How to protect LED from overvoltage

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
Since this thread has grown more than I expected at the beginning lets make a brief summary of what I need. I need to uniformly (to about 5% accuracy) illuminate a surface of 30x50 cm from a height of 25 cm using some LEDs (8 or 9 LEDs). On these surface will be placed 4 MPPCs that need to be tested and calibrated. I need that only 20 photons to hit the MPPCs at one instant so a very small (but still unknown) current is needed to go through the LEDs. The LEDs will be driven by a signal generator that can output from +5 to -5 V. I only need 8 pulses of light in a row. The pulses are to be spaced by 580ns (so 1.7MHz).
In order of priority I need:
  1. Uniformity of illumination (about %5)
  2. Ability to pulse light in short bursts separated by 580ns (1.7MHz)
  3. Enough simplicity that someone with little knowledge about electronics like me can make (or if complex you have to be willing to help me a little bit)
  4. Protection against overvoltage (but only if it doesn't interfere with the above conditions)
Could someone give me his/her best guess about the best design? I was thinking about 9 LEDs in parallel with some potentiometer in series to individually tune the luminosity.
 

Sensacell

Joined Jun 19, 2012
3,785
Here is one idea.

It's a constant current driver, adjusting the pot should give 0-50 ma current.
The pot is 50 ohms to properly terminate the output of the signal generator and cable.
You want to be sure the impedance matches the cable/generator to prevent distortion of the pulse shape.


This is only one channel, but you could add more by selecting pots with a value that would yield a 50-ohm total input impedance.
For example, 10 channels of 500 ohm pots in parallel will give you 50 ohms.

You need to build this on a ground plane, with short clean wiring. You are building an RF circuit, so you need to wire it like one.
solder-less breadboards and the like are going to give crappy performance.
 

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Alec_t

Joined Sep 17, 2013
15,119
I need that only 20 photons to hit the MPPCs at one instant so a very small (but still unknown) current is needed to go through the LEDs.
What sort of LEDs will you use? Normal white light LEDs incorporate a phosphor, the decay time of which is likely to be orders of magnitude longer than you would want. Even at a LED current of 1 microamp, with a 1 nanosecond pulse, I would guess gazillions of photons would be emitted (I could be wrong :) ).
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
What sort of LEDs will you use? Normal white light LEDs incorporate a phosphor, the decay time of which is likely to be orders of magnitude longer than you would want. Even at a LED current of 1 microamp, with a 1 nanosecond pulse, I would guess gazillions of photons would be emitted (I could be wrong :) ).
The LED I was thinking to use is this: Datasheet
I think it is an ordinary white LED. Do you happen to know a LED better suited given my requirements?

As far as the light yield is concerned, do not worry: generations of particle physicists have done this kind of measure with just a simple LED and resistor. You have to consider that I will use a diffusor slab, that the LEDs are quite far from the MPPCs, that the MPPC active surface is very little (order of squared millimeters), that the quantum efficiency of the MPPC is big compared with a PMT but still not 100%, etc...
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
Here is one idea.

It's a constant current driver, adjusting the pot should give 0-50 ma current.
The pot is 50 ohms to properly terminate the output of the signal generator and cable.
You want to be sure the impedance matches the cable/generator to prevent distortion of the pulse shape.


This is only one channel, but you could add more by selecting pots with a value that would yield a 50-ohm total input impedance.
For example, 10 channels of 500 ohm pots in parallel will give you 50 ohms.

You need to build this on a ground plane, with short clean wiring. You are building an RF circuit, so you need to wire it like one.
solder-less breadboards and the like are going to give crappy performance.
Thank you very much for the circuit design. It is simple enough that with a little effort I think that I can understand it completely!
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
Read post #23.

White LED's use phosphor down-conversion from a blue or UV LED.

These will smear your pulse out in the time domain.

Use a monochrome LED.
Thank you for the hint. I chose white LEDs because I was lazy and I didn't want to put any time in trying to match the MPPC sensitivity spectrum with the LED emission spectrum. But now I think that I should try to do that.
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
@Sensacell
I have tried to simulate your circuit for 1 LED and for 10 LEDs with the QUCS program.
It seems OK, but just to be on the safe side I would be very grateful if you could take a look at these pictures and tell me if I implemented it correctly:
Single LED (Voltage Sweep)
Single LED (Resistance Sweep)
10 LEDs (Voltage Sweep)
10 LEDs (Resistance Sweep)
Schematics
where "Voltage Sweep" means that I have swept the Signal Generator voltage from 0 to 5 volts and "Resistance Sweep" means that I have swept the potentiometer resistance from 0 to 50 or 500 Ohms for the single LED or 10 LEDs cases respectively.

The only thing that still I am a little bit insecure about is that by sweeping the potentiometer resistance over its entire range the current going through one LED is changing only by less than 1%. Is that enough to account for the nonuniformity among all the LEDs?
 

Sensacell

Joined Jun 19, 2012
3,785
Have another look at my schematic diagram.

Your diagram is not correct, the ends of the POT should go to GND and the incoming signal, the "wiper" of the pot taps the divided voltage potential between the ends- this voltage feeds the transistor base.

Doing a voltage sweep works because it emulates what the pot is supposed to be doing, were it connected properly.
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
Have another look at my schematic diagram.

Your diagram is not correct, the ends of the POT should go to GND and the incoming signal, the "wiper" of the pot taps the divided voltage potential between the ends- this voltage feeds the transistor base.

Doing a voltage sweep works because it emulates what the pot is supposed to be doing, were it connected properly.
You were right. I didn't fully understand the role of the potentiometer. Now I think I got it right.
Because in QUCS there is a model for the potentiometer but it seems that it is not included in the standard release, I re-simulated the circuit with Proteus 8.

Would you be so kind as to take a look at these circuits and check them?
Single LED (Voltage Sweep)
10 LEDs (Potentiometer tuning)
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
@Sensacell
Sorry to bother you again ...
Do the values of the resistors need to be modified if I were to use this green LED?
Can I use a 2N4124G instead of the 2M4123? In case do I need to modify something?
(Our provider's shipping time for the 2M4123 is really long)
 
Last edited:

Sensacell

Joined Jun 19, 2012
3,785
Nice, looks good.

The circuit generates a constant current, it doesn't care about different LED Vf. Green will be fine.
The schematics show 100 pf capacitors, way too small.

I would add a 0.1uf capacitor to each stage on the 10 LED driver.
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
Thank you for the hint. By "each stage" do you mean to place a capacitor on each channel as near as possible to the LEDs like this?
Before I edited post #31 but maybe you had already read it so I will ask again:
Can I use a 2N4124G transistor instead of the 2M4123? In case do I need to modify something?
(Our provider (monotaro) shipping times for the 2M4123 are really long and our other provider (RS Japan) doesn't even has it in stock)
 

Sensacell

Joined Jun 19, 2012
3,785
2N4124 should be fine.

Yes - you have the right idea with bypass capacitors now. General rule: there can be too few, but not too many.
They just help reduce the high frequency impedance of your power line, improving performance- place them close to the loads.

Note that you can lower the current range by increasing the value of the 82 ohm resistor.
 

Thread Starter

LordOfThunder

Joined Jun 27, 2018
80
2N4124 should be fine.

Yes - you have the right idea with bypass capacitors now. General rule: there can be too few, but not too many.
They just help reduce the high frequency impedance of your power line, improving performance- place them close to the loads.

Note that you can lower the current range by increasing the value of the 82 ohm resistor.
Thank you very much for everything. Now I can finalize and build the circuit!
 
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