sensing a small voltage change and converting to an on/off signal

[NeX]

hi everyone, sorry to bother you all with this idea but it seems to have gone beyond my knowledge.

i have a light, this light is on all the time, but it pulses very briefly where it gets a bit brighter. i am using a light dependent resistor to try and detect these pulses and turn them into an on/off signal that i can supply to my micro processor.

now what i have tried is to use a transistor, but that didn't work because the light is on all the time, there was a small voltage getting through which caused the transistor to be partly on. i tried putting a resistor between the LDR and the transistor but that stopped the transistor from working at all (it was a veriable resistor so i could test all resistances)

then i tried a reed relay, the LDR kept the reed relay on all the time, so again i used a resistor to bring it down until it was just off, but this didn't work either, the change in voltage is just too small.

if i run the LDR straight to an LED, it is possible to see the pulse, so it is making a difference, but it seems to be so small that i cant detect it with anything and turn it into a square wave.

i tried a darlington pair for the transistor, but the transistor was still semi on.

i am sure this should be a simple fix, but i am at a loss. any input would be really appricated, thanks everyone!

 

[ericgibbs]

hi everyone, sorry to bother you all with this idea but it seems to have gone beyond my knowledge.

i have a light, this light is on all the time, but it pulses very briefly where it gets a bit brighter. i am using a light dependent resistor to try and detect these pulses and turn them into an on/off signal that i can supply to my micro processor.

now what i have tried is to use a transistor, but that didn't work because the light is on all the time, there was a small voltage getting through which caused the transistor to be partly on. i tried putting a resistor between the LDR and the transistor but that stopped the transistor from working at all (it was a veriable resistor so i could test all resistances)

then i tried a reed relay, the LDR kept the reed relay on all the time, so again i used a resistor to bring it down until it was just off, but this didn't work either, the change in voltage is just too small.

if i run the LDR straight to an LED, it is possible to see the pulse, so it is making a difference, but it seems to be so small that i cant detect it with anything and turn it into a square wave.

i tried a darlington pair for the transistor, but the transistor was still semi on.

i am sure this should be a simple fix, but i am at a loss. any input would be really appricated, thanks everyone!

hi,
Try a capacitor coupling from the LDR and its load resistor to the base of the transistor, you will have to add a biasing resistor the base.
By using a capacitor it will be AC coupled and the DC due to the steady light will be blocked.

 

[NeX]

hi,
Try a capacitor coupling from the LDR and its load resistor to the base of the transistor, you will have to add a biasing resistor the base.
By using a capacitor it will be AC coupled and the DC due to the steady light will be blocked.

interesting, i can give it a try but will the capacitor's charge and discharge timing change the timing that the light flashes? as its the timing i am trying to detect and i need it to be as perfect as possible.

thanks for your help!

 

[MikeMl]

Is the lamp AC or DC powered? Can you insert a low value resistance (few Ohms) in series with the lamp? That will make the lamp slightly dimmer overall, but you should be able to capacitively-couple the step change during the time it gets brighter.

 

[NeX]

i suppose i should straight out say this, the lamp is a display, i can't control its brightness or interface with it (that i know of) other than using the LDR,

it would be great if i could dismantle the display and take a trigger for the micro processor straight from one of the transistors in the display, but i really don't think that is possible, and it would still have the same problem of being on, and more on.

 

[NeX]

thanks, the capacitor idea worked..... kinda, as it is my test LED is on constantly, but as soon as i touch the negative leg of the capacitor it starts flashing exactly how i want, if, whilst holding the negative leg i touch any other part of the circuit the light goes out completly.

does this mean that i have the capacitor the wrong way round? (the negative leg of the capacitor is fed by the 5v from the LDR and the positive leg is going to the transistor base)

or does it mean that the capacitor is not big/small enough? (electrolitical 100uf 25v cap)

thanks everyone for your help, this is so close to working and then i can move on to the micro proccessor programing

 

[ericgibbs]

thanks, the capacitor idea worked..... kinda, as it is my test LED is on constantly, but as soon as i touch the negative leg of the capacitor it starts flashing exactly how i want, if, whilst holding the negative leg i touch any other part of the circuit the light goes out completly.

does this mean that i have the capacitor the wrong way round? (the negative leg of the capacitor is fed by the 5v from the LDR and the positive leg is going to the transistor base)

or does it mean that the capacitor is not big/small enough? (electrolitical 100uf 25v cap)

thanks everyone for your help, this is so close to working and then i can move on to the micro proccessor programing

hi,
Post a diagram showing how you have the LDR and transistor connected .

