# IR TXR - RXR circuit

Discussion in 'The Projects Forum' started by sankar88june7, Apr 12, 2009.

1. ### sankar88june7 Thread Starter New Member

Apr 12, 2009
1
0
pls can anyone answer the doubts concerning the following circuit:

http://www.electronicsforu.com/elec...04-Oct04.pdf&title=Anti-Theft Alarm For Bikes

1) i know the transistor bc547 is there to provide enough current and voltage to drive IR TXR , but how do you find the exact value of current and voltage?

2) what is the use of the resistance , 390 ohm and the capacitance 100uf at pin 2 of TSOP ?
3) what exactly is the voltage value of output of RXR when you say output is high and low?i.e what voltage comes at pin 2 of lm 311
4) whats the use of the 1uf capacitor at pin 3 of TSOP?
5) and whats the real use of the comparator LM 311, cuz as far as i can see,........when RXR o/p is low, comparator o/p is low
and when RXR o/p is high, comparator o/p is high
6) also whats the design of IC 7805...i mean whats the need of the capacitors an 1 and 3 in its design

p.S- there is a small mistake in the circuit , negative input pin of 311 is no: 3 and positive one is 2 ....

2. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
1. You look up the specifications for the IRLED. The transistor is driven into saturation, so the resistor is the most significant element in determining the current.
2. The IR receiver should switch between 5 volts and 0 volts. Its spec sheet will give operating conditions.
3. See #2.
4. Probably to slow response and prevent false triggers.
5. Read the text that explains the circuit. A data sheet for the LM311 will help.
6. Get a data sheet for it - in fact, get a data sheet for all the devices used in the circuit. You will learn something.

3. ### SgtWookie Expert

Jul 17, 2007
22,199
1,801
I have no idea why the author used T1 in a voltage follower configuration. He must've had his reasons. The current through IRLED1 will be the sum of the current flowing out of R4 and R5 through T1's emitter. Without knowing the Vf of IRLED, that will be difficult to calculate.
Apparently, it's a filter for the TSOP1738's Vcc. I have no idea why he used such a large value of capacitance. Find a datasheet for the TSOP1738; perhaps it'll explain it there.
R7 and R8 form a voltage divider across Vcc, which is 5v.
R7 is 2.2k, R8 is 1k. Pin 2 is connected to the junction of R7 and R8.
V(Pin2) = (5v/(R7+R8)) x R8 = (5v/(2.2k + 1k)) x 1k) = (5v/3.2k) x 1k = 1.5625v.
Probably to keep the output from transitioning too quickly. Check the datasheet for the TSOP1738 for proper usage.
You'd have to look at the TSOP1738's datasheet; perhaps it's output does not go from rail to rail. Don't know offhand.
The caps are to remove transients from the supply. Look at a 7805 datasheet, you'll see caps on both the input and output.

Good catch.

I wouldn't suggest actually using this circuit. It would be quite a draw on the motorcycle battery, causing it's life to be shorter than usual.

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,349
730
Seems a bit overcomplicated to use a Flip Flop and Transistor, and relay.

An SCR at the output of the comparator would accomplish the same task, wouldn't it?

The 1uF suppresses transients, and the 100uF and small resistor to Vcc of the 1738 are for "Soft Start", so it doesn't go into alarm when turned on.