Current meter

[timelessbeing]

Hi everyone. I bought an inexpensive 20A hall effect sensor board on eBay, and then decided to build a simple meter out of it. I breadboarded the circuit, and it appears to work. What do you experienced guys think about it? Is the design "sound"?

The voltage regulator runs a bit hot (12V input), but I measured 60mA on the output which is within spec. It has automatic thermal shutdown anyway.

I chose such low resistor values to overcome the current flowing through the voltmeter (which only has 20kohm). I considered using an opamp to buffer that end. It would be more elegant, but it added more components and complexity, and 50mA seemed okay to me so I decided bugger it. The 1/4W resistors are only dissipating 1/10W. I could still raise the value of those resistors a bit without sacrificing too much accuracy. (I only wanted 5%) Thoughts?

If I wanted to read the AC signal with a scope, I could simply probe the V(o) since it has much higher impedance right?

Thanks for reading.

 

[Reloadron]

Your current sensor part number threw me a little, is this sensor what you actually have? Your drawing shows an AC712 but I believe what you have is an Allegro Micro Systems ACS712T in the -20 to 20 Amp range. Should that be the case with 0 Amps current the output will be about 2.5 Volts. The sensitivity of the sensor is about 100mV/Amp so a range of -20 Amps to 20 Amps is .5 to 4.5 Volts. Would that be what you have?

Ron

 

[timelessbeing]

Yes, ACS712T. Thanks for catching my typo.

 

[ronsimpson]

Your 47/47 ohm resistors are causing some of your heat.
These two resistors look like 24 ohms to 2.5 volts. It will cause a 0.1% low reading.
If you used 470/470 it would look like 240 ohms to 2.5V and with a 20k meter it will cause a 1% low reading.
Your meter is not 1% now.
I was just thinking of changing the 47 to something higher.
The current sensor can drive the meter with out the op-amp.

 

[timelessbeing]

These two resistors look like 24 ohms to 2.5 volts.

I'm not sure what you mean by this. Can you please explain?

The datasheet says that the sensor needs at least 4.7kohm of output resistance, so the meter alone would overload it. I can't use a 4.7kohm resistor divider, because the meter current would affect it. That is why added the opamp buffer.

0.1mA of meter current, through 470R, adds 47mV of offset. That will show up as roughly 1/2 an amp of error. On a 20A scale, that's 2.5% error I think.

Do you think the heat is a problem? What about 100R ... that would reduce the current through the stack to 25mA, or 63mW per resistor. Good compromise?

 

[ronsimpson]

These two resistors look like 24 ohms to 2.5 volts.

A resistor divider using two 47 ohm resistor like you drew looks just like one 23.5 ohm resistor going to a 2.5Volt source.

The datasheet says that the sensor needs at least 4.7kohm of output resistance

So your 20kohm meter is just fine. The load needs to be above 4.7k and I think your meter is 20k.

You say your meter is 20k ohms and 0.1mA. So that takes 2 volts.

I think the censor outputs 2.5V at 0 current. The other end of the meter is at 1/2 of 5 volts which is also 2.5V. So at 0 current the meter will have no voltage (no current).
At 20A the censor will output 0.5V and the other side the meter is 2.5V so the meter sees 2 volts.

 

[KeepItSimpleStupid]

Also, flip your caps on the regulator. Put the 10 uf at the input. Don't use V- unless it's a negative voltage. Use COM or an inverted triangle with a number in it.

 

[timelessbeing]

Oh FFS. I can't believe I missed that. The sensor will drive the meter just fine. Major brain malfunction there.

A resistor divider using two 47 ohm resistor like you drew looks just like one 23.5 ohm resistor going to a 2.5Volt source.

I'm not familiar with this analysis technique. How does it work, and what is the significance?

I think the censor outputs 2.5V at 0 current.... At 20A the censor will output 0.5V and the other side the meter is 2.5V so the meter sees 2 volts.

The sensor's sensitivity is 100mV/A. At +20A, it will output 2.0 + 2.5 = 4.5V. The other side of the meter has 2.5V, so the potential difference is 2.0V which is full scale. That was the design intention.

Also, flip your caps on the regulator. Put the 10 uf at the input.

Can you explain why? I copied the application circuit from the data sheet exactly.

Don't use V- unless it's a negative voltage. Use COM or an inverted triangle with a number in it.

The voltage @ V- could be positive or negative. I used '-' to indicate that it must be a lower potential than V+.

