# Minimum analogue arduino voltage signal

Discussion in 'Embedded Systems and Microcontrollers' started by Rickpercy87, Nov 26, 2015.

1. ### Rickpercy87 Thread Starter New Member

Nov 20, 2015
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0
Hello Im using a microwave emitter for my project. The signals being sent to the arduino will be below 200hz when they are being sent to the arduino.

My question is that, will my amplifier be strong enough? I'm not sure how to apply the maths correctly. Sensor emits micro volts, I'm getting about 40 milli volts on the output. Will the arduino be able to work with 40mV?

Also looking at the schematic I assumed the output of the amplifier would be about 5V P/P. Due to the voltage divider. If anyone can have a look and clarify it to me I'll be ver6 greatful.

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
You are not making sense.

https://www.arduino.cc/en/Reference/AnalogRead

Are you using 5 volt Arduino or 3.3 volt Arduino?

If I recall right, Arduino standard is 10 bit (0-1023), but it can also be setup for 12 bit (0-4096).

Just reading the article I linked, the 5 volt Arduino in 10 bit mode will do 4.9 mV steps, so receiving 40 mV is not a problem.

3. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,968
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Each stage of the linked amp has a gain of about 100 (assuming '1m' denotes 1 MegOhm!). Given an input signal of, say, 5uV you should therefore expect an output of about 100 x 100 x 5uV = 50mV.
The datasheet should tell you the signal level its ADC can handle.

4. ### Rickpercy87 Thread Starter New Member

Nov 20, 2015
16
0
Cheers, that's just what I'm looking for with the 4.9mV. I'm still really new to arduino, I'm probably in over my head Haha.

How did you work out the gain is 100 per stage? I'm aware of how to work it out with transistors but not in this circuit with op amps. I thought the 1M was mainly for determining the operating frequency along with the 2.2nF cap. How is it that you guys know these things but my college tutors don't?

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
ASSUMING!!!!!!!!!!!!!!!!!!!!!!!!

The schematic in datasheet shows 1m resistor on each op amp. Assuming those are 1 MegaOhm resistors.

Then.

The first op amp has input resistor of 10 kOhm. Calculating gain:
Gain = Feedback resistor / Input resistor = 1,000,000 Ohm/10,000 Ohm= 100

The second op amp has input resistor of 8.2 kOhm. Calculating gain:
Gain = Feedback resistor/ Input resistor = 1,000,000 Ohm/8,200 Ohm= 121.9 call it 122

In conclusion. If the assumptions are correct, then you have gain on first op amp of 100 and gain on second op amp of 122.

6. ### Rickpercy87 Thread Starter New Member

Nov 20, 2015
16
0
Yeah that's what I thought at first, but wondered why not just use another 10k instead of the 8.2k. They are 1 megaohm, it gives a gain of about 12,200, then? Wow.

I hope it works, I imagine there will be all kinds of distortion with a gain so big.

Thanks for clearing things up for me