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Axle

Infrared Receiver

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Hi,

Im relatively new to electronics and i am hoping that someone can assist me. Ive built a simple infrared reciever circuit that flashes a LED when a remote is pointed at the receiver. Im using a photodiode connected to a single transistor.

While there are several circuits available for download, im looking to build one from scratch and better my understanding of transistors. Ive looked around but the answers are still eluding me...
I just need a nudge in the right direction and ill do some research from there.

I have 2 questions:

1. How do i extend the range of the circuit?
I have tried using a darlington connection to increase the current amplification i get from the photodiode but the range has not improved.

2. Im looking for the LED to flash at a constant brightness whether or not im close to the photodiode or 20cm away. Can this be done simply?
From my understanding of how photodiodes operate, the more infrared light it recieves the higher the current. In my little amplifier the LED is ultra bright at close ranges and barely visible further away.

Hope someone has some ideas...
Thanks

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Hi Axle,
Welcome to our forum. ;D
You didn't attach the schematic of your circuits so it is difficult to see how you connected them.

A photo-diode can be revese-biased with a voltage through a high-value resistor and doesn't conduct without IR radiation. With IR radiation on it then it conducts (in reverse the current is called leakage current) and creates a voltage drop across the resistor. An amplifier circuit is needed for it with a high input resistance, usually using an opamp.

A photo-diode can also have no bias voltage and then it is used as a tiny solar cell, creating a tiny voltage when IR radiates on it. Then it is amplified.

The IR photo-diode needs a filter to block light from activating and saturating it.

It should be easy to make a circuit to flash an IR LED pretty far away. Your TV remote is an example.

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A carrier will help to transmit further


No it won't. It doesn't even need a carrier frequency. The sound can directly modulate the light or IR beam.
A carrier is used with IR so that other sounds can also be transmitted at different carrier frequencies for stereo. Also, a high carrier frequency allows a highpass filter to remove buzz from AC operated lights.

A carrier is used for radio because a high frequency radio signal goes much farther than shouting.

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Hi,

Thanks for the replies. I have attached the circuit im using.

I get the following behaviour,

From what I understand from transistors, the input impedance of the above circuit is about 200k as the transistor im using has a hfe of 200. The photodiode produces a current of 2ua at normal ambient lighting so I get about 0.4ma through the 1k resistor. When I hold the remote close to the photodiode it produces a current of about between 10-15ua and I get a about 2-3ma through the 1k resistor.

However if I move the remote further away and press the buttons the current I get remains at 0.4ma.

AudioGuru, From reading your response I think this might be because the impact of the ambient light overides the signals from the remote as I move away from the photodiode. In other words the ambient light produces a current of 2ua irrespective of how close or far I am and pressing a remote button when im far away from the photodiode results in this value only changing from 2ua to say 2.1ua - which doesn’t change the current I get through the 1k resistor by much. Is my reasoning correct?

Thanks again for all the replies. Its much appreciated.

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Hi Axle,
Your circuit doesn't show the polarity of the photo-diode, or the collector and emitter of the transistor. It looks like it is an emitter-follower. Right?

Your simple circuit responds to any kind of light or IR.
IR remote control systems modulate the beam with a high frequency carrier then the detector has an electronic filter that is tuned to the high frequency, ignoring most other light or IR. Their detector also has a high gain amplifier for the high frequency.

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Hi Audioguru,

Yes its an emitter follower. Im not sure of the polarity as both pins on the diode are of equal length. However i have connected it in the circuit such that if no remote signal is present it provides 1ua of current. If a remote signal is present it provides between 10 and 15ua of current

Thanks
Axle

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Now all that is needed to extend the range is to cover the photodiode with undeveloped film or something else to block visible light but pass IR, then make a tuned amplifier that is tuned to the frequency of the carrier of the remote. Rectify the received, filtered and amplified signal and use it to drive the LED.

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Hi,

As an aside (should i start this as a new post ?) i came across high pass filters while doing some research. Sounds interesting and i came up with the attached circuit as a possible filter.

I believe that most remote controls use a frequency of about 36khz and based on this ive selected the values of the capacitor and R2 so that it attenuates all frequencies below 30Khz.

Are these calculations correct? I dont think so as i dont get any voltage when a remote button is pressed. Vout remains at zero.

Any ideas?

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Hi Axle,
The filter is still part of your IR receiver so keep it in this same post.

The 0.0001uF capacitor and 55k resistor make a highpass filter with a cutoff frequency of 29,090Hz when it has a very high impedance load. It is very simple so will have a very gradual cutoff slope, 14.5kHz will be reduced to half, 7.3kHz will be reduced to 1/4 and 3.7kHz will be reduced to 1/8th, etc.

It doesn't pass DC through the capacitor so the transistor at its output will do nearly nothing. If the transistor is biased correctly with a DC voltage divider then the divider will load-dwn the filter and the photo-diode. You would need to rectify the resulting high frequency output to measure an output anyway.

Modern TVs and VCRs use an IC that is an IR detector. It has a small black case that filters out light, a very sharp bandpass filter, an amplifier with automatic-gain-control and lots of gain for high sensitivity.
http://www.datasheetarchive.com/datasheet.php?article=3376817

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Hey,

Thanks for the info. Unfortunately these modules arent available here in South Africa. Ive tried quite a few electronics shops and the closest thing they have is the two pin photodiode im using. :(

Im stumped at the moment as im not connecting the filter to anything, im just checking the ouput voltage (Vout) relative to ground and i still get a zero reading on my multimeter. If my calculations are fine then surely when i hold the remote close to the photodiode i should get some reading across R2?

Keep well

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Don't you have modern TVs and VCRs and parts for them in South Africa? Don't people throw them away and you can remove their parts?

Modern remotes send an IR beam in bursts of 38kHz pulses. The output of your filter will have 38kHz pulses with a high impedance. Do you have an oscilloscope or high frequency milli-voltmeter to measure them?
A multimeter is made to measure 50Hz or 60Hz AC power line frequencies. It won't measure frequencies as high as 38kHz.

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Hehe ,

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

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

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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

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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.

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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


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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

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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.

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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

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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>

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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

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