Re: Basic Circuit Question
- From: John Popelish <jpopelish@xxxxxxxx>
- Date: Fri, 28 Oct 2005 10:00:08 -0400
faceman28208@xxxxxxxxx wrote:
I agree that it should work. The diode isn't a perfect one way current blocker. It requires that there be a fraction of a volt used up across it, before conduction takes place, but a 9 volt battery should have plenty extra to provide this voltage and also the more than 3 volts needed to do the same thing for the LED, with enough left that the resistor has some voltage across it, to limit the current (via ohm's law) to about (9 - 0.6 - 3.7)/2670 = .0018 A. That almost 2 mA should be producing a visible indication. Try turning the diode around and see if you are confused about which way it conducts. Remember that the diode will conduct when the end with the strip is more negative.I am trying to debug a kit that I bought that isn't working. I am trying to test various links on the board with a an LED and resistor soldered together.
When I hit the put the ends of the LED/resistor at a point on the board with a direct link to the power supply, it lights up (so the obvious isn't a problem).
If I move my LED/resistor pair to the next in the chain, a diode I get:
(-V9) ------- (- Diode +) ---- (- LED + ) --- (Resistor) --- (+9V)
and the LED no longer lights. Should I expect it to light up, suggesting there is a problem with the Diode? Or is there something more to diode that I am unaware of (I just thought they were one-way current blockers -- suggesting the above should work).
The Diode is a TN4001. Resistor is 267Ohm. Foward voltage on the LED is 3.7.
.
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- Basic Circuit Question
- From: faceman28208
- Basic Circuit Question
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