Infrared Receiver

[Axle]

Hehe ,

Modern Tv's and VCR's we thankfully do Have! 

Im not sure why im struggling but the all component resellers i got in contact with didnt have these modules. It seems they dont import certain components if there isnt a big enough demand for them.

Great idea. Perhaps ill call a TV/VCR repairs shop and find out if they can direct me to a shop that might have these modules or if i can purchase an broken TV/VCR. Im not sure where people would dump there old TV's though and rummaging through dump sites looking for components doesnt sound too appealing...

It didnt occur to me the problem could be with the multimeter im using. I dont have an oscilloscope and the multimeter im using is a pretty simple one so it most likely the issue.

If i dont get one of these modules i think i should put this project on hold for a while. Building one from scratch seems quite complicated and given my lack of experience and limited knowledge its probably a bad starting point! I have been defeated ... (for the time being at least)

Just wanted to say thanks again for all the help.

Keep well.
Axle

[Axle]

Hi,

Me again.

Ive made some progress since the last time i posted.

Ive contacted one of the local component sellers and they indicated that they may import the Vishay IR module going forward. If they do, and i think they will, then i should be able to get my hands on one of them in the near future.

In the meantime i havent given up completely. Im still trying to learn as i go along and i think i have made some progress.

Ive sorted out the range issue now. The IR filter was the key. Ive also sorted out my problem of having the same LED brightness no matter how far or how close i am to the reciever. I used a high gain (Ibx 40000!) darlington configuration to ensure that i get about 3v dropped across R1 irrespective of distance.

So thanks for the help in getting me this far. I have a few more questions though and would appreciate any ideas on these.

1. I dont have an oscilloscope so im not sure how my input signal looks. I would assume it is a train of pulses with a frequency of 38Khz?

2. My reading on V1 (See attached) is above 3V when a button is pressed. How come i can see this voltage if i cant see the voltage at Vout (after filtering)? Is it that at V1, im seeing the voltage due to the dc bias?

3. After filtering how would the output signal look? Another train of pulses from 0-3V?

Anyone with some ideas on these questions?

Thanks

[audioguru]

Ive made some progress since the last time i posted.

Good.

1. I dont have an oscilloscope so im not sure how my input signal looks. I would assume it is a train of pulses with a frequency of 38Khz?

Yes.

2. My reading on V1 (See attached) is above 3V when a button is pressed.

Each IR pulse turns on the darlington transistor so its emitter rises to +4V max, then the lower end of the LED would rise to +2.2V or less. A DC meter would measure half which is 1.1V or less. Your LED is missing its cathode line and is drawn backwards.

How come i can see this voltage if i cant see the voltage at Vout (after filtering)? Is it that at V1, im seeing the voltage due to the dc bias?

You have a highpass filter that blocks DC. You need a lowpass filter that smooths and passes DC.
A lowpass filter is a series resistor feeding a capacitor to ground. Then the smoothed DC output is across the capacitor.

3. After filtering how would the output signal look? Another train of pulses from 0-3V?

You cannot get 3V because the darlington plus LED voltages reduce the 5V to only 2.2V or less, then the average of the pulses is 1.1V or less. After filtering if the output has a very high load resistor value then it will be 0V and 1.1VDC. The filter will smooth the pulses into DC.

[Axle]

Been really busy these past 2 days so apologies for not responding sooner.

Thanks for the info. Every bit helps me to improve...

As per the datasheet you posted,  im looking to build the complete circuit in three stages.

Stage one
is the amplifier which is the darlington configuration.

Stage two
(as per the datasheet) is then the filter to ensure that only the signals of interest are passed. Hence ive gone for a high pass filter to remove any dc and ambient noise. Given that remotes operate at 36-38Khz, ive gone for the component values below.

Stage three
converts the pulse to a 0-5V pulse. I thought of using a comparator for this.

While all the stages sounds simple in principle im really struggling  with the filter aspect of things. From my readings things shouldnt be this complicated! I just need to make sure of the following :

For the circuit i posted before this,

When calculating the cuttof frequency f3b=1/(2piRC), do i included the emitter resistor of the darlington pair for R in the formulae? I really think my calculations are incorrect for my filter! Ive used the following, R1=10K, R2=90k and C1=0.0001u (capacitor code 101)

Thanks

[Axle]

Hey again,

Still me trying to get things working and to enhance my understanding.
So far i think this project has definitely given me the opportunity to learn quite a few things.