 

[KeepItSimpleStupid]

Well, looks like your right. I seem to remember the other way around. Thanks. I guess this applies to the capacitors that are near the regulator and not the bulk DC caps.

 

[ronsimpson]

A resistor divider using two 47 ohm resistor like you drew looks just like one 23.5 ohm resistor going to a 2.5Volt source.

I'm not familiar with this analysis technique. How does it work, and what is the significance?

Compare, a voltage divider (1/2) from 5V using 47 ohm resistors AND a resistor of 23.5 ohm to 2.5 volt source.
With no load both circuits measure 2.5V.
Short the output to 5V and/or 0V.
>The current from the second circuit is 2.5V/23.5ohms.
>The current for the first circuit comes from two resistors.
>>One resistor will have 0 volts so its current is 0V/47 ohms=0
>>The other resistor will have 5V/47 ohm.

2.5/23.5 = 5/47

So both circuits (under those conditions) look the same.

I am trying to show what error the resistors introduce. And how much power loss the 47 ohm resistors cause. Heat in the voltage regulator is almost an issue.

 

[Reloadron]

I think what you should consider is going to the data sheet and look at the applications with a focus on application #3. Note what they are doing with the pair of 100K resistors. Your sensor has an output span of 0.5 to 4.5 Volts for an input current of -20 to 20 Amps. The sensitivity is right about 100 mV / Amp. With 0.0 Amps the Vout is right around 2.5 Volts. The -20 to 20 Amps actually represents a 40 Amp span.

You mentioned earlier that your current drain was about 60 mA. The majority of that (about 53 mA) is a result of the 47 ohm resistors in there. The idea being to get the common side of your meter to about 2.5 Volts. I see what you are doing but question the method over what they do in application #3 of the data sheet.

While the sensor will sense AC or DC current all of this only pertains to DC current. Take a look at Application #2 in the data sheet. A good example of detecting the AC peak current.

I agree with Ron in that this is not a good way to go about things and will induce errors you want to avoid.

Ron

 

[timelessbeing]

I like their use of 100 ohm resistors, but changing the gain will render my meter useless.

 

[ronsimpson]

I like their use of 100 ohm resistors, but changing the gain will render my meter useless.

What resistor? The resistors that make the 2.5V?
With the 100//100 ohms that add 50 ohms to the meter and the meter at 20k the error is 20,050/20,000 or very little.
'useless' ??? Mechanical meters are not known for accurate readings. Add the errors of the censor and the 100 (50) ohm error is a tiny part of the total error. If you replace the mechanical meter with a DVM that is 1,000,000 ohms then 1,000,050 is not far away.

 

[timelessbeing]

I meant that the meter I bought would be useless in terms scale, not error.

I matched the sensor module, circuit, and meter so that it would be easy to read. 1.0V=10A, 2.0V=20A etc.
If I changed the gain, not only would I need a meter that reads different voltages, but I won't get the nice 1:0.1 proportion which allows me to use the existing meter 'face'. I wanted this to be simple; It's supposed to be a quick hack, not a huge project.

AC is not part of the intent. With AC, RMS values are more useful anyway, and I don't even want to go there. I will mostly use this for testing automotive applications, but if I really want to see some AC, I can just stick some scope or DMM probes directly on the sensor output, right?

 

[KeepItSimpleStupid]

If you bought a 0-2V meter, then the 20K is probably 20,000 ohms/volt which basically means that the impeadance seen by whatever your measuring depends on voltage. Measure 2V, it sees 40K. There will be a resistor inside the meter unless, of course, you bought a surplus CURRENT meter with that scale with some value of current as full scale: e.g. 50 uA

Now, what is usually done is a current meter is used and you take a variable resistor and voltage source in series and adjust for 1/2 scale. Measure the variabe resistor and you get the meter's resistance. Now, you can calculate the proper series resistance (with trimmer) to set the full scale voltage.

 

[timelessbeing]

the impeadance seen by whatever your measuring depends on voltage

It's a galvanometer (see drawing).

 

[ChrisP58]

Since your current sensor can drive the meter directly, you don't need the opamp on that side of the meter. But I would still use it, but move it to other side.

Use a pair of matching high value (10K to 50K) resistors to make your 2.5V point, but feed that to the + input of the op-amp. That way you have the very low current through the resistor, but you have the near zero ohm output of the opamp driving the meter.

I would further suggest putting a trim pot between the two resistors and feeding the opamp with the wiper. That will give you a way of trimming the meter to show zero at zero current.

 

[timelessbeing]

That sounds really good. I will do that next. Thanks!