light detection circuit

Thread Starter

Hyperian

Joined Apr 25, 2008
13
I'm using a photodiode in parallel with a resistor in series with a LED to detect when light hits the photodiode.
The photodiode goes from 7kohms (with light) to 5Mohm (without light) and I'm using a resistor of 1.1Mohm.

I am planning on replacing the resistor (since it's big) with a transistor. If i remember my classes right, i should use a NMOS. any advice on the setup?
 

SgtWookie

Joined Jul 17, 2007
22,210
I'm using a photodiode in parallel with a resistor in series with a LED to detect when light hits the photodiode.
The bolded portion really doesn't make sense. A schematic would be invaluable here.
The photodiode goes from 7kohms (with light) to 5Mohm (without light) and I'm using a resistor of 1.1Mohm.
That sounds reasonable.
I am planning on replacing the resistor (since it's big) with a transistor.
Why? You can get 1M resistors in 1/10 W, and even smaller if need be.

If i remember my classes right, i should use a NMOS. any advice on the setup?
Let's start with a schematic on what you have now. If the physical size of your existing resistor is a problem, that is very easily solved. Adding a variable like a transistor would mean that you'd have to add more resistors to control the transistor - so your project would escalate in cost and complexity while plummeting in reliability.
 

Thread Starter

Hyperian

Joined Apr 25, 2008
13
I was trying to avoid having to draw a schematic cause i would have to do it with paint but here it is

Well i can get my hands on a NMOS easier than a specific type of resistor that my school might not provide.

I was thinking that the gate of the NMOS would be connected to a lower value resistor which connects to the power supply to get a specific voltage on the gate to allow for the NMOS to act as a resistor. (i hope i'm right it's been a while)
 

SgtWookie

Joined Jul 17, 2007
22,210
OK, I see the problem.

You need the photodiode and resistor in series to create a voltage divider network.
If the photodiode is on the + side, when the photodiode is in a low resistance state (light), the output of the divider will be high; when the photodiode is in a high resistance state (dark), the output of the divider will be low. However, the amount of current output available will be very small; in bright light it will be roughly 5V/7kOhms, or about 0.7mA. This would be enough to bias on an NPN-type transistor, which could then sink the current from your LED; which would need a current-limiting resistor.

See the attached.
 

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Thread Starter

Hyperian

Joined Apr 25, 2008
13
interesting, but i'm trying to replace the 1.1Mohm resistor with a transistor biased in the linear region so it will have a resistance of around 1Mohm.
 

SgtWookie

Joined Jul 17, 2007
22,210
It's easier to get your hands on resistors than it is to get your hands on transistors.

Do you have a particular transistor in mind? Without knowing what the actual specifications of the transistor are, it will be exceedingly difficult to recommend a way to bias it.

Meanwhile, it will require a pair of resistors to properly bias the transistor.

Your call.
 

Thread Starter

Hyperian

Joined Apr 25, 2008
13
mm i thought with BJTs you need 2 but with MOS you need 1? anyway i was thinking of something like this

i think i saw something like this before but i cant simulate it in any of the circuit programs since i cant find the components i need, what circuit simulating program do you use?
 

Audioguru

Joined Dec 20, 2007
11,249
The photo-diode should be reverse-biased. It leaks a very small current when it is in light. It needs a very strong light to develop enough current for the transistor to light the LED.

I corrected the circuit.
 

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