Phototransistor switching freqeuncy

dl324

Joined Mar 30, 2015
16,846
What is the pulse amplitude driving the LED? What does the signal look like on the cathode of the LED?

Operating the photo transistor at a higher voltage will increase it's sensitivity and drive capability.
 

Thread Starter

Alasttt

Joined May 13, 2015
68
910k is too much and signal impendence is so high everything falls to noise.
Yes hence why I had 5.6k to reduce this, Ive used 33k and its better now but still way off. Also the wave is the opposite of what it should be.33k.PNG
and the output is only 400mv peak to peak.
 

GopherT

Joined Nov 23, 2012
8,009
Change the current limiting resistor to 100 to 330 ohms (something way less than 1.5k). That assumes you are powering it from 5V.
 

Thread Starter

Alasttt

Joined May 13, 2015
68
We will flip the signal once you get a strong one. 33k is fine for now. Now change the resistor on the LED
I have changed the resistor on the LED it has made an improvement to the peak to peak voltage. I dont know why but I get a good output from the emitter. Even though it is connected to ground. I think it may be a pnp or npn thing.

output from collectorCollector 33k.PNG
output from emitteremitter 33k.PNG
 

Thread Starter

Alasttt

Joined May 13, 2015
68
Take a photo of your setup and post it.
95228e60-6b68-43de-8f90-e34708602bdf.jpg
The left is the infrared emitter, with a resistor to ground. 100 ohms. The other resistor going into it is to clean up the input square wave, and is connected to the other pin.
The phototransistor left pin is the flat edge (emitter), it goes to bottom rail which is 0. The other pin goes to 5v via the 33k. The output is from the emitter in this pic but I have moved it to the collector,
 

GopherT

Joined Nov 23, 2012
8,009
View attachment 98305
The left is the infrared emitter, with a resistor to ground. 100 ohms. The other resistor going into it is to clean up the input square wave, and is connected to the other pin.
The phototransistor left pin is the flat edge (emitter), it goes to bottom rail which is 0. The other pin goes to 5v via the 33k. The output is from the emitter in this pic but I have moved it to the collector,
Now one from the top with better light. And the whole board.
 

dannyf

Joined Sep 13, 2015
2,197
at this frequency the output of the phototransistor is nothing like a square wave.
Typical reaction time is in the sub-us territory so 4Khz shouldn't be a problem - the remote controls for example run at ~10x of that.

Your issues are with something else in your circuit.
 

dl324

Joined Mar 30, 2015
16,846
Right sorry could you explain what that means ?. My power supply is two crocodile clips one to positive and one to negative, how do I decouple this ?
Put two caps across the power supply; 0.1uF for high frequency decoupling, and an electrolytic for low frequencies.
 

GopherT

Joined Nov 23, 2012
8,009
To flip your signal. You can go back to something like we started with. It was just easier to get you on the same reference as everyone else (me) is used to seeing an NPN in a switching circuit.

image.jpg
 

Thread Starter

Alasttt

Joined May 13, 2015
68
Ok, now aim the two devices (IR LED and the phototransistor) right at each other. Like you want to shoot from your gun down the barrel of your enemy's gun. The remote control for your tV works very poorly if you stand on the side of your TV. Point right at it!
I have been pointing them at each other, even though in the pic they look upright !,point.PNG
Thats what im getting. Two things, it needs to be cleaned to make it look more square, and it also needs to be flipped.Data.PNG

Thats the input.
 
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