Choose the right components for an MMIC amp

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:

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.

Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn't run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).

• L = \$\frac{V \text{$\Delta $t}}{\text{$\Delta $I}}\$ this is the basis for the equation you saw in the app-note.

• \$\text {$\Delta $I}\$ is something you define, not determine.

You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case \$\text {$\Delta $I}\$ would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.

Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ (and hence the duty cycle D = \$\frac {V_o} {V_ {\text {in}}}\$). But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L ~ \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).

Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support.