor) and proceeding clockwise, There is a voltage drop of 0.3V for the germanium diode, followed by a drop of 12V for Vc, followed by a rise of 24V for Vcc, and then a drop of 0.7V for the silicon diode. The sum is then: -0.3V - 12V + 24V - 0.7V = 11V. This remaining 11V will be the potential drop across the final component in our loop, the resistor, bringing us back to point C where we started. Thus the potential drop across R is 11V.

ok fair enough but what about the voltages ( A,B,C and D) across the different points or is it the same for the voltage drop across R?

Thanks again for all the help

 

[gneill]

ok fair enough but what about the voltages ( A,B,C and D) across the different points or is it the same for the voltage drop across R?

Thanks again for all the help

After you've established all the individual potential differences, then you can go back and sum them from the ground node to the individual nodes (you can work in either direction around the loop).

 

[wtfman]

After you've established all the individual potential differences, then you can go back and sum them from the ground node to the individual nodes (you can work in either direction around the loop).

Wait can you show me an example with either point A or B. I am not sure if I fully get what you meant.

 

[gneill]

Potential at point B with respect to ground: Starting with the ground node and heading to point B: Go from ground through Vcc through diode 1: +24V - 0.7V = 23.3V.

 

[wtfman]

Potential at point B with respect to ground: Starting with the ground node and heading to point B: Go from ground through Vcc through diode 1: +24V - 0.7V = 23.3V.

So at C would be 23.3V + 11 V = 34.3 V with respect to ground?

 

[gneill]

So at C would be 23.3V + 11 V = 34.3 V with respect to ground?

Almost. The resistor has a voltage drop as you pass through it in the direction of current flow. That is, going from left to right through the resistor you have an 11V drop.

 

[wtfman]

Almost. The resistor has a voltage drop as you pass through it in the direction of current flow. That is, going from left to right through the resistor you have an 11V drop.

So C is 23.3V - 11 V = 12.3 V and D would be 12.3V - .3V = 12V

:)

 

[gneill]

Just so.

 

[wtfman]

Just so.

Thanks I really appreciate all the help :)