Need analog laser driver circuit

Thread Starter

somiBabu

Joined Jan 13, 2017
34
Hi i need a analog signal laser driver circuit. i have a laser with 5mw power. if i am using npn transistor sine is converted to square wave so i need as it is my input wave form as output. is any one have it. i need to send OFDM signal through laser
 

DickCappels

Joined Aug 21, 2008
6,187
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?
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
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?
Hi sir,
basically i need a circuit which is less than 500 KHz bandwidth. Diode is very cheap laser threshold is 2 V maximum is around 5.5 V Red wavelength. for free space communication project
 

DickCappels

Joined Aug 21, 2008
6,187
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)?
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
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)?
Minimum is 18 mA 1.8 voltage, maximum is 35mA , 6 v(linear region)

Minimum is 18 mA 1.8 voltage, maximum is 35mA , 6 v(linear region)
hi cn u help me
 
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DickCappels

Joined Aug 21, 2008
6,187
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.

upload_2017-2-11_23-12-1.png
 

DickCappels

Joined Aug 21, 2008
6,187
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.
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
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.
IS ANY REPLACEMENT FOR LT1028?
IN FIG WHAT IS IN BOTTOM TRAN 0 3 1e-6 1e-6 and input sinewave (?)
sorry for bothering?
 
Last edited:

DickCappels

Joined Aug 21, 2008
6,187
What is it about the LT1028 that is a problem, I want to make a recommendation that will be helpful.

Can you get Linear Technology parts easily? They are the only ones I can model, and testing them in the model improves your chances of getting the circuit to work correctly the first time. (One can order from the Linear Technology website.)

The circuit was drawn and tested with the LTSPICE circuit simulation software.

The sine wave generator was used to verify the linearity and gain of the amplifier at different frequencies. The cryptic note "TRAN 0 3 1e-6 1e-6" are instructions to the simulator telling it the kind of simulation to perform, the length of time the simulation is to run, at what point in the simulation it should start recording the voltages within the circuit and the maximum time slice to use in the simulation.
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
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?
 

DickCappels

Joined Aug 21, 2008
6,187
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?
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.

Yes, you can do it with one transistor if it is just a linear amplifier, but it will not be very accurate. I do not understand the part about it acting as a differentiator. Maybe I misread your requirements as needing a linear driver, now it sounds like you just need an on/off pulses, but the mention of differentiation has me confused.

Give us a more complete description of your needs and you will likely get something that satisfies those needs.
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
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.

View attachment 120357
Thank you for the reply sir.
Can you please roughly explain the working principle of this circuit? I'm facing difficulty understanding the role of each component in the circuit given.if possible plz upload the transient analysis. i am very thank full for that
 
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DickCappels

Joined Aug 21, 2008
6,187
Q1 is the voltage to current converter. Its emitter voltage is set by U1, and the emitter voltage divided by the value of R6 sets the emitter current. R6 sets the number of amps per volt for this stage. The collector current is more than 99% of the emitter current over the range of less than18 am to greater than 36 ma. R7 on the base of Q1 makes it unlikely that Q1 will oscillate by itself and also limits current to D1 should the control signal go below 1.8 volts.

U1 sets the emitter voltage of Q1 to -1/2 the voltage at the left side of R3. A small capacitance (15 pf plus a little stray) across R2 reduces the gain by 3db above 1 MHz which prevents oscillation at higher frequencies.

U2 sums together currents from R4 and R5, and outputs -10 volts per milliamp of total input current. R5 supplies a constant 234 microamps and R4 provides from 1.8/10k to 6.0 volts/10k (180 micoramps to 600 microamps) so the total output voltage as a function of input voltage is -4.2 volts to -7.8 volts into the left side of U3.

That range of voltages applied to the left side of R3 results in a range of currents through Q1 of 17.5 ma to 32 ma. The gain (amps per volt) can be adjusted by varying R4 without affecting the offset. The offset can be adjusted without affecting gain by varying R5.

An inexpensive and easily obtainable amplifier that can probably be substituted for the LT10281 is the LM318. You might have to change the values of C1 and C2 to prevent the opamps from oscillating.

Remember bypass capacitors from each power supply pin to ground.
 

Thread Starter

somiBabu

Joined Jan 13, 2017
34
Sir 12V is directly applied to the Diode. Is it right ,. how it will survive for that much voltage. laser diode mostly (mine) tolarable upto 6V.
 

DickCappels

Joined Aug 21, 2008
6,187
Ideally, a curve that shows voltage as a function of current will help us understand the situation.

Can you show me the diode's datasheet, or at least show me the specification for voltage?
 

DickCappels

Joined Aug 21, 2008
6,187
That looks like the diode will have a little more than 2 volts across it, which is what I expected. It is a close call, but when you use it in the circuit, you might find that the extra voltage drop across a 100 ohm resistor will limit the current to less than the 36 you specified.

To answer your earlier question, "Sir 12V is directly applied to the Diode. Is it right ,. how it will survive for that much voltage. laser diode mostly (mine) tolarable upto 6V.": Don't worry, the collector is a current source and it will only put out the current that your input waveform calls for. The diode will only see as much voltage as it needs to have for the current you send to it. The voltage across the diode will only be a little more than 2 volts and any extra voltage will automatically be "swallowed" by Q1's collector.
 
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