How can I measure this voltage and current?

[baileychic]

How can I measure this voltage and current?

It is 400V peak-to-peak @ 1 Hz and it is connected to an R load. Current in R varies between 100uA and 1.5A and I need to measure both voltage and current.

What sensor can I use? I will be using 24-bit ADC to measure the current.

I will be using a PIC16F877A.

 

[KlausST]

Hi,

What sensor?
* a shunt (resistor) for current measurement
* a voltage divider (two resistors) for voltage measurement
This is how I'd do it.

24 bits is overkill.

Klaus

 

[baileychic]

Can I drop the voltage to say 5Vp-p range using a voltage divider and then use a diode and a capacitor to charge the capacitor to the peak value of scaled signal which will be 5V - diode drop. Let's say it is 4.3V, then 4.3V gives a voltage reading of 400Vp-p using ADC of PIC.

How about current measurement? How can I use a shunt resistor (parallel) resistor across R load?

Should I use low-value R say 0.01R in series with the load and measure the scaled voltage across the 0.01R (scaled using another set of a voltage divider, diode, and cap)?

I need to measure like 100uA, 101uA, 108uA...

1uA resolution.

If I scale the current (1.5A) to 5V range then what resolution ADC I have to use to get 1uA resolution?

For voltage measurement is it better if I first convert the 400Vp-p voltage to say 200Vp signal using a diode and a 200uF 300V capacitor and then use a voltage divider to get 5V DC for ADC input?

 

[KlausST]

Hi,

I don't recommend to use analog peak measurement method.
* it's higher hardware effort
* it's not precise
* it's slow

But without knowing waveform and expected precision and calculation method (RMS, peak, peak-to-peak, rectified average....)

Klaus

 

[baileychic]

It is 400Vp-p 1 Hz square wave.

 

[KlausST]

Hi,

Maybe a dumb question: If you know it's 400Vpp ... why do you want to measure it?

Do you expect 50% duty cycle?

Think about this:
400Vpp square wave, assumed 50% duty cycle, can be:
* -200V to +200V --> 200V RMS, 0V average, 200V full wave rectified average
* 0V to 400V --> 282V RMS, 200V average, 200V full wave rectified average
* 1000V to 1400V --> 1217V RMS, 1200V average, 1200V rectified average

So if you really want to measure Vpp then you should do so. Measure maximum, measure minimum and subtract both values.

Klaus

 

[baileychic]

Why should I measure minimum? If I use a diode to convert 400Vp-p signal to 200Vp signal (200 - 0.7V) and use a 200uF 300V capacitor and charge it then it will charge to 200 - 0.7V and if I scale this voltage to 5V DC for ADC then 5V represents 200Vp or 400Vp-p.

Why should I measure -200V when it is a perfect square wave?

Now, tell me how to measure the 100uA to 1.5A current with 1uA resolution.

 

[albbg]

I think you should evaluate to complicate the circuit and use two linear optocouplers (one for the voltage the other one for the current) in order to avoid the ADC (and possibly other components) to be destroyed in case of measurment resistors failure. Or, at least you should use some transzorb to protect the ADC inputs.

Furthermore be careful to the power dissipation of the shunt (in series with the load) since it can be quite high (up to 5V*1.5A=7.5W).
If you convert 1.5A in 5V (even if should be better to leave some margin to account for eventual spikes), at the minimum current flow the voltage to measure will be 333 uV (and an equivalent load resistence of 4 Mohm) that means you could have to remove some noise to have a precise measurement.
About the number of bits, you need (1.5A-100uA)/1uA=1499900 voltage levels that is log2(1499900)= 21 bits that means 24 bit are enough for the resolution you specified.

I think the best method is to do all the calculations via software excluding some noise removal that should be done in hardware

 

[baileychic]

You mean I have to use Linear optocouplers like IL300 between scaled 5V DC and ADC input?

 

[KlausST]

Hi,

Why should I measure minimum?

Because you said you want to measure peak-to-peak.
The one "peak" is the maximum, the other "peak" is the minimum.

Why should I measure -200V when it is a perfect square wave?

Is it a perfect square wave? Only you know.

About IL300 and 24 bit ADC:
IL300 has a -44% .... +55% inital accuracy (transfer rate)
IL300 has a +/-0.25% transfer gain linearity this is 0.5% in total. Or a 1:200, which means less than 8 bits linearity.
IL300 is specified with a tiny measurement range of 1:10 (1mA ... 10mA).
Use any cheap 10 bit ADC .... it has better accuracy and precision.

I really recommend to do the specifications and calculations on your own.. with realistic values.

Klaus

 

[FvM]

If you want 22 to 24 bit accurate current measurement (1.5 A range, 1uA resolution is about this order of magnitude), you'll surely won't use IL300. I would neither use it for an 1% accurate voltage measurement, but it's less critical. If the input is known AC and isolation a requirement, a current transformer can work. Otherwise shunt with isolated ADC or a recent AD/DA isolator like AMC1200.

You may also want to use automated range switching with less ADC resolution.

You are apparently assuming a resistive load, in this case you get higher accuracy by measuring 400 Hz fundamental instead of square wave magnitude. If the load isn't purely resistive, you might get a problem with square wave. But I don't know your specification.

 

[baileychic]

The load is resistive only but I will get another confirmation about it from the client.