Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Circuit design help........................................

Status
Not open for further replies.

kdg007

Full Member level 3
Joined
Jan 6, 2012
Messages
162
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,298
Activity points
2,420
this the circuit i built (transmitter and receiver)for my project,As you can see i kept 1 white LED.Instead of the voltage at R1,i kept frequency generator to give square wave.
**broken link removed**
the above picture is the waveform from the lm339 comparator when i used only 1 LED with the single transistor.
**broken link removed**
this is the waveform i got from using 1 LED,the signal is not proper.y is it so ? am i missing something ?
**broken link removed**
any suggestions would help.thanks
 
Last edited:

Actually i can't understand your question. What is photo1 and photo2? What s the first signal (in photo1 upper signal) and the bottom one?
 
  • Like
Reactions: kdg007

    kdg007

    Points: 2
    Helpful Answer Positive Rating
To check for correct operation of your driver circuit, you would preferably measure signals from the driver itself as a fisrt step, e.g. voltage across R7.

A weakness of your darlington driver is the lack of a base-emitter resistor for Q2. A sufficient low ohmic resistor (e.g. 100 ohm) helps to switch of Q2 fast. The square wave frequency can't be seen rom your waveform, so it's impossible to guess if Q7 switching speed matters.

I also think that the photodiode cicruit isn't well designed. The shorted comparator output is just a drawing error, I presume. Operating the photodiode fowward biased is the least sensitive of possible photodiode circuits and also affected by temperature drift.
 
  • Like
Reactions: kdg007

    kdg007

    Points: 2
    Helpful Answer Positive Rating
i redesigned the receiver circuit again.. you are right,there is a mistake according to the calculations.I kept 100k instead 1M..now its working fine.had to tune the input terminals properly :).thanks
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top