I'm planning to build an Pyranometer amp based on the AD620 In-Amp, at this point I would ask your knowledge on the following details:
a) Apart from using a 1.5v dry battery with a (huge?) voltage divider, there is another way to simulate the microvolt/milivolt signal level for the INA input? (I want to test the device by comparing the input Vs the voltage output using a Oscilloscope and/or millivoltmeter).
b) Should I use a Low-Pass filter (cut-off freq around 40 Hz or less, to leave out the 50/60 Hz AC Hum) after the INA-amp or between the Pyranometer and the INA-amp?
Active pyranometers use AFAIK a transconductance amplifier, which effectively short circuits the solar cell.
There is NO voltage across the cell in this case.
Any solar generated current is compensated for by the transconductance amplifier, and that compensation is measured by an A/D.
Passive pyranometers use a resistor to "shortcircuit" the cell, and the resulting very low voltage is measured with an opamp. This is AFAIK a compromise (no electronics in the pyranometer).
Leo..
Wawa:
Active pyranometers use AFAIK a transconductance amplifier, which effectively short circuits the solar cell.
There is NO voltage across the cell in this case.
Any solar generated current is compensated for by the transconductance amplifier, and that compensation is measured by an A/D.
Not my case.
Wawa:
Passive pyranometers use a resistor to "shortcircuit" the cell, and the resulting very low voltage is measured with an opamp. This is AFAIK a compromise (no electronics in the pyranometer).
Leo..
According to the manufacturer, the pyranometer will output about 70mV at about 1000W/m2.
They don't mention nothing about using a resistor to "shortcircuit" the cell..
yv1hx:
They don't mention nothing about using a resistor to "shortcircuit" the cell..
A passive pyranometer is nothing more than a $1 solar cell with a $0.10 burden resistor across
(resistor is inside the instrument).
The price you pay is for the design of the case/filters/etc. and calibration.
The datasheet mentions a 1ohm output impedance (the burden resistor?),
so you could use a 1ohm resistor in the voltage divider to simulate that.
Maybe better to use a regulated supply instead of a battery.
An 68ohm:1ohm divider, connected to a 5volt phone charger, outputs 1/(68+1)*5= ~72.64mV.
Leo..