Hi,
If i had a diode and resistor in series with each other, in parallel with another set of diode and resistor in series. The resistors are 100kΩ and both the diodes are pointing in the same direction.

How would you add the resistances?

In finding the overall current, I've tried guessing the equivalent resistance as 0,50,100 and 200. None seem to give the right answer, but 70kilo ohms will. I'm not sure how to obtain this quantity though.

(This is part of a wheatstone bridge question with five resistors, this particular branch will become R5, the fifth resistor where normally it would be a voltmeter/ammeter for a balanced bridge. The objective of the question if to find the overall current )

Thanks for the anticipated help.

 

Is the applied voltage AC or DC and if DC will it be applied such that the diodes are forward biased or not. A complete diagram could be useful as sometimes these are trick questions that require no math.

Lefty

 

I have attached the diagram of the diode-resistor arrangement. The applied voltage is DC. I read somewhere that for semiconductors, you add the parallel branches, but if that is so, then the effective resistance of this particular branch is 200 kΩ. Is this right?

Thanks for the prompt reply!

 

As drawn, your diodes are reversed biased, so there wouldn't be any current in the branches under the DC conditions you indicated.

 

I think, IF the diodes were biased in the forward direction, and enough current was flowing to cause the diodes to conduct, then you could subtract 0.7 v. from the supply volt. and then divide by the resistance in that branch, to solve the branch current.
But as was stated, these diodes are NOT conducting. Therefor zero current.

 

Yes, if the drawing is correct, then it is most likely a trick question to test your knowlege of semiconductor current flow. As the diodes are reversed bias there will be no current flow through either resistor.


Lefty

 

Basically the diodes aren't resistances, they are voltage drops. In any circuit you want to analyse figure the .7V drop into it, then work from there.

In other words you take the power supply voltage, subtract 0.7VDC from it, and calculate the current from there.

 

If the current through R5 = 0, I end up with R1 and R3 in series but in parallel with R2 and R4, which gives me the overall resistance of 75 ohms that I needed for the right answer.

Thanks a lot!

 

Hi.

first apply the add voltage method to both of the resistance-diode sets connected in series. Then you will apply add current method to the parallel branches.

 

If the current through R5 = 0, I end up with R1 and R3 in series but in parallel with R2 and R4, which gives me the overall resistance of 75 ohms that I needed for the right answer.

Thanks a lot!

Your diagram doesn't reflect what you just wrote.