# 100uA analogue meter conversion to 1A?

#### neazoi

##### Advanced Member level 5
I have a 100uA analogue (needle) meter and I want to convert it to 1A so I can read currents to 1A. When the current is 1A it should read 100uA, when it is 0.5A it should read 50uA and so on (linear conversion).

How can I do it without posing great resistance to the DC, so as not to loose voltage fed to the series circuit?

(DC can be anything from 1.2v to 20v and the device draws 500-800mA)

#### BradtheRad

##### Super Moderator
Staff member
Normally you install a shunt resistor across the meter. The shunt carries most of the current. Only a tiny amount goes through the meter (for yours it should be 100 uA max).

A few inches of wire could be all you need. Make secure (or soldered) connections to the meter terminals because the meter coils are low ohms. If it carries full current it burns up.

neazoi

### neazoi

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#### BigBoss

##### Advanced Member level 5
A sufficiently small resistor can be a solution with a OPAMP ( low offset value ). If you design the OPAMP circuit carefully ( Gain ) you can scale the current very precisely.

#### Easy peasy

##### Advanced Member level 5
If we assume your 100uA meter is 100 ohm R, then this gives 10mV to drive it to 100uA and full scale

therefore to measure 1A, we need a shunt resistor of 10mV / 1A = 10 milli-ohm ( 100, 1 ohm resistors in parallel for example )

Cu wire is OK but as it heats its R goes up and you lose accuracy, you are 30% out as you go from 25C to 100C in the wire

Say you buy a nice stable 10 milli-ohm shunt resistor, the current to be measured goes thru this, then take 2 smaller wires from the ends of the shunt and take them to the meter ( right way round ) now the shunt will drive 10mV at 1A and the meter will read very close to true ( actually the external ckt needs to provide 1A + 100uA = 1.0001 amp ) - this is an error of 0.01% - so we can ignore it.

good luck ...

neazoi

### neazoi

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#### neazoi

##### Advanced Member level 5
If we assume your 100uA meter is 100 ohm R,
How do I measure it? Is it enough to measure it with the multimeter? I worry because the multimeter may cause huge deflection as you measure the resistance and may burn the precious meter?

#### FvM

##### Super Moderator
Staff member
How do I measure it?
With an instrument or circuit that is guaranteed not to exceed 100 uA. Might be a multimeter with manual range select, or a combination of voltage source, series resistor and mV meter.

Regarding shunt temperature coefficient, consider that the meter has also copper coil and respective copper T.C. A thick copper shunt that doesn't heat up and tracks the meter temperature will probably give best overall accuracy.

neazoi

### neazoi

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#### neazoi

##### Advanced Member level 5
With an instrument or circuit that is guaranteed not to exceed 100 uA. Might be a multimeter with manual range select, or a combination of voltage source, series resistor and mV meter.

Regarding shunt temperature coefficient, consider that the meter has also copper coil and respective copper T.C. A thick copper shunt that doesn't heat up and tracks the meter temperature will probably give best overall accuracy.
Which scale should I select in the manual meter?

#### c_mitra

##### Advanced Member level 5
Which scale should I select in the manual meter?
Use a DVM- they are cheap, easily available and reasonably accurate. If that is autoranging, you need not worry, otherwise select the 1K (most will say 2K, because 1.999 display) range.

Still simpler way to get the resistance:

set up wit a 1.5V cell, a couple of resistors and a 1K linear pot.

Estimate a current of 100uA and see the meter showing a deflection of 100uA.

Now put a pot across the meter and adj so that the reading becomes half. Remove the pot and measure the resistance. That will be equal to the meter resistance.

neazoi

### neazoi

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#### betwixt

##### Super Moderator
Staff member
And don't forget you can use a combination of shunt and series resistances. A shunt to bypass most of the current and a resistor in series with the meter to limit how much of the remainder you measure.

It really depends on how much voltage you can afford to lose in the shunt. If for example it is in the input of a voltage regulator, you can probably afford to use a much higher value because the regulator will keep the output voltage steady anyway. If it is after a regulator or in say a battery feed, you probably can't afford the voltage drop so the value has to be as low as possible.

