I'm going to make a few assumptions here, due to what I think is right. If you fill in more information, I can give a more specific answer. I'll try to include the math that shows the general case.

The first step will be to estimate the small-signal impedance of the diode. Create a piecewise-linear I-V model for your laser diode. The source data would ideally be in a datasheet, but you may need to measure the curve yourself. It should look something like this:

Find the slope of the active region. In this drawing, that would be approximately 100uA/20mV. Take the inverse of that, and for **THIS** diode the impedance would be 200 Ohms.

**Assumption:** The RF input signal is 10 MHz FM with 25 kHz deviation (f_min is roughly equal to f_max)

Now lets calculate what capacitor value we need for the bias-t circuit. The capacitor provides a low impedance path for the RF signal to load (diode), and a high impedance path for DC. When sizing the capacitor, we need to make it large enough to provide a RF short. I'll shoot for a 1:100 impedence ratio. That means that our capacitor needs an impedance of 2 Ohms at our minimum frequency.

```
Xc = 1/2*pi*f*C; C=1/(2*pi*f*Xc)
C = 1 / (2 * 3.14 * 10MHz * 2Ohms) = 8nF
```

Our minimum capacitance is 8nF, assuming a 1:100 ratio. You can make the capacitance larger, but larger capacitors may have worse parasitic characteristics. Use a good ceramic (C0G/NP0) capacitor here.

Next, we need to calculate the inductor value. We want the inductor to act like an open circuit at RF frequencies. Let's design the inductor to have a 100:1 impedance ratio to the diode at RF. We then need our inductor to have an impedance of 20,000 Ohms at the minimum frequency.

```
Xl = 2*pi*f*L; L = Xl / (2 * pi * f)
L = 20,000 Ohms / (2 * 3.14 * 10MHz) = 318uH
```

From these calculations, our minimum inductance is 318uH, which is fairly large. At some point, the stray capacitance in the inductor will begin to look like a short circuit. If you purchase an inductor, look for the self-resonant frequency as the upper limit that it can be used. You would need to comprimise the impedance ratio to find a viable inductor.

Inductors are more complicated to select than capacitors. Taking a Digikey search for "fixed choke", select and apply the parameters in this order:

- Select all saturation currents greater than your maximum diode current. (apply)
- Select all self-resonant frequencies greater than 2-3 times your maximum frequency. (apply)

As you decrease your inductor size, less RF power will go to the laser diode. You can compensate this by increasing your input power, but your efficiency will suffer. As with all engineering, you need to decide which trade-offs you make in your design. If you post more information, I can tailor the answer to your data.

This is about as simple as it gets - and it meets your spec.

Replace zener with variable reference voltage.
Transistor shown as bipolar can be FET.

That circuit comes from guess where :-)

Another.Diagram a bit small. Original was missing. Does high side monitoring which can be nice.

Our friends at Silicon Chip say at

- Basically, this circuit is a conventional switchmode regulator adapted for constant current output and is specially designed for stepper motor drivers - although it could be used for other applications as well. The circuit works as follows: IC1 (LM2575T) and its associated components (D1, L1, C1, etc) operate as a switchmode power supply. Normally, for constant voltage operation, the output is connected - either directly or via a resistive divider - back to the feedback input (pin 4) of IC1.
- In this circuit, however, Q1 senses the current flowing through R1 and produces a corresponding voltage across R3. This voltage is then fed to pin 4 of IC1. As a result, the the circuit regulates the current into a load rather than the voltage across the load. Only one adjustment is needed: you have to adjust VR1 for optimum stepper motor performance over the desired speed range. The simplest way to do this is to measure the motor current at its rated voltage at zero stepping speed and then adjust VR1 for this current. The prototype worked well with a stepper motor rated at 80O per winding and a 12V nominal input voltage. Some components might have to be modified for motors having different characteristics.

## Best Answer

Yes it will work, but if it gets too much current you may burn it out, to little current and it will not lase. Max current is usually 10 to 15 percent above its min lase current.(depends on laser)

some good info here and an interesting chapter here and full table of contentsthey also have some driver references and schematics

I learned a lot from that site about how to drive some re-purposed red laser diodes that I had laying around.

I personally would start off with the lowest current possible, move it up till it lases and then calculate 10% above lasing current and make that your safe constant current, pulsed current can be higher.