PIN Photodiodes - Please shed some light....

Thread Starter

Dan_xd45

Joined Oct 13, 2014
5
I am tinkering around with try to simplify a design for a ballistic chronograph that was developed by David Collins in the Nuts and Volts Magazine in June of 2009. Basically, I am using two frames, where each frame consists of (3) IR LEDs (QEE123) and (3) PIN Photodiodes (QSE773). I am trying to develop the entire project using a 5V DC power supply. So, I reviewed the data sheet and noticed that the peak emission for the LEDs is at 20mA and am therefore using (3) parallel branches of a 150 ohm resistor in series with the LED to yield roughly 24mA.

My problem lies in understanding how the the receiver circuit should be set up. In the article, he uses 4 of these photodiodes in parallel, reverse-biased where all the cathodes are run to Vcc and all the anodes are joined and used as the input into a series of op-amps.

The theory behind this (as I understand it - please correct me if I'm wrong) is that under normal circumstances, the emitter is active and will emit IR signal to photodiodes, allowing current to pass through. However, we are looking for when a projectile crosses through causing a small, quick disruption and want to somehow identify and amplify ONLY the voltage drop to produce a TTL signal that will be used to trigger a timer (one frame to start and one to stop). However, with these photodiodes in parallel, doesn't this pose an issue for inconsistent values depending on where the bullet crosses (left to right)?

Any help would be greatly appreciated. I'm attempting to attach Collin's emitter, receiver and amplification circuits in a .jpeg

Collins.jpg
 

MrCarlos

Joined Jan 2, 2010
400
Hello Dan_xd45

There are some things to understand in that design.
You have 4 IR LED’s (QEE123) connected in series.
the electrical characteristics of these IR LED’s are:
VF = 1.7 Volts.
IF = 20 mAmp.
This means that when the IR LED is crossed by a IF current, at its terminals will be a VF voltage drop.

Since you have 4 IR LED’s connected in series, and also a resistance of 150 Ohms.
When this circuit the IF current flows there will be a voltage drop at the ends of:
(VF x 4 = 6.8V) + (IF x 150 = 3V) = 9.8 Volts Total.
Therefore the U3 (LM317 IC) must be able to provide enough voltage to that circuit that the IF current flow.
Unless they are not polarized to give light-fully.
Note that the IC U3 is connected as a constant current source.

Now the PIN photo-diodes: (QSE773).
Probably they are reverse biased. Something to read:
http://en.wikipedia.org/wiki/PIN_diode
But it could also be that well work in your circuit. Reverse biased.
Study document E02 handbook SI Photodiode.PDF. attached.
An IR LED must be exactly in front of an IR sensor.
I have not seen the magazine by David Collins. But I guess He describes the dimensions to which they belong, (separation), each group of emitter-sensor.
Nearby a group of other, probably the circuit does not respond.
Very remote one group from another. probably the meter, connected to the called terminal Pulse X is saturated.

The distance between each transmitter-receiver group is critical in this circuit
Have You some data about ?

This is a case to study more consciously.
Sorry, the file I try to Upload is so big.
 

Thread Starter

Dan_xd45

Joined Oct 13, 2014
5
Thanks for all the responses. I will try to be a little more clear. My circuit is a little different than the one shown above. I have my IR LEDs in parallel in lieu of series because I am trying to get by with a 5V power source and the forward voltage on the 3 IR LEDs exceeded my source voltage. I have a picture of one of my frames attached below. This set up yields roughly 25mA to each of the IR LEDs and the photodiodes were placed in line with the IR LEDs. On my receiver circuit, I can't figure out why with the hookup shown in the second picture, I am getting roughly 4.88V when measured with a voltmeter across the open portion of the circuit pictured and can pass my finger through the frame and get a voltage drop (which is good). I just don't understand why I can't get a measured voltage across the resistor when the open portion doesn't exist.
 

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Sensacell

Joined Jun 19, 2012
2,567
Ok, I still don't get it? I see the frame, assuming the bullet passes through the center of this frame, is it supposed to detect a bullet passing anywhere through this frame?
The signal would be vanishingly small and overwhelmed by any modulated ambient light, like most man made light sources.
With all that AC gain, you will only get a nasty 120 Hz mess, unless you run this in total darkness and shield the input wiring very carefully?

Or is the bullet supposed to cross the four sensors in sequence, 1,2,3,4? along the linear axis of the sensors?

This is an interesting project, I want to understand how it works!
 
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