I'm having a little trouble with an isolation amplifier. I'm using an HCPL-7800 isolation amplifier, in the circuit format provided by the manufacturer as a test circuit, and feeding it low voltage DC (0-200mV input). The gain of the amplifier should be flat at about 8.029 nominal, but at low input voltage (1mV) I'm seeing a gain of 25, and at my max input voltage (180mV) I'm seeing a gain of 7.9 (linear drop). I've almost exhausted every resource on this. I've laid out the board according to the manufacturer's recommendations, used the proper soldering temperatures, input resistance, etc. I've even been in correspondence with Avago, who claims it's properly set up, but they themselves have no idea why it isn't working correctly, and their demo board is a near copy of my circuit (different resistors, different input range). They have however given up on this issue and stopped responding. Does anyone have any insight as to what would cause this phenomenon? Thanks All
]1
Amplifier – Solving Isolation Amplifier Gain Problems
amplifiergainisolation
Related Solutions
I don't know where you got the numbers from. Your circuit should give you (1+51.1/11.9) * 2V + 2V out with 4V applied, which is 12.59V out, and 2V out with 2V applied. You can easily run the simulation in Circuitlab if you want (flip your power supplies first, they are reversed).
You want a gain of 5 (2V change = 10V on the output) so the resistor ratio must be 4:1. Say you use 40K and 10K.
For offset to 0 for 2V in, note that the gain is -4 (as compared to +5) from the inverting input so you need to apply 5/4*2V = 2.5V to the 10K resistor, as shown below:
simulate this circuit – Schematic created using CircuitLab
As a general rule of thumb, AC mains voltage should always be isolated by a transformer or some type of isolation device. There should also be some way to limit both current (fuse) and voltage (TVS or MOV) on each line.
Depending on your plans for this application and where you intend to sell it, you will likely have to go through some type of UL or CE certification. Take a look at IEC 61010. That document covers many of the basic tests required.
When making decisions about required isolation, you have to consider the environment(lab bench, outside, can it get wet?), the enclosure, clearance distances, and what happens when things go wrong at a minimum. Say a resistor or diode in the design fails short, does your design start a fire? If something did catch fire, could the fire spread out of the box? Could someone be shocked by touching the box if a wire came loose? These are things you have to think about before trying to measure mains voltage. They also determine the amount of isolation required.
To answer the original question which was about cost. I've tried both approaches. Both work but I usually prefer digital isolation. In older products, isolation was usually done in analog space. The front end was typically a differential op amp with several large resistors in series with the inputs with protection diodes followed by an ISO124 for galvanic isolation. An ISO124 is not a cheap part ($19.85) but it is tried and true. It does have limitations such as bandwidth and offset voltage that have to considered. It is also bipolar. Many MCU AtoD's cannot handle any voltage below 0V. Since you are measuring an AC signal, you may have to use a seperate AtoD.
The really cheap isolation op amps are usually intended for measuring current across a shunt and have a small input range. Although you can divide down the input signal using precision resistors, signal to noise ratio will be impacted. Also gain may vary from channel to channel. This may or may not be an issue for your application. It depends on the accuracy you are trying to achieve. The small input range has usually been an issue for me so I haven't tried this approach although it could work.
I've also tried digital isolation. Digital isolators are plentiful and cheap. I've personally used digital isolators for CAN bus, I2C, and SPI. Typical front end is usually some type of filter circuit followed by an opamp, and then AtoD. The AtoD is isolated from the MCU using the digital isolator. Since the AtoD is upstream of the isolator, the quantization noise of the isolator does not have to be considered. This produces a more accurate measurement with a wider bandwidth.
One final point. Safety always trumps cost and should be your primary concern.
Best Answer
This isolation amplifier is really designed for a very specific application- measurement of motor phase current on a VFD using a shunt. That's why it has a very low full scale voltage of +/-200mV (negating the need for an op-amp in that application), and why you need a ~55:1 attenuator for your application.
You may be able to use it for your ~+/-10V application but you will need to account for the input offset voltage and input offset voltage drift. The input-referred (to your divider) effect of the +/-3mV maximum offset voltage over temperature is about +/-170mV, so less than 2%, which is adequate for the intended application.
You can trim off the offset voltage at the output since the part is quite linear, however you cannot easily compensate for the drift with temperature, which is maximum +/-10uV/°C or about +/-550uV/°C at the input to your attenuator. So if you null it at 25°C and the temperature changes to 50°C that could represent as much as about 14mV change at the input to the attenuator.
That's still pretty good, and this is a pretty inexpensive isolation amplifier (due to the high volume application) so this is not a bad way to go if it meets your requirements, but each one would have to be individually nulled. This can be done digitally if the amplifier output is digitized by an MCU or with a trimpot with a suitable reference voltage and circuit. There are obvious advantages and disadvantages each way (for example, data stored in EEPROM has to be written there and it could be lost at some point in the life of the product).
Remember gain is delta-Vout/delta-Vin so you can't get accurate gain measurements by assuming Vout(0) = 0. It will be as much as +/-24mV differential voltage at the output.