help need:38khz ir transmitter using multivibrator

[marcbarker]

what is PV phote dioade and 'PC mode' ?

Sorry... Photovoltaic silicon diode operating in PhotoConductive Mode

BPW41 is one if I remember rightly..

 

[aruna1]

ok guys i made it.it can generate 38khz.i tested it with HRM3800 infrared reciver.here is the diagram
**broken link removed**
but now my problem is its working distance is very very low.only about 2cm:cry:

can someone help me to amplify its working distance?

 

[Nigel Goodwin]

You need a current limiting resistor in series with the LED, and the other resistor values are still too low.

Have you measured the frequency it's outputing?.

 

[Hero999]

I would disagree that the resistor values are too low. I think that given a supply voltage of 3V lower values are necessary.

 

[aruna1]

You need a current limiting resistor in series with the LED, and the other resistor values are still too low.

Have you measured the frequency it's outputing?.

can you suggest a value for current limit resistor?

i didnt measure frequency(i dont have any frequency measuring instruments.i guess it gives nearly 38khz cos it trigers the ir reciever

 

[Hero999]

What's the forward voltage and current?

Use the following formula:
[latex]R = \frac{Vin-Vf}{If}[/latex]

Assuming Vf = 1.2V and If = 32mA, I'd go for 56R in series with D1.

I'd even consider making the other resistor values higher, not lower as Q2's base current must be 1/10 Ic to be able to drive D1.

A more sensible option is to add another buffer transistor to reduce the impact of loading on the oscillator frequency.

 

[marcbarker]

Nice to see your multivib working. Ingenious how you utilised the hFE of the transistor to limit the LED's current, instead of using a current limit resistor. How do you know you've not fried the LED?

Perhaps you might shelve the multivib now and use a 555 solution instead, is only one capacitor and 2 resistors.

 

[Hero999]

Perhaps they don't sell 555s where the original poster lives, Vietnam?

 

[aruna1]

Nice to see your multivib working. Ingenious how you utilised the hFE of the transistor to limit the LED's current, instead of using a current limit resistor. How do you know you've not fried the LED?

Perhaps you might shelve the multivib now and use a 555 solution instead, is only one capacitor and 2 resistors.

Perhaps they don't sell 555s where the original poster lives, Vietnam?

ahh guys problem is with power supply.ne555(which is available here) requires minimum 4.5v.and LMC555(which requreis minimum 1.5v) is not available.and i also chked for max756(dc booster) so i can boost 3v to 5v but that ic is not available.
any idea how to make a inductor less booster?(inductor require with max756 (22uH) is also not availble

 

[Hero999]

How about a CMOS oscillator?

A CMOS HEX inverter can work down to 2V.

You can use two gates for the oscillator and parallel the remaining four gates to drive the LED.

 

[aruna1]

How about a CMOS oscillator?

A CMOS HEX inverter can work down to 2V.

You can use two gates for the oscillator and parallel the remaining four gates to drive the LED.

that seems good.i will begin my search now.thanks

 

[marcbarker]

At 3 V, the available output current of (paralleled) cmos gate might be low enough for the gate itself to act as an 'LED current limit resistor'.

Or it might not be enough current, check a cmos gate datasheet. The original multivib solution is growing on me, if it works.

 

[Hero999]

I should have made myself more clear, you need the 74HC04, CD4xxx logic isn't much use for LEDs, especially at 3V which is already pushing the lower supply voltage limit. I'm assuming this is to be powered from two 2 alkaline cells so you really need it to work down to 2V.

With a supply voltage of 5V the maximum output current is 25mA.

I don't know what the maximum output current is with a supply of 2V but four parallel outputs should be enough to power an LED.

Note that the maximum ground current is 50mA which buts a limit on the LED current.

 

[marcbarker]

hi,
If all you need is a beam break <..> detected by a IR diode followed by an 'ac' coupled amp, with a simple bandpass circuitry is all that is needed.

That seems like the easiest solution. AC couple a raw photodiode into a narrow band amplifier and detect the envelope with a single-diode AM detector.

Since it's a continuous tone, increasing sensitivity is a real doddle - just reduce bandwidth and increase gain as required. A 567 PLL is a great value way of narrowing bandwidth.

This project is really easy!

 

[aruna1]

That seems like the easiest solution. AC couple a raw photodiode into a narrow band amplifier and detect the envelope with a single-diode AM detector.

Since it's a continuous tone, increasing sensitivity is a real doddle - just reduce bandwidth and increase gain as required. A 567 PLL is a great value way of narrowing bandwidth.

This project is really easy!

well transmitter is the hardest (with voltage restrictions) for me.

 

[Hero999]

What about the receiver?

Does that have a minimum voltage requirement too?

74HC4046 can be used as a tone decoder.

You can use a spare NOR gate or two as an amplifier for the IR detector.

I've drawn the circuit, just in case the book is removed from Google. I thought about copying it but the moderator would probably have removed it for copyright reasons.

You'll need to change the values for 38KHz, see the datasheet.

The Forrest Mims circuit scrapbook - Google Books

 

[marcbarker]

74HC4046 can be used as a tone decoder.

You won't get a much better circuit than Hero999 had posted. Easier to test too.

R3 and R4 set the minimum and maximum frequency limit extents of the PLL oscillator (2.1 kHz, 2.7 kHz).

The 'PLL lo-pass filter' (R1 & R2) also affects detection bandwidth a lot. You'd narrow the PLL bandwidth, if the PLL was nuisance locking onto any stray signals that the reciever was picking up. That could happen if you were increasing the detection sensitivity by preceding the PLL with a lot of amplification. Making them variable resistors if need be you can tune by ear, instead of calculating new values.

Remember, if you are using the same power supply for the transmitter, to ensure that the supply rail between the LED driver and photodiode receiver are sufficiently decoupled. If not, your 'beam signal' (that you want to break) will just bypass the IR link and travel through the power rails!

 

[Hero999]

It's probably easier to tune the transmitter than the receiver.

Yes, supply decoupling is essential: a 100nF capacitor must be connected across each IC.

 

[marcbarker]

Yes, the centre frequency is easiest to choose in the TX. Especially also if an LC circuit is used in an amplifier.

Yes, supply decoupling is essential: a 100nF capacitor must be connected across each IC.

I wasn't talking about '[IC] supply decoupling', I was talking about "supply decoupling"....

If the (audio!) signal is finding it's way through the supply rails and then through an IC's (imperfect) supply rail rejection, sensitive amplifiers can amplify it as though it is a recieved signal. Adding 0.1's across IC pins doesn't do much, apart from lull you into a false 'sense of security' before you design a PCB without breadboarding it first.

 

[aruna1]

What about the receiver?

Does that have a minimum voltage requirement too?

74HC4046 can be used as a tone decoder.

You can use a spare NOR gate or two as an amplifier for the IR detector.

I've drawn the circuit, just in case the book is removed from Google. I thought about copying it but the moderator would probably have removed it for copyright reasons.

You'll need to change the values for 38KHz, see the datasheet.

The Forrest Mims circuit scrapbook - Google Books

reciever doesnot have any voltage limitations.its fixed and powered through AC mains.

so about above circuit.its the reciver right? buti need a transmitter.
thanks anyway