InfraredChannel

[QuickStrike]

I have been trying to build a wireless infrared channel.

The distance is only about 4-5cm

Problems:

Can't seem to get the Receiver to work, I have tried all 4 different connections of very simple setup and nothing. In some I get 5V and some I get 0V. However, for the one that I get 5V the signal goes to zero when I interrupt the channel with my finger. But I don’t get any square waves the signal looks like a DC signal.

I know I need an amplifier and a voltage comparator but first I need to get the communication running.

Also I have seen many circuits using a transimpedance amplifier (converts and amplifies current to voltage), why not simply put a resistor in series to the photodiode and amplify the voltage at the node of the resistor and photodiode (sfh203FA) connection. (See figure).

 

[Someone Electro]

the ambient light is blinding the phototrasistor.Put an IR filter on it so that it only gets IR light (the confuguraton where it got 0V wen you put your finger on is corect)

Its much easyer to use IR reciver modules its a 3 pin thingy that outputs positive wen it gets IR light that is the freq. that is tuned for (from 30kHz to 40khz)

You simply conect the power suply on the two pins and an LED whith an resistor.That way the LED shod light up wen it recivers IR light modulated at the rigt freq.

 

[QuickStrike]

Phototrnasistors are slow, I want to use photodiodes coz they are fatser, I want in the range of megaherts.

Someone Electro wrote : "the ambient light is blinding the phototrasistor.Put an IR filter on it so that it only gets IR light (the confuguraton where it got 0V wen you put your finger on is corect)".

The photodiode (sfh203FA) I'm using already has IR filter.

 

[Nigel Goodwin]

The photodiode (sfh203FA) I'm using already has IR filter.

Check it using visible light, you will almost certainly find it's very sensitive to visible light - IR filters reduce the sensitivity to visible light, not prevent it. You should modulate your IR beam, and AC couple your receiver.

 

[QuickStrike]

I will modulate it but first I want to receive the signal that I'm putting at the gate of the mosfet of the transmitter.

i.e I'm putting 1Mhz at the gate of the mosfet from signal generator and I just want to see a 1Mhz signal at the receiver. Then I'll put amplifier and voltage comparator, modulators etc..

what u mean by : "AC couple your receiver".

Can u tell me which of the 4 configuration of receivers from the attachment, is the correct one?

 

[QuickStrike]

I have put a black plastic tube joining LED emitter and detector and I still get a 5V look alike a dc voltage..when I squeeze the tube in the middle the signal goes to 0V.

 

[QuickStrike]

It is kind of working now.

At these resistor values I get an amplitude of:

468 Ohms - 0.15V
384 Ohms - 0.12V
1K Ohms - 0.31V
1.5K Ohms - 0.61V
3.3K Ohms - 1V
5.1K Ohms - 1.5
10K - Looks like a sine square wave
100K goes flat

For any of the values above when frequency is increased the square shape shrinks and become to look like a triangle, and loose amplitude and width. (Means the signal is not stable).

Now is when I'm confused, do I try to amplify the 1.5V or amplify the current?

How do I make the signal stable? i.e. for the signal not to loose shape and amplitude.

Now, do I just connect an amplifier or voltage comparator?

 

[Someone Electro]

meaby the photodiode cant switch that fast.

use an simple Op amp amplefyer

 

[Pommie]

You appear to be modulating your signal at 1MHz. Thats why you see a constant signal - the frequency response of your components must not be high enough. Change your modulation to around the 30kHz range and you should see a signal. You could try pointing a IR remote at your receiver and see if you get a signal from that.
HTH

Mike.

 

[Someone Electro]

why do you use such an high freq.?

 

[QuickStrike]

The photodiode I'm using can switch on and off at 5ns

It does work at 30 KHz very well.

So how do I make it work for 1 MHz?

It is no use connecting an amplifier or a voltage comparator at this stage, coz the signal looses shape and width above 1MHz, I have to make it work for 1 or 2 MHz first, then connect an amplifier or and a voltage comparator.

 

[Someone Electro]

why not use 30kHz?It works fine at that freq.

why do you want to have it at 1 Mhz?

 

[Nigel Goodwin]

what u mean by : "AC couple your receiver".

Use capaitors to couple the stages of the receiver amplifier, for a start you should have a capacitor coupling the signal from the photodiode.

