# Explain why DMM reads 00.0 on negative side of circuit

Discussion in 'General Electronics Discussion' started by stspringer, Jul 11, 2019.

1. ### stspringer

114
9
May 10, 2019
Hello All,
Another question any help appreciated. Thanks in advance

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2. ### Martaine2005

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May 12, 2015
That's actually really confusing.
Can you explain that again please?
All I can think of is the LED is acting as it should, a diode.
But as I couldn't decipher your question, or where you are probing, it's just a guess.

EDIT: LEDs don't have positive and negative legs.
EDIT2: Put the resistor before the LED and try your probing again. See if you get the opposite results on your DMM.
Martin

Last edited: Jul 11, 2019
3. ### stspringer

114
9
May 10, 2019

"LEDs don't have positive and negative legs."
Yes they do. Positive is the anode negative is the cathode. https://www.westfloridacomponents.c...w-to-tell-which-lead-is-positive-or-negative/

My picture shows the negative leg of the led. I put the black probe from my meter in the negative power rail. I put the red probe in the hole on the negative led leg. My meter reads .474 which is the resistance of the 470 ohm resistor. This is all done on the negative side of the circuit as shown in the picture.

Now with the black lead still on the negative power rail if I place the red probe on the positive side of the circuit on the other leg of the resistor my meter reads 00.0 which means there is continuity, but I was wondering why it would not read .474 ohms like the positive side did.

I hope you are clear on what I am saying about the more positive side or more negative side of the circuit, which I point out in the picture.

"EDIT2: Put the resistor before the LED and try your probing again. See if you get the opposite results on your DMM.
Martin"

The resistor is bridging the negative and positive sides of the circuit on the breadboard see the picture.

Last edited: Jul 11, 2019
4. ### BobK

7,682
1,686
Jan 5, 2010
Are you measuring Ohms on a powered circuit? That is a no-no.

Bob

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5. ### Martaine2005

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754
May 12, 2015
That will measure the resistor only.
Sorry you lost me. Please edit the circuit picture with connections from your DMM.
Also, LEDS as you already stated have anodes and cathodes. Which tells you which leg to connect to the positive and negative of a power source.
Martin

6. ### stspringer

114
9
May 10, 2019

No, it is not Powered Bob I learned that from you and others

7. ### Martaine2005

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754
May 12, 2015
That never even crossed my mind. Well spotted.

8. ### stspringer

114
9
May 10, 2019
No, it is not Powered

9. ### Martaine2005

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754
May 12, 2015
Can you clarify post 5 please.

Martin

10. ### stspringer

114
9
May 10, 2019
The circuit is not powered. I will try to explain it better.

My circuit, the side with the short red jumper wire, going to the anode leg of the led, I am calling this the more positive side.
My circuit, the side with the small black jumper wire feeding the other leg of the resistor to complete the circuit, I am calling the more negative side.

The resistor is the bridging the breadboard center channel to complete the circuit.

If I keep my black meter probe on any negative power rail, and then I place my red meter probe on the more "positive side" on the "negative cathode side" of the LED "the side with the short red jumper wire" my meter reads .474 ohms.

If I keep my black meter probe on any negative power rail, and I then place my red meter probe on the more "negative side" the side with the short black jumper wire" on the other resistor leg my meter reads 00.0 to 00.3

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Last edited: Jul 11, 2019
11. ### Martaine2005

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May 12, 2015
Is this what you are doing? Excuse the terrible pic

Martin

12. ### stspringer

114
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May 10, 2019
Pretty much please see my new pics. The resistor is hard to see in my new pics but it is in line with the negative leg of the led across the breadboard gap and fed by the small black jumper wire.

13. ### Martaine2005

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May 12, 2015
Ok we got there in the end.
What you are reading is perfectly normal. You are measuring the resistor. Then you are just probing a wire between the resistor and power source, it's a dead short for the DMM.

Martin

14. ### stspringer

114
9
May 10, 2019
Why do you say it is a dead short? Isn't 00.0 on the ohm setting meaning continuity? Also there is no power going to this breadboard.

15. ### Martaine2005

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May 12, 2015
Yes, continuity means a loop without a break.
What you are doing *is* just putting the meter probes together.

Martin

16. ### stspringer

114
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May 10, 2019
Ok so here is where I am confused. Why am I getting a good resistor reading .474 ohms only on one side of the resistor, if the resistor is the bridge to complete the circuit?

17. ### Martaine2005

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754
May 12, 2015

This Is continuity.

Martin

18. ### stspringer

114
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May 10, 2019
Yes but I don't see "understand" where I have a break. Why is the resistor braking my path?

19. ### Martaine2005

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754
May 12, 2015
You don't have a break. You have very low Ohms reading. A short Or a link or electrically connected.

Martin

20. ### stspringer

114
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May 10, 2019
So in my case it would be electrically connected, right? I mean the circuit works when I apply power