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Topic: Photo Diode - Amplifier (Read 3496 times) previous topic - next topic

Hi,

I want to build a Geiger counter which measures gamma detection with the help of photo diodes.I have found a circuit which I have attached to this question.

I have a few questions:
Can I add multiple photo diodes to it and if yes how? Can it simply put them in a line?
Do I use K1 to interface with the Arduino and do I need a voltage divider?

I am sorry for this "dumb" questions but I am new and just try to learn, so it would be very nice if you could help me. :-)

King regards,

Greenality

The image I have found is out of the pdf which I have found at this location:
www.ecosia.org/url?url=http%253A%2F%2Fxa.yimg.com%2Fkq%2Fgroups%2F10603698%2F1658129756%2Fname%2FElektor-rivelatore%2Bgamma%2Bpin%2Bdiode.pdf&v=0&i=1&q=Measure+Gamma+Rays+with+a+Photodiode+Radiation+detector+using+a+BPW34&p=0

jremington

PIN photodiodes will not work if connected in series, and they may not work well if connected in parallel. Get the circuit working with a single photodiode before you try to make modifications. It is not straightforward to interface that circuit with an Arduino, because you will need a pulse amplifier, like the pulse stretching circuit shown in the article you linked.

68tjs

#2
Feb 04, 2014, 06:16 pm Last Edit: Feb 04, 2014, 06:18 pm by 68tjs Reason: 1
Optical PIN diode are equivalent to a current source. They can not be connected in series.
Voltage sources adds in series, current sources adds paralleled.

But I don't understand  for what reason  "they may not work well if connected in parallel".
The  limitation I see is the minimum reverse voltage per diode, for a PIN diode -3V is the minimum, -5V is optimum and bandpass can decrease if to many diodes are parralleled (to many capacity).
As a result you may be forced to increase Vcc of the amplifier and verify that the output never exceeds the Vcc arduino ie 5V.

-> PIN diode can work below -3V but it is diode model/sample dependent.


I agree when you write "Get the circuit working with a single photodiode before you try to make modifications". This is the safest method to obtain a first result.

pito

#3
Feb 04, 2014, 06:39 pm Last Edit: Feb 04, 2014, 06:50 pm by pito Reason: 1
You may wire more diodes in parallel, maybe an increasing the voltage will help to eliminate the higher capacitance (when diodes in parallel).
I would not connect arduino input directly - you need another circuit to create larger pulses (as the pulses you get are 0.1V peaks or less). You may use the second circuit with LM311 comparator, powered from 5V, and without the T2 transistor - wire the output from R6 directly to arduino - and count the pulses.
PS: I can confirm it works with diode, I did it long time back with other diode (SHF-206) and with similar amplifier (Jfet frontend). Mind the number of pulses coming from the diode is _very_ small.. You cannot compare it with geiger tube.. :smiley-slim:


jremington

Quote
But I don't understand  for what reason  "they may not work well if connected in parallel".
As pito indirectly pointed out, when photodiodes are connected in parallel, the capacitance of the inactive diode will reduce the amplitude of the pulse signal generated by the active diode.


I would not connect arduino input directly - you need another circuit to create larger pulses (as the pulses you get are 0.1V peaks or less). You may use the second circuit with LM311 comparator, powered from 5V, and without the T2 transistor - wire the output from R6 directly to arduino - and count the pulses.


Hi,

at first i want to thank you all for your answers which were quiet helpful.

And is it possible that you (pito) may make a circuit diagram or something for me, a quick handwrite would be enough for me, to make me happy ;-). It would be very nice if you or someone could do that for me because I am one of the more visual types. :-)

King regards,

Greenality

68tjs


the capacitance of the inactive diode will reduce the amplitude of the pulse signal generated by the active diode.

What matters is not if a diode is active or not but the reverse bias of the diodes is sufficient.
As the bandwidth is sufficient there is no reduction in amplitude, that is why I said minimum -3V and -5V optimum for reverse polarization.
Since it is very difficult to measure the bandwidth of a transimpedance without expensive hardware, advice I can give is to increase the number of diodes one by one and I think that on this point we agree.

jremington

Quote
As the bandwidth is sufficient there is no reduction in amplitude,

That is simply nonsense.

pito

#8
Feb 05, 2014, 12:28 am Last Edit: Feb 05, 2014, 12:42 am by pito Reason: 1
Quote
And is it possible that you (pito) may make a circuit diagram or something for me

Both circuits are depicted in the pdf above.. Just connect the LM311 one to the output of the diode amplifier, and do power the LM311 with 5V instead of 9V  (power it from arduino's 5V). Connect output of the LM311 (let R6 there) to an arduino input. Read the pdf article.
Use a 10turn potentiometer as the P2 - the comparator trigger level to set will be difficult otherwise. The "issue" with this kind of detector is the amplitude of the pulses from the amplifier differs based on the particle's energy. So proper setting of the P2 will be the key here..

68tjs


Quote
As the bandwidth is sufficient there is no reduction in amplitude,

That is simply nonsense.

OK for the nonsence.
If the input signal is DC, ie  its frequency is zero, capacity has no effect and the input signal is amplified in the same way, whatever the value of the capacity.
If the frequency of  input signal is infinite, capacity is a short-cut for the input signal and Ve = 0 -> Vs =0.

If you don't think  it is a bandwidth problem what is it ?

pito

#10
Feb 05, 2014, 01:23 pm Last Edit: Feb 05, 2014, 01:36 pm by pito Reason: 1
Look at this - this is your amplifier simulation, where the C3 (simulating the input diode capacitance) is swept from 1pF-30pF, step 3pF. The impulse from the current source is 50nA large (simulating the detection of a particle), and from 1.000us to 1.020us (20ns duration).
With high impedance sources the input capacitance plays a role, indeed..

pito

and the output pulse, input capacitance swept as above.

pito

#12
Feb 05, 2014, 01:34 pm Last Edit: Feb 05, 2014, 01:42 pm by pito Reason: 1
and the output pulse again, all settings as above..
It shows the output pulse with ~30pF diode capacitance is an half of the amplitude with diode capacitance of ~1pF..

Thanks to all of you, especially pito. :-)

King regards,

Greenality

pito

My simulation shows the LM311 is not fast enough to shape a 20uS long pulses from the diode amplifier. Here is a simulation with a modification of the comparator I did with an output to arduino. You have to use a fast comparator (high speed one).
The decoupling capacitors not shown for clarity.

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