You're using a 1MHz modulation signal, so you only need to pass 1MHz, DC coupling it is only swamping the reciever with visible light. Your receiver should also incorporate a high pass filter, again you only need to pass 1MHz signals.

Can u tell me which of the 4 configuration of receivers from the attachment, is the correct one?

Either of the right hand ones, depending which you want.

 

[QuickStrike]

Actually I need to pass an FSK signal (1 MHz and 2 MHz).

Would it be possible to just connect a voltage comparator and maybe with a feedback might prevent the signal from changing shape and width when increasing frequency.

 

[Nigel Goodwin]

Actually I need to pass an FSK signal (1 MHz and 2 MHz).

Fairly obviously a high pass filter will pass 2MHz as well as 1MHz!.

However, you don't need to have a high pass filter as high as that, just high enough to remove all the low frequency rubbish - in particular 50/60Hz modulation from lights.

1MHz to 2MHz seems rather a large chnage for FSK?, seems to me to introduce more difficulties for no good reason?.

Would it be possible to just connect a voltage comparator and maybe with a feedback might prevent the signal from changing shape and width when increasing frequency.

Don't use such a high frequency, it's pointless using a 1MHz FSK shift - try using something like 1MHz and 1.1MHz (or even less).

 

[QuickStrike]

Ok I'll use 1 and 1.1MHz

But, could be possible to just use a voltage comparator. Without the need to use a high pass filter.

The whole point of my project is so I don't use any capacitors.

Since the distance is only 3-4cm I probably don’t even need an amplifier. No?

So that is y I want to try to make it work with just a voltage comparator. So I was hoping to build a circuit using the voltage comparator that can receive a 1MHz signal and give a digital output.

 

[Nigel Goodwin]

Ok I'll use 1 and 1.1MHz

But, could be possible to just use a voltage comparator. Without the need to use a high pass filter.

The whole point of my project is so I don't use any capacitors.

Since the distance is only 3-4cm I probably don’t even need an amplifier. No?

So that is y I want to try to make it work with just a voltage comparator. So I was hoping to build a circuit using the voltage comparator that can receive a 1MHz signal and give a digital output.

It appears to me to be VERY important to remove visible light interference, and as any artifical light will produce a large 50/60Hz signal out of your photodiode, plus ambient natural light will swamp it's output, DC coupling is just asking for problems. A single capacitor from the output of the photodiode to the comparator should cure those problems, blocking the DC 'swamping' by ambient light, and (by a suitable choice of value) acting as a high pass filter to reject 50/60Hz. As 60Hz to 1MHz is such a massive difference a single small capacitor should be enough.

I don't see the point in trying to do a project without capacitors?, capacitors are an essential component in almost any electronic circuit, and you already have capacitors as supply decouplers! - or at least you SHOULD!.

 

[QuickStrike]

So I don't really need an amplifier for my case, (i.e 3-4 cm distances) right?

But how can I make the voltage comparator convert current to voltage and at the same time compare the signal with a voltage reference to give a digital output or a square wave output? Can this be done?

 

[Nigel Goodwin]

So I don't really need an amplifier for my case, (i.e 3-4 cm distances) right?

It depends, try it and see, put a scope on the output of the photodiode and see what you get.

But how can I make the voltage comparator convert current to voltage and at the same time compare the signal with a voltage reference to give a digital output or a square wave output? Can this be done?

I don't really see what you're asking?, the series resistor and photodiode provide a voltage output - feed it via a small capacitor to one input of the comparator (which will probably need biasing with a pair of resistors). Feed the other input from a similar potential divider but with a preset resistor so you can adjust the bias point - adjust it so you get a reliable output from the comparator.

Don't forget a pullup resistor on the comparator output - many comparators require them.

 

[QuickStrike]

Very concerned about this problem:

At Node A, I measure the amplitude of the square wave at 30-200 KHz and it is 5V, however when I pick up the frequency to 3MHz the amplitude goes down to 50mV and it looks like a triangle rather than a square wave.

How can I keep the signal from loosing shape and amplitude at 3MHz or higher frequencies?

I’m thinking maybe a transimpedance amplifier will probably work better since it will amplify the current to voltage rather than just using the voltage at A.


Not too concerned about the following questions:

At C there is a DC offset and at D, the DC offset if even bigger, can this be solved?

Also at C & D there is an overshoot when signal goes high and also when goes low, can this be solved?

Note: I'm not using the same voltage comparator and mosfet shown below.