current measuring question

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aliyesami

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i thought i posted this but not seeing my post so excuse me if i am posting twice.
I am trying to measure a current consumption of my circuit and i am putting the voltmeter in series on the positive terminal of the power supply and getting a 1.4mA rating.
The issue is that the circuit is using a display and the display manufacturer says that it should draw 40mA with backlit on.
on full backlit power its only drawing 1.4Amp

am i measuring it wrong ?
 

Hi,

We don't see the circuit, so all we can do is guess.

Maybe you are measuring at a 12V line and your circuit uses a SMPS to step down voltage.
The backlight usually needs 2...3V.

And/or then back light is not fully ON.

Klaus
 


Probably. Kindly conform that you have completely isolated positive terminal of power supply from the circuit and only and and only one path exist from the positive terminal of your power supply to your circuit that must be your ammeter.
Also make sure you have selected a proper display range in your ammeter.
Also check the calibration of your ammeter against a known/calibrated resistance and power supply.
 


probabably there is a typo or you have written what was carried out.

a "voltmeter is series with a positive terminal"...

1.4mA and then 1.4 Amp...


you can rephrase your question if it was so.
 

A current meter in series with a circuit has a voltage drop that probably reduces the current draw of the circuit. The voltage drop caused by a current meter is called The Burden Voltage Drop in its spec's.
Maybe your circuit uses a power supply voltage of only 3V and your current meter causes the supply voltage to drop to 2V then the current in the circuit is much less than normal.
 

The question you did not answer was; Was the display backlight visibly fully on, when you measured the current?
If not, it's the problem Audioguru describe.

The current shunt in the meter has a voltage drop. Most backlights I've seen has a Vf between 4 and 5V, normally around 4V5. If you reduce the voltage below this, the LEDs will not go on and draw current at all.
I know, since I've been fooled by this myself.
 

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