Ive attached a circuit of a filter and im hoping that someone can verify that my voltage and current calculations are correct.

Please look at the attached. Here goes:

Situation: No Remote Signal
---------------------------

1. Dark current through the photodiode is 2ua. (This is what i measure with a multimeter)
2. The capacitor looks like a open circuit given there is no signal?
3. Hence the current through R1 is 1ua and the voltage drop across V=IxR=0.4V

Situation: Remote Signal
------------------------

1. Current through the photodiode is now 10ua. (This is what i measure with a multimeter)
2. The signal has a frequency of 38Khz (38000Hz) so the capicitor has a impedance of z=1/(2*pi*f*C) which in this case is about 1/(2*3.1415*38000*(1/1000000))=4ohms.
3. So the signal sees R1 in parralell with (4ohms + R2)?
4. Given the impedance of R1 and the (capicitor + R2) is almost the same the current splits equally between the two branches. As such 5ua goes through R1 with a resultant voltage drop of 1V across it?
5. The drop across the capicitor is small and we can ignore it?
6. The drop across R2 is then the same as R1 which is 1V?


I Would really appreciate any ideas from anyone. Pretty Please....

Thanks a million
Axle

[audioguru]

Situation: No Remote Signal
---------------------------

1. Dark current through the photodiode is 2ua. (This is what i measure with a multimeter)
2. The capacitor looks like a open circuit given there is no signal?
3. Hence the current through R1 is 1ua and the voltage drop across V=IxR=0.4V

Correct.

Situation: Remote Signal
------------------------

1. Current through the photodiode is now 10ua. (This is what i measure with a multimeter)
2. The signal has a frequency of 38Khz (38000Hz) so the capicitor has a impedance of z=1/(2*pi*f*C) which in this case is about 1/(2*3.1415*38000*(1/1000000))=4ohms.

Correct.

3. So the signal sees R1 in parralell with (4ohms + R2)?

No. The signal from the photodiode is loaded with 4 ohms (the reactance of the 1uF capacitor) in series with R2, and they are in parallel with R1 so the total load is about 100k ohms.

4. Given the impedance of R1 and the (capicitor + R2) is almost the same the current splits equally between the two branches. As such 5ua goes through R1 with a resultant voltage drop of 1V across it?

Yes, half the signal is wasted.

5. The 38kHz drop across the capacitor is small and we can ignore it?

The value of the capacitor is way too high. It will take a long time to charge. You could use a 0.001uF or less capacitor and the 38kHz drop across the capacitor will still be small.

[audioguru]

Your question #3 was correct.

[Axle]

Thanks AudioGuru.
Much appreciated.

[Axle]

Hi Everyone,

Me again. Ive been playing around and it seems that even with an IR filter, range is still limited. It seems the signals are far to small as we move away from the reciever.

Ive come across differential amplifiers and they might be what im looking for to boost range. I essentially want to amplify signals of about 1mv to something i can use and i think the attached circuit can do the trick. I believe the attached is a very simple 'op-amp'.

Ive included a current mirror active load and grounded one of the inputs to make a high gain amplifier.

Just a few questions:

1. Will this amplify the small 1mv signals to something useful?
2. Given no Collector resistor what is the gain of such a configuration?
3. The signals into Q9 is likely very small (few mv only) so should there be a dc bias to ensure the transistor Q9 is always on? Should the bias flow to Q8 as well or should it just remain grounded?

Thanks
Axle

[audioguru]

Why make an opamp when they are inexpensive?
A TL072 is a dual low noise opamp. It can be set with a gain of 100 for each opamp for flat response to 40kHz and a total gain of 10,000.

An IR receiver module for a TV has all the circuitry that you need>

[nickey]

this might help for parts

im not sure where u live but if it is in the johannesburg region there is a place called screenvision in langlaagte that has almost every electronic component known to man

hope this helps u