Hi Folks,

I have a voltage divider to measure my PV panels. The max voltage on the array would be around 450V so I've worked with 500V.
The voltage divider will, at 500, give me 3.4V which is max for my MCU. All is well.

Because of the high voltage, I have opted for a HCPL-8740 which is an opto-isolated op-amp. Datasheet

I have wired it to what I thought would have worked but have some weirdness in it. It is done (loosely) to the example circuits they have in the datasheet.

Seeing as my divider does not need any amplification, need a gain of 1:1. All I am looking for is a linear following of the input but with the isolation.


When I build this up on breadboard with a 5k multi turn pot as the divider, odd things happen. Well, odd to me.
When the input is at zero, the output is at 2.8V as the input rises past 10mV the output starts to climb and eventually saturates at about 4.5. The input though is still less than 1V. I was expecting a 1:1 here.

The connector labelled DC-DC convertor is just that, the programme I am using does not have it so I make-do It's an isolated 5v in 5V out.
VCC for the whole thing is 5V, coming in on pin 3 on CN5 (sorry, did not mark it)

The "fuse" and connectors around it are a hall effect current sense and have no connection to the opamp.

I know it is my misunderstanding of this but I don't know why - I thought it would be simple

thanks,
Crispin

 

This part was designed for measuring the voltage drop across a low-value current sense resistor. There are isolation amps available that are more suited to your application.
Having said that:
The output of the HCPL-7840 is differential. The outputs have a common mode voltage of about 2.5V nominal. You have to follow this amplifier with a differential amplifier, as shown in the applications. The gain of the HCPL-7840 is 8, ±5%. See DC Electrical Specifications on p.8 of the datasheet for both of these specs.
Input current is relatively high, so your 10k divider impedance might be a problem. You might be able to minimize the problem by adding 10k in series with pin 2.
I would look for an amplifier which is more suitable for your needs.

 

Thanks.

I did not realise the second op amp in the example was needed (I know I know) but thought it was to further increase the gain.

Any tips on what might be better placed for the application?

I had not noticed the input current would be high - I assumed rolleyes that all op amps were high impedance. It seems overkill for me to put something else in front of it?
All I want is an isolated but linear output (failed on a normal opto isolator as well)

 

Why do you think you need an isolation amplifier? I would think that a simple voltage divider would allow you to connect directly to your MCU.
The divider you show on your schematic will give you 5.05V out with 500V in.