I need to modulate a diode laser with a RF source. I've found parts on Minicircuits that can do this, but I'd like to have the Bias-T and amp on my own PCB. So, I need to design a bias-T circuit that will mix my constant current source with the RF source to drive the laser.

I'm having a hard time choosing the capacitor and inductor values for the Bias-T, since I don't know the load impedance of the diode laser. How can I find this out? Would it just be better to put a 50 ohm resistor in parallel with the diode and then design the Bias-T for a 50 ohm load?

Update: After some thought, it seems reasonable to use the I/V characteristics of the diode, after turn-on, to approximate the resistance. This yields a very low value - 5-10 ohms. This, however, doesn't approximate the load while the diode is off - I assume this would be very high. But, the purpose of the Bias-T would be to bias the diode so it's always on, so I presume just the on characteristics would be sufficient. Sound reasonable?

Update 2: My design needs to supply ~200 mA to the diode. It's a 405 nm diode, so the operating voltage is ~4-5.5 v. I have a wide bandwidth, so anywhere from a Fc of 5 Mhz to 50 MHz will probably work. My transmit bandwidth is about 2 MHz.

  • \$\begingroup\$ Which laser are you using? I'm also interested in this. \$\endgroup\$
    – tyblu
    Jan 22, 2011 at 19:38
  • 2
    \$\begingroup\$ tyblu: My research is into underwater freespace optical communication, as such we're using blue/green lasers. Currently the cheapest way to get a blue/violet diode laser is to salvage them from BlueRay DVD players. A 6x burner diode will give you about 200 - 300 mW optical power. 12x burners are closer to 600 mW. Now that the pico projectors are coming down in price 445 nm (1 W) are more available. A good place for info is laserpointerforums.com, though if you just want to buy straight up diodes, this guy is quite good: hightechdealz.com \$\endgroup\$
    – gallamine
    Jan 24, 2011 at 15:08
  • \$\begingroup\$ What's the frequency of the RF signal? \$\endgroup\$
    – W5VO
    Feb 17, 2011 at 6:45

1 Answer 1


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: Figure from Wikipedia Diode Modeling article

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)

Bias-T circuit. From Wikipedia

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:

  1. Select all saturation currents greater than your maximum diode current. (apply)
  2. 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.

  • \$\begingroup\$ Awesome explaination, W5VO. Is the 100:1 ratio a rule-of-thumb, or are you just picking some reasonable? My design needs to supply ~200 mA to the diode. It's a 405 nm diode, so the operating voltage is ~4-5v. I have a wide bandwidth, so anywhere from a Fc of 5 Mhz to 50 MHz will probably work. My transmit bandwidth is about 2 MHz. \$\endgroup\$
    – gallamine
    Feb 21, 2011 at 17:40
  • \$\begingroup\$ At 100:1, your loss from the components is going to be about 1%, or mostly negligible. You may need to adjust your rule for the inductor to maybe 10:1, or whatever you can get. It should just impact your efficiency, and maybe your matching with the amplifier. \$\endgroup\$
    – W5VO
    Feb 21, 2011 at 18:46

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