10V | o--------o | | .-. .-. | | | | | |10k | |10k '-' '-' | | Vout-o---|<---o | | .-. .-. | | | | | |10k | |1k '-' '-' | | o--------o | === GND
This would be electrically the same as the schematic niftydog posted.yes, but there is also a separate input to the top right at +10V. and another ground at the bottom left, the two sides of the H are not connected.
In this case, find the value of R such that the voltage at the right middle node is .6 volts greater than the left middle node.The question also asks for the value of R where the diode would stop conducting.
Am I going nuts? The circuit I see has the diode reverse biased. If this is the case, Vout=5 volts.This would be electrically the same as the schematic niftydog posted.
What you must do for the first question is replace the diode with a .6 volt source (such as a battery; positive on the right) and solve the network for the voltage at the left middle node. This assumes that the diode voltage can be assumed to be .6 volts independent of the current in it.
In this case, find the value of R such that the voltage at the right middle node is .6 volts greater than the left middle node.
OK, but my question was primarily meant for The Electrician.No you are correct, sorry. I have been working on the same circuit with a 20k resistor instead of 1k after figuring out that part.
For the 1k I got Vout = 5 volts and after a little help I got Vout = 5.715 volts when there is a 20k resistor in the place of the 1k resistor. Can someone verify that?
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