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diode Experiment in the Kybett, Boysen "All New Electronics self teaching guide"


dacart

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Forgive me if this is the wrong area for this but here goes.
I just started going through the book and have arrived at the diode experiment which is a basic circuit that calls for a 10 volt PS (I'm using a 9 volt battery) to a 1M pot to a diode.  You are supposed to measure the current between the + of the PS and the pot and the voltage between the + and - leads of the diode.  I cannot get a reading for current at all.  I know I'm doing something totally noob-like but I cannot figure it out. ???

I have made sure that the anode and cathode are in the right direction and, in fact, get voltage readings, just not any current  :'(

The pot is wired with pin 1 going to +  term of the battery. Pins 2 and 3 are shorted to each other and then go to  the diode and then on to the - term of the battery.

TIA

Dan Carter

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I'm assuming that when you write LED you're talking about the diode (it's not an LED FWIW.)  The diode is wired "forward i.e the negative side is going towards the neg. term on the 9volt.  With the 9 volt it is definitely wired correctly.  I switched to using resistors rather than a pot and while my voltage reads seem good, I still can't get ANY reading for current.  I am placing the red lead lead of my multi-meter before the resistor and the black lead after the diode where the circuit goes to neg.  Do I need to go to ground?  I am very confused if so because there is no indication for ground on the schematic, just the circuit going from the pos. to neg. battery terms.

Sorry again for the total noob question .

Dan Carter

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It sounds like you connected your current-meter directly in parallel with the battery, instead of in series.
Then you killed the battery (because a current-meter is a dead short) and maybe blew the fuse in your current-meter.

But like I said before, if you turn down the pot then the current becomes infinite and might blow the pot or also kill the battery.

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