Difference Between Isolated Amplifier vs Linear Optocoupler?

Thread Starter

johnyradio

Joined Oct 26, 2012
434
I understand that an isolated amplifier uses capacitors for isolation, and a linear optocoupler uses an LED and photodiodes.
But, assuming unity gain, why would i choose one over the other?

I'm guessing the main difference is that the isolated amplifier offers some gain/buffering.

Here's an example of an isolated amplifier:
TI AMC1311x High-Impedance, 2-V Input, Reinforced Isolated Amplifiers
https://www.ti.com/lit/ds/symlink/amc1311.pdf

Here's a linear optocoupler:
Ixys/LittelFuse LOC110 Linear Optocoupler
https://www.littelfuse.com/media?re...ttelfuse-integrated-circuits-loc110-datasheet

Personally, I'm interested in DC applications for the iso, but AC is also interesting.
 
Last edited:

Thread Starter

johnyradio

Joined Oct 26, 2012
434
I found an answer on SE. Key points:
  • Isolation amps are good for measuring accuracy over a wide dynamic range, with at least 1,500 volts isolation, so you can measure line voltage at 600 VAC with no problem, and measure the current as well.
  • Optocouplers are faster and cheaper.

Copied in full:

Isolation amplifiers can have a high cost. Analog Devices AD202KN is $15, good to 400 HZ at best, and the AD210KN which samples at 500 KHZ is about $30 each. Compare that to the $1.00 USD for many opto-couplers.

Also the isolation amplifiers have a 'conversion' time, as they are transformer coupled. Some INA series use capacitive coupling but the delay to integrate the signal back to analog is still there.

If you need a low cost and very fast responding isolated loop a fast opto-coupler cannot be beat. They can be a thousand times faster then a isolation amp, because they do not have to convert the signal to PWM, then intergrate the signal to convert it back to analog.

The response time of a fast opto-coupler is in the nS range, while a isolation amp may take several uS to 100 uS.

Isolation amps are good for measuring accuracy over a wide dynamic range, with at least 1,500 volts isolation, so you can measure line voltage at 600 VAC with no problem, and measure the current as well.

What isolation amps are NOT good for is servo-loops. They greatly increase the time it takes to correct for a change in voltage, current or load. It can make the circuit 'ring' as it settles down or worse yet the circuit may oscillate because the feedback is too slow.

Also isolation amps often need a modest low-pass filter at the output to remove any conversion noise, well up in the 100 KHZ to 500 KHZ range. Some need buffers as they may not have a substantial drive current. The AD202KN and AD210KN can only source or sink about 1 mA.

https://electronics.stackexchange.com/a/365401/65001
 

schmitt trigger

Joined Jul 12, 2010
921
I have used both. The optocoupler solution requires a pair of additional opamps, NOT from the same package, and some ancillary components. The quoted optocoupler speed is very dependent of said opamps slew rate, if you plan on isolating +/- signals.
 

Ian0

Joined Aug 7, 2020
10,035
Don't forget the third option - especially if you are measuring mains voltage - a transformer: Needs no circuitry (hence no power supply) on the measuring side and is cheap. Only disadvantage is the amount of space it requires.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
Don't forget the third option - especially if you are measuring mains voltage - a transformer: Needs no circuitry (hence no power supply) on the measuring side and is cheap. Only disadvantage is the amount of space it requires.
but works only for AC, not DC
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
Simple opto's with just one photodetector are more subject to LED non-linearity, aging, and temperature drift.

Devices with two photodetectors correct for all that. They come in diode output and transistor output flavors. The LOC110 in my question has diode outs. As @schmitt trigger mentioned, require some additional external circuitry.

1704382565847.png 1704382597868.png

They run about $2, not counting the extra circuitry. Does anyone know a cheaper way to get high-linearity, for a very low-current iso? My application is secondary feedback on an isolated DC-DC converter. It's for a battery charger, so i think about 5 mV linearity over 5 volts full scale is acceptable. Not sure how to convert that to % (is it 5 mV/5, or 5 mV/1?).
 

Ian0

Joined Aug 7, 2020
10,035
? My application is secondary feedback on an isolated DC-DC converter. It's for a battery charger, so i think about 5 mV linearity over 5 volts full scale is acceptable. Not sure how to convert that to % (is it 5 mV/5, or 5 mV/1?).
so why don’t you do it like the rest of the world does, by comparing the output with the reference on the isolated side, then transferring the error signal across the isolation?
if you start introducing delays into the feedback loop, you will make it oscillate.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
so why don’t you do it like the rest of the world does, by comparing the output with the reference on the isolated side, then transferring the error signal across the isolation?
if you start introducing delays into the feedback loop, you will make it oscillate.
How to transfer the error signal across the isolation?
 

Ian0

Joined Aug 7, 2020
10,035
The FOD2742 is an optical error amplifier. This is what you mean, correct?
https://www.mouser.com/datasheet/2/308/1/FOD2742B_D-2313634.pdf

View attachment 311706

How can i know it's linearity? Is it related to CTR?

How about drift at higher temps? Datasheet states
temperature coefficient of 50 ppm/C
Yes, that will do, but it is nothing other than a combination of a fairly ordinary optoisolator and a TL 431.
It isn’t particularly linear, but had it ever bothered you how linear the open loop gain of an op-amp might be?
 

Ian0

Joined Aug 7, 2020
10,035
Why do you mention the open loop gain of an op-amp?
because the circuit is identical to an op amp. The TL 431 input is its inverting input, the non inverting input is connected to a 2.5V reference, and the output is the output of the opto.
Any ideas about calculating linearity or drift at higher temps?
look At the TL431 datasheet.
You might need to know the speed of the opto to calculate the phase shift for the feedback.
 

Ian0

Joined Aug 7, 2020
10,035
Take the standard op-amp gain equation:
gain = A/(1+AB)
where A=open loop gain and B=feedback factor
How much difference does the actual value of A make provided that it is large?
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
Take the standard op-amp gain equation:
gain = A/(1+AB)
where A=open loop gain and B=feedback factor
How much difference does the actual value of A make provided that it is large?
This went over my head.
Open loop gain is gain of the op amp without any feedback, correct?
Gain is calculated based on the formula you gave.
How do we calculate linearity from that?
Unclear what you're implying about linearity.
 

Ian0

Joined Aug 7, 2020
10,035
This went over my head.
Open loop gain is gain of the op amp without any feedback, correct?
Gain is calculated based on the formula you gave.
How do we calculate linearity from that?
Unclear what you're implying about linearity.
The non linearity is about the same as the difference between gain calculated by the normal method (R1/R2) and the gain calculated from A/(1+AB) where B=R2/R1

If you want 10V output and have a 2.5V reference then the gains is 4, so B=0.25
If you assume that A=100000, then the actual gain is 3.9998.
Bearing in mind that you might be using 1% tolerance resistors at the best, the real gain due to resistor error will vary from 3.92 to 4.08.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
434
Is there such as thing as secondary side regulation based on a simple binary feedback, ie "you're too high" and "you're too low", without indicating by how much? It seems that would be less vulnerable to imprecision in the feedback, and would only require an analog comparator, rather than error amplifier.
 

Ian0

Joined Aug 7, 2020
10,035
Is there such as thing as secondary side regulation based on a simple binary feedback, ie "you're too high" and "you're too low", without indicating by how much? It seems that would be less vulnerable to imprecision in the feedback, and would only require an analog comparator, rather than error amplifier.
Yes. Some of the Power Integrations Link Switch devices do it that way. It does tend to lead to more ripple on the output, but in some cases that doesn't matter.
 
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