 

[NeX]

here is a diagram, sorry for the quality it was made in MS Paint,

 

[canadaelk]

Sorry to say this: It won't work.
- The base of the transistor is floating
- The pot is floating too. That means the left side of the electrolytic is allways at the + rail, regardless of the restance of the LDR
- The LDR response is propably to slow and should be replaced with a semiconductor device
- The LED is in parallel to the relay. From what I understand, from your earlier postings, the relay is not necessary. Also, the coil should be bridged with a "snubber" circuit (diode or capacitor), to prevent the back-EMF when the relay disengages to destroy the transistor
I would go with a comparator IC, very controllable and straight foreward
E

 

[NeX]

do you mean that the circuit i have drawn wont work? because it is working, but only when i have my hand on the negative leg of the capacitor. what is there in the way of semiconductors that detect light changes?

could you explain your post as i don't quite understand it, do you mean that i should ground the other side of the pot?

thanks

 

[ericgibbs]

do you mean that the circuit i have drawn wont work? because it is working, but only when i have my hand on the negative leg of the capacitor. what is there in the way of semiconductors that detect light changes?

could you explain your post as i don't quite understand it, do you mean that i should ground the other side of the pot?

thanks

Hi,
Look at this sim image and your circuit with mods.
It works OK, providing that the light pick up drives the LDR hard enough.!

 

[NeX]

Hi,
Look at this sim image and your circuit with mods.
It works OK, providing that the light pick up drives the LDR hard enough.!

ok cool, but is that second transistor a different kind?

grounding the other side of the pot worked, but now that i have neaten it up on a board its stopped working again. its probably my wiring so i will check it again

 

[NeX]

ok i tried it and everything works perfectly thanks!

but what i don't understand is why. i am new to electronics and everything i know is self taught so there is a lot of holes in my knowledge, i know the theory, like what a capacitor does and what a transistor does etc,

but i don't understand why the input to the transistor needed a resistor to 5v and a resistor to gnd, why cant it be "floating"?

 

[12345678901234567890]

Hi,

Looking at your schematic with the input floating, your signal in essence is coming through the LDR, through the capacitor into the base to drive the transistor.

However the capacitor may not be discharging as was mentioned earlier by another poster, so the capacitor continues to rise positive in voltage for every input.
The signal must always be applied between the base AND emitter of a transistor amp. in CE mode.

Your signal was not grounded, but just adding positive potential to the capacitor at every LDR change in resistance.

The circuit that was posted by the poster who had base resistors and the pot grounded, was demonstrating how to bias the transistor amp stage at just cutoff so a required small change of the LDR was picked up as a base to emitter signal input to this amp, to cause the transistor to conduct as a switch to be able to be processed by the transistor, for the output.

I hope this helps some..
Someone else could explain it better.

 

[NeX]

i think i understand what you mean, i was trying to use the transistor in a sort of school boy way. i am used to being able to use things in series, eg, battery to switch, switch to light, light to battery.... all those components are passive and can sit inline. but the transistor, and the capacitor can't be used like that because of the way they work, so they need drain points, and balancing is that right?

i think i get it, and i think i know what i would have to do next time so thats good thanks for taking the time to explain

 

[12345678901234567890]

See if this helps explain a little bit of how his circuit input works, to detect a change in the LDR's resistance.

This tutorial is not using a capacitor as original circuit, but just showing the concept of what the LDR signal is acting like to the input transistor.

R4 represents the LDR as it changes in resistance due to light pulse changes.

Hope this makes it more clear to understand.

His input transistor is being used as a voltage controlled switch.

 

[NeX]

wow, thanks very much for that, i think i get it now, but there is still so much i don't know about electronics!