Problem with resistor

[Karthik rajagopal]

Hi all,
I recently bought a 0.5ohms 5watt resistor for my over current protection circuit . As per the calculations , the resistor must provide a voltage drop of 0.5v when 1A of current flows through it
But when I measure it with my multi meter, it reads 0.05v. This becomes 0.1v for 2A. I even measured the readings with another multimeter to make sure that nothing is wrong with my meter and I got the same reading .
Please help me to solve this problem .

 

[Ylli]

Three possibilities exist.
1. You are measuring the current incorrectly.
2. You are measuring the voltage drop incorrectly.
3. You have a 0.05 ohm resistor, not a 0.5 ohm resistor.

Ohm's Law will not be violated!

 

[Nanren888]

Was it getting hot?

 

[Bluejets]

This becomes 0.1v for 2A

3. You have a 0.05 ohm resistor, not a 0.5 ohm resistor.

E(volt)=I(current)R(resistance)

W = I squaredR...would hardly get hot.

 

[davenn]

Hi all,
I recently bought a 0.5ohms 5watt resistor for my over current protection circuit . As per the calculations , the resistor must provide a voltage drop of 0.5v when 1A of current flows through it
But when I measure it with my multi meter, it reads 0.05v. This becomes 0.1v for 2A. I even measured the readings with another multimeter to make sure that nothing is wrong with my meter and I got the same reading .
Please help me to solve this problem .

to stop us all guessing ...
SHOW us the circuit

 

[Karthik rajagopal]

Three possibilities exist.
1. You are measuring the current incorrectly.
2. You are measuring the voltage drop incorrectly.
3. You have a 0.05 ohm resistor, not a 0.5 ohm resistor.

Ohm's Law will not be violated!

I checked the current drawn by the circuit using my multimeter this time which I din't do the last time. I found out that the reading of my ammeter display was incorrect .
Can anyone help me in calibrating the voltmeter ammeter display ? I have enclosed the image.
I tried adjusting the variable resistor at the back of the display, but this did not resolve the problem .

 

[Ylli]

This looks like a DROK unit. Have you got a model number on it? or at least tell us the voltage and current ratings for it.
https://www.droking.com/cs/2015/08/11/drok-dc-voltmeter-ammeter-dual-display/

 

[Karthik rajagopal]

This looks like a DROK unit. Have you got a model number on it? or at least tell us the voltage and current ratings for it.
https://www.droking.com/cs/2015/08/11/drok-dc-voltmeter-ammeter-dual-display/

 

[Ylli]

"Measuring current: 10 A (direct measurement, built in shunt)."

Why are you using an external 0.5 ohm resistor?

 

[Karthik rajagopal]

"Measuring current: 10 A (direct measurement, built in shunt)."

Why are you using an external 0.5 ohm resistor?

I was actually trying out the over current protection circuit which I found in the internet .
The 0.5 ohms was a part of this circuit.

 

[Ylli]

Calibration will require comparing the units readout with that of a know calibrated meter. That reference meter normally should be 10x more accurate than the unit you are trying to calibrate. You may not need it to be calibrated that closely.

 

[davenn]

The 0.5 ohms was a part of this circuit.

WHICH part

 

[Ylli]

It's R9.

 

[Bluejets]

I found out that the reading of my ammeter display was incorrect .

R9 is shown as 0 but there would I imagine be some notes or whatever that are not shown to give the required size.

So I gather it is now working the way it is supposed to...??

Don't know why you are fiddling with external ammeters when you originally said you are looking for a voltage of 0.5V,
so..........measure that with your multimeter.

According to the circuit shown there is no load adjustment just whatever flows through P1 whatever that is. I'm thinking there is more or alterations to this circuit we are not being told about.

https://www.droking.com/cs/2015/08/11/drok-dc-voltmeter-ammeter-dual-display/

Don't know about others but for me this just brought up a page of googlede gok ( rubbish) sorry meant googelde gook

Looked at what appears to be the same V/A meter the Op showed above. This one was on Ebay and according to them, ................
NO current shunt included......

