Circuit Analysis (diode)

[quantum_enhanced]

I am using a new textbook that displays circuits in a way I have never seen. Am I percieving this right? Here's how I transformed the circuit in my textbook:

Additionally, I understand that i=1mA, however, I cant seem to understand how V=5V.

Help?

 

[clombard1973]

Looks good but ....

I am using a new textbook that displays circuits in a way I have never seen. Am I percieving this right? Here's how I transformed the circuit in my textbook:

Additionally, I understand that i=1mA, however, I cant seem to understand how V=5V.

Help?

Hello,

Yes your drawing of the circuit is correct. You can assume that the positive side of the -5Vdc supply is connected to the negative side of the +5Vdc supply.

Now typically one assumes a voltage drop across the diode of about 0.7Vdc. If that were the case, then i would be equal to 0.93 mA and not 1 mA. If that were the case summing the voltage drops around the loop would yield a drop at V that would be -4.3 Vdc (-5 + 9.3 + V = 0).

If though you were given that i is 1 mA then the text must be assuming the diode is ideal. An ideal diode has no voltage drop across it and simply acts to let current flow in one direction but not the other.

If that is the case then when you sum the voltage drops around the loop you should get a voltage at V that is -5Vdc not +5Vdc (-5 + 10 + V = 0). Basically the easiest way to see this is if the diode is ideal then there is no voltage drop across it. Therefore at V with respect to ground you are simply measuring the voltage across the -5 Vdc supply. So the voltage at V must be -5Vdc.

If your text has an answer in it that says +5Vdc then it is a typo as that is the voltage at ground with respect to V and not V with respect to ground.

"Ground" is just an arbitrary place within the circuit that is chosen as a "zero volt" location to take voltage measurements with respect to. Where you placed ground in your second circuit is always how I have seen it and therefore V, with the diode being ideal and having no drop, is at the negative side of a 5 Vdc supply with respect to the positive side; so it is definitely -5Vdc.

Take care,
Craig (Hi)

 

[quantum_enhanced]

Thanks for your response. I hate when the textbooks don't have the right answers in the back of the book, especially during the initial stages of learning a new concept!