# UV light sensor op amp explanation

Discussion in 'Sensors and Actuators' started by aris, Jan 21, 2014.

1. ### aris

2
0
Jan 21, 2014
Hello,

My programming background is pretty good but I am new to electronics. I decided to build a micro-processor based UV light integrator to measure the amount of UV light cast at a point over a period of time. This is for screen printing.

I have narrowed the sensor choice down to 2 and the datasheets on both have a circuit example. These devices generate a very low voltage in the order of 100mV or so. I understand that I will have to scale their voltage to the 0-5V range so I can feed it to the microprocessor analog input for ADC.

Would someone be kind enough to to help me understand the two different circuits? I know that the left one is a non-inverting arrangement and the right one is an inverting arrangement.

For example:
- Is there a class to better describe each circuit?
- I do not understand the use of the capacitor on both.
- why is the photodiode connected this way on each circuit?

Thanks in advance for any replies!

2. ### KrisBlueNZSadly passed away in 2015

8,393
1,267
Nov 28, 2011
Hi there and welcome to Electronics Point

I should start by warning you that I have no experience with UV photodetectors.

I've just done some quick research on photodiodes in general.

Photodiodes generate a current, called photocurrent, in response to incident light. They can be used in two modes: photovoltaic, and photoconductive.

Both of those circuits operate the diode in photovoltaic mode. In this mode, the diode generates the photocurrent itself, rather than allowing current to flow from a voltage source. This gives good accuracy and linearity, and eliminates the dark current because there is no (or very little) voltage across the photodiode.

In the diagram at top left, for the UUVI-01 device, the photodiode's cathode is connected to the 0V rail. Photocurrent is converted into a positive voltage by 1 megohm resistor R1. This voltage is proportional to the amount of incident light.

This voltage is buffered by the op-amp, shown in the dotted box marked OPTIONAL, which is connected as a voltage follower. The op-amp is needed because the photocurrent is very small, the voltage across R1 is also small, and the circuit has a very high impedance. This means that the signal at the point marked OUTPUT is very susceptible to interference, and capacitance.

C1 at 0.1 µF will have a huge effect on the circuit's response. Even though it's a fairly small capacitance, the current in that part of the circuit is tiny, and the delay caused by the capacitor could be in the order of seconds!

The second circuit, at top right, for the GUVA-S10GD device, is the recommended one. The anode of the photodiode is connected to the 0V rail, and the cathode is fed into an inverting transresistance amplifier implemented with an op-amp.

A transresistance amplifier converts a small current at its input into a larger voltage at its output.

In this circuit, photocurrent from the diode's cathode is negative. It tries to pull the op-amp's inverting ("-") input negative, and the op-amp responds by making its output positive, causing an equal opposing current to flow through R, the 100 megohm feedback resistor. This keeps the op-amp's inputs at the same voltage. This is the essential behaviour of an op-amp with negative feedback.

Therefore the photocurrent is converted into a proportional, positive voltage at the op-amp's output. The circuit is similar to the inverting amplifier circuit shown below it, but Rin is not present, because the input signal from the photodiode is a current, not a voltage. Rin is used when the input is a voltage; it converts the input voltage into a current. It is not needed when the input signal is a current.

The output voltage can be calculated from the photocurrent using Ohm's Law as applied to R. V = I R, where
V is the (positive) output voltage, in volts;
I is the photocurrent from the photodiode, in amps;
R is the resistance of R, in ohms, i.e. 1e8.

For example, a photocurrent of 10 nA will produce an output voltage of +1V.

This circuit has capacitor C, 1 pF, in parallel with the feedback resistor. This capacitor is needed because of the relatively high capacitance of the photodiode, which can make the circuit oscillate. The capacitor does not affect the calculations for conversion of photocurrent to output voltage.

The op-amp used MUST have extremely high input impedance. It should be a JFET-input op-amp such as an LF411A. These generally have large input offset voltages, and the offset voltage may need to be trimmed. Many single op-amps can be trimmed in this way; see the data sheets for details.

The op-amp circuit must be located as close to the photodiode as possible, to minimise capacitance (which slows response, and can cause instability) and noise pick-up. The circuit should be constructed to minimise capacitance.

You may need to enclose the whole assembly in a tin can, connected to 0V, for shielding, with the photodiode behind a hole or protruding through a hole.

Best quality components should be used. Use film or mica capacitors, not ceramic! Use good quality metal film resistors.

Both of these circuits should be powered from split supplies. Older op-amps perform best with higher supply voltages, e.g. ±15V, but newer types may be designed to operate at lower voltages. See the data sheets. The negative rail doesn't need to be as high a voltage as the positive rail; -5V is enough. Decoupling capacitors must be located near the op-amp, with short connections to the pins.

If a wider output voltage swing is needed, use a second amplifier. Trying to change the gain of the transimpedance amplifier will affect its response and could cause instability.

You will need to clip the voltage going into the ADC so it doesn't exceed the device's supply rails.

You can find more information in Linear Technology's Application Note #399. This application note recommends their LTC6244HV dual op-amp. Or use Google.

Last edited: Jan 21, 2014
3. ### aris

2
0
Jan 21, 2014
First of all, Kris, thank you for the fantastic reply.

You have explained a lot of things that were not at all obvious to me and you have supplied me with the starting points for a good long study/research couple of weeks. It was getting quite hard to go beyond the basic op amp material without this.

I can see that your advice to use high quality components is especially relevant. The tiny currents these photo diodes give out and the amount of amplification that would have to be applied has also had me concerned about noise and the associated complications in software. Not having measuring/monitoring tools to actually assess the signal they give out makes it harder.

The reason I started asking about all this was because the GUVA-S10GD is available in a couple of modules, the UVM-30A is one of them and it outputs 0-1V which maps to UV index 0-10 linearly. But the UVI-01, which has better detection characteristics for my application, does not come in a module so I would need to make one.

The conclusion is that I will get the module, start the programming and at the same time work on correctly amplifying the UVI-01. Maybe I will get lucky enough to have everything working well enough to simply switch sensors when I am ready.

Again, thank you for your advice. If I am successful with the integrator, I promise to post the info and code online.

I used to live in Australia but now I live in Northern France where the UV is not quite as strong

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Relevant info: the Arduino has a mode where the reference for the analog input will read 0-1.1V in the range instead of the default 0-5V. This is switched using analogReference() and supplying INTERNAL as the parameter (INTERNAL1V1 on the Mega). With this, I can use the module which outputs 0-1V and have the signal sampled at almost full resolution. Keep this in mind: "After changing the analog reference, the first few readings from analogRead() may not be accurate."

Last edited: Jan 21, 2014