 

[Ylli]

I think the '.5' is obliterated by the R5 markings. In any case, R9 is the only reasonable place for it.

The link I posted was for the Drok units, but the OP's is apparently an eHUB. Looks the same....

 

[Karthik rajagopal]

R9 is shown as 0 but there would I imagine be some notes or whatever that are not shown to give the required size.

So I gather it is now working the way it is supposed to...??

Don't know why you are fiddling with external ammeters when you originally said you are looking for a voltage of 0.5V,
so..........measure that with your multimeter.

According to the circuit shown there is no load adjustment just whatever flows through P1 whatever that is. I'm thinking there is more or alterations to this circuit we are not being told about.



Don't know about others but for me this just brought up a page of googlede gok ( rubbish) sorry meant googelde gook

Looked at what appears to be the same V/A meter the Op showed above. This one was on Ebay and according to them, ................
NO current shunt included......

It's actually 0.1 ohms which is hid by the R5 marking.
The ammeter is a part of my power supply to which I connected the above over current protection circuit in series. The output voltage of my power supply comes out via the resistor and my ammeter which are connected in series. Full circuit explanation :

 

[Bluejets]

So now the 0.5R becomes 0.1R and to top that, all is working because you read your meter incorrectly..??

I checked the current drawn by the circuit using my multimeter this time which I din't do the last time. I found out that the reading of my ammeter display was incorrect .

I think you need to start again and tell us what your problem is.
At the moment it appears to be in the panel volt/ammeter you bought from your local eHub ..??
Can you supply a link?

Did you read this bit....???

Looked at what appears to be the same V/A meter the Op showed above. This one was on Ebay and according to them, ................
NO current shunt included......

 

[Karthik rajagopal]

So now the 0.5R becomes 0.1R and to top that, all is working because you read your meter incorrectly..??



I think you need to start again and tell us what your problem is.
At the moment it appears to be in the panel volt/ammeter you bought from your local eHub ..??
Can you supply a link?

Did you read this bit....???

Sorry for the confusion . The circuit that I sent uses a 0.1 ohms resistor which was replaced by me as 0.5 as I didn't get a 0.1 ohm resistor . My issue about the voltage drop across the resistor got resolved and it was due to my faulty ammeter display which I had bought online from amazon .
So, I tried calibrating my meter display using the variable resistor present on the backside of the board(which says ADJ_I) which did not fix the problem .
More over my voltmeter ammeter display has a built in shunt (mentioned in the picture that I had sent above ).
I need help in calibrating my meter.
If any kind of information needed plz ask me as I am not sure what information will exactly be needed by you to resolve my problem .
Thanks

 

[Bluejets]

You pretty much answered your own question right at the start.

As per the calculations , the resistor must provide a voltage drop of 0.5v when 1A of current flows through it

So as this is true, the other part of the provided calculation is also true.(E=IR and W=IsquaredR or V*A)
i.e. a 0.5R resistor set up with 0.5v drop across it, must have 1A of current flowing through it. Not difficult to do.

Voltmeters are fairly accurate, even a cheap one so the calibration could be done quite easily yourself.
If 0.5R resistor makes calculation difficult, then use a 0.1R or a 1R or 10R, just modify the equation to suit.

Cannot comment on your Amazon meter as there would be very little detail available for it I would imagine.
Perhaps it is faulty. Try a new one , they're cheap.

 

[Karthik rajagopal]

You pretty much answered your own question right at the start.


So as this is true, the other part of the provided calculation is also true.(E=IR and W=IsquaredR or V*A)
i.e. a 0.5R resistor set up with 0.5v drop across it, must have 1A of current flowing through it. Not difficult to do.

Voltmeters are fairly accurate, even a cheap one so the calibration could be done quite easily yourself.
If 0.5R resistor makes calculation difficult, then use a 0.1R or a 1R or 10R, just modify the equation to suit.

Cannot comment on your Amazon meter as there would be very little detail available for it I would imagine.
Perhaps it is faulty. Try a new one , they're cheap.

Thank you so much for helping me out.