Hi sir,Your needs might be answered with a very simple circuit, but we need to know more. For example, what frequencies are present in your OFDM signal (what kind of bandwidth do you need)?
What are the expected peak maximum and minimum values do you expect through the diode?
Can you show us the diode?
Minimum is 18 mA 1.8 voltage, maximum is 35mA , 6 v(linear region)We are getting closer...
What will the maximum and minimum currents be? What is the range of your analog signal (assuming the signal is DC coupled, what voltages correspond the maximum and minimum currents)?
hi cn u help meMinimum is 18 mA 1.8 voltage, maximum is 35mA , 6 v(linear region)
IS ANY REPLACEMENT FOR LT1028?Yes, you should connect your laser diode in place of D2.
D1 is a "normal" small signal PN diode and is only there in case the control voltage goes below ground.
The first part, "Why two op amps?" Not knowing anything about your plans (plan to make 1 or 100,000?), or detailed requirements of the circuit (Available parts, available power supplies and space, signal source impedance, DC accuracy, ease of calibration, etc.) I provided a circuit that is likely to work the first time and can be easily calibrated to to your needs.THANK YOU
WHY WE NEED TWO OP AMPS .
CAN WE USE ONLY TRANSISTOR FOR THIS PROJECT, WHAT ARE THE DIFFICULTIES?
I HAD ONE THING IS IF U GIVE INPUT OF SINE IT WILL COME AS SQUARE WAVE IN CASE OF ONLY TRANSISTOR . IS IT ACTS AS DIFFERENTIATOR?
Thank you for the reply sir.This should do it. The power supply is ±12V. Remember to use bypass capacitors from each power supply to ground, a .01 uf ceramic and a 100 uf electrolytic. If your control signal never goes below ground you can omit D1. D2 is a placeholder for your laser diode.
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