# Basic Circuit Question

Discussion in 'Electronic Basics' started by [email protected], Oct 28, 2005.

1. ### Guest

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.

2. ### John PopelishGuest

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.

3. ### petrus bitbyterGuest

You'd see some visible light with this chain although much less then you
would without the diode. So I see three possibilities for the error:
- Bad contact (soldering for instance)
- Diode is reverse connected
- Diode is defective.

petrus bitbyter

4. ### Guest

would without the diode. So I see three possibilities for the error:
Me do a bad soldering job? ....Definite possiblity.
Definitely not the case.
I'm going to walk over to radio shack and get a replacemnt for \$0.57

Thanks all.

5. ### Rich GriseGuest

You have one end of the diode marked "+" here. Did you know that the
band on a diode is usually the cathode (and with the 4001, always),
which in schematics is sometimes marked "+", which that end will be
_inside a power supply_. Outside the power supply, the diode will
conduct when the banded end is more negative than the blank end by

Hope This Helps!
Rich

6. ### ehsjrGuest

Are you saying that you disconnected whatever was connected to the +
side of the diode and then connected your LED/resistor tester?
You said you moved the *pair*. That is a little ambiguous. I assume
you left the resistor end of your tester connected to +9, and moved the
LED end to the + side (non-banded end) of the diode.

Yes, it should light up if the diagram is correct. And if
it is correct, it does suggest a problem with the diode. It
could also be a bad solder joint, or bad connection from the
tester to the circuit.

Or is there something
How do you know the resistor is 267 ohms?
(The implication is that you have a multi-meter.
If you do, it makes a better teste device than a
tester made from a resistor and a LED)

Check the diode again. It should be a 1 N4001, not a T N4001

Ed

7. ### Jasen BettsGuest

I'm guessing that you've got either the diode or the battery in backwatds,
for the LED to light the banded end should be towards the negative battery
terminal.

if the diode is between the battery and the rest of the circuit to protect
the circuit from a reversed battery then either it or the battery are
reversed.

if so it may be faulty or the faullt may lie elsewhere.

if not it may be behaving correctly.

Bye.
Jasen