Brian.

#### neazoi

##### Advanced Member level 5
And don't forget you can use a combination of shunt and series resistances. A shunt to bypass most of the current and a resistor in series with the meter to limit how much of the remainder you measure.

It really depends on how much voltage you can afford to lose in the shunt. If for example it is in the input of a voltage regulator, you can probably afford to use a much higher value because the regulator will keep the output voltage steady anyway. If it is after a regulator or in say a battery feed, you probably can't afford the voltage drop so the value has to be as low as possible.

Brian.
It is there to measure the current a small transmitter consumes. The tuning of the transmitter depends on the durrent, which has to be set somewhere between 500-700mA (not critical).

I measured my meter and it has 1360 Ohms resistance (it deflects to about 60uA when measured on the 2K scale of my multimeter).
If I calculate it right it needs a shunt of 0.136 ohm.

I can connect in parallel three thick film 5W 0.33R for a total of 15W 0.11R
Alternatively I can use one heatsinked 10W 0.1R
Alternatively I have a ceramic 5W 0.15R

Which way should I go better and what drop will I have for say 600mA?

#### Easy peasy

##### Advanced Member level 5
how to measure: 1.5V cell or power supply, 10k pot wired as rheostat, put in series with the meter, wind rheostat down to min while observing meter, when at FSD, measure volts across coil, R = V / 100uA

As to copper shunt - this will not help with tempCo of observed current, both will go up with temp, but the very small watts in the coil will not heat the coil as the shunt will be heated from the main current, hence small error related to coil current.

#### betwixt

##### Super Moderator
Staff member
Easy Peasy is quite right but unless you are using very thin copper wire, a current of 1A will not cause much self heating.

I would use two 0.5Ohm resistors in parallel (or four 1 Ohm ones) and add a 1K variable resistor in series with the meter. Use another meter to measure the current and adjust the pot to calibrate the 100uA one. With 0.25 Ohms in series with the supply you would only lose 0.25V at 1A load.

Brian.

#### neazoi

##### Advanced Member level 5
And don't forget you can use a combination of shunt and series resistances. A shunt to bypass most of the current and a resistor in series with the meter to limit how much of the remainder you measure.

It really depends on how much voltage you can afford to lose in the shunt. If for example it is in the input of a voltage regulator, you can probably afford to use a much higher value because the regulator will keep the output voltage steady anyway. If it is after a regulator or in say a battery feed, you probably can't afford the voltage drop so the value has to be as low as possible.

Brian.
Hm... I think this is better. No measurements, nothing to worry.
Use whatever power resistor you have (the lowest value the better for lower voltage drop) and then calibrate with the series rheostat to the current drawn by the lab power supply.
The scale should be linear, i.e. if you measure 1A for 100ua, you should measure 0.5A for 50uA.

Am I getting this right?

#### betwixt

##### Super Moderator
Staff member
Yes, as simple as this:
Code:
Feed ---------> shunt resistor ---------> load
----> pot ---> 100ua Meter ----
As long as the voltage dropped across the shunt is high enough to make the meter reach FSD, you can trim the calibration with the pot. It should be linear.

Brian.

#### c_mitra

##### Advanced Member level 5
If I calculate it right it needs a shunt of 0.136 ohm.
Assume that the meter coil is 1360 Ohms. Just use a 620-680 Ohm in series. Now the meter coil is about 2k.

Now use two 0.1 Ohm resistor in series. They will be 0.2 Ohm - put this 0.2 shunt in parallel with the meter (including the 620-680 resistor).

You have now a meter that reads 0-1A. At 1A it will drop about 0.2V but you can put it in the input side of the regulator.

#### dick_freebird

##### Advanced Member level 5
Depending on your plan, your accuracy needs and other
externalities, maybe one of the Hall sensors (like LEM)
and a scaling resistor could be a way to go. These want
power in addition to the sensed current though.

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