Impedance Matching a capacitive source

Question

I am trying to work out how to match a current source coupled with a capacitor to a 50 Ω load for maximum power transfer. Specifically I have a 6pf PIN photodiode which driven by a 2.5 ghz optical signal. At 2.5 Ghz the PIN diode has an impedance of -10j Ωs.

$$\begin{align} Z_c &= \frac{1}{jwC}\\ &= \frac{1}{j2\pi f C} \\ &= \frac{1}{j2{\pi}2.5 \cdot 10^9 \cdot 6 \cdot 10^{-12}}\\ &\approx -10j ~ \Omega \end{align}$$

^{simulate this circuit – Schematic created using CircuitLab}

I have seen impedance matching but usually they are matching a non reactive source impedance, e.g. 50 Ω, to some complex load. In this case the source has a purely complex impedance of -10j and I want to match this to 50 Ω by using some transformer method (Single stub, 1/4 wave or combination or what ever). One conceptual problem I have is that that the load and sources should be complex conjugates of each other and If I manage this then I would end up with a purely inductive impedance and the circuit load and source would be a simple LC resonator and no power transfer would occur ? I haven't done this for quite a while and would appreciate some guidance.

I have seen some approaches which match a mixed complex load with real and imaginary parts, but not a matching a purely complex generator to a purely real load, ie 50 Ω.

Answer 1

Why not use a CommonBase device, letting the reverse-biased photodiode dump its current into the emitter. Try the 2SC5646A, a 12.5GHz device.

For 2.5GHz bandwidth, you need about 80 picoseconds collector tau. With 1PF total capacitance on the collector, you can use 82 Ohms Rcollector.

Run the CB stage on 3 volt, have 82 ohms in emitter to GND, have 82 ohms in collector to +3volts. Use cermet pot, 1KOhm to base, and bypass the base to GND with 1,000pF tiny tiny SurfaceMount capacitor. Use a ground plane under the circuit.

Buffer the collector, tied to base of a second 2SC5646A, its collector to +3volts, and emitter to GND thru 150 ohm resistor. Emitter is your output; do not short it.

Adjust the pot's wiper for 820 milliVolts across the 82_ohm Rcollector, thus 10mA flows in the CommonBase first device. Voltage gain will be GM * Rc; GM will be Ic/0.026 or Ie/0.026, thus GM is 0.4. Product of GM * RC is 0.4 * 82, or Av = 32X (about 30dB).

Use a 2-sided PCB, and under the first device's collector node, use a Dremel to remove the GND foil, giving a useful boost to the bandwidth.

Answer 2

Just think about what you are asking - you have a photodiode that has a parasitic capacitance of 6 pF. That photodiode produces currents representative of the optical data it receives. The 6 pF at 2.5 GHz has an impedance of about 11 ohms so, what you really want (forget about max power transfer) is an amplifier with an input impedance that is significantly smaller than 11 ohms. This low input impedance dwarfs the impedance of the parasitic capacitor and therefore, if resistive, maintains the band width of the data signal.

This is accomplished by using a TIA (trans impedance amplifier): -

However, these are not easy to design at 2.5 GHz but I don't know what other options you have. If your optical signal is closely bounded to a small part of the spectrum around 2.5 GHz then using reactive components may indeed be your best bet. But, if you have a wide bandwidth digital signal (as I suspect) you only realistically have the TIA option so look at Texas Instruments for really fast op-amps.

Answer 3

Your model of the diode is too simple. If it were really a capacitor in parallel with a current source, you could connect an inductor in parallel with it that resonates at 2.5 ghz. This combination has a very large impedance, and so it can basically be ignored. The circuit is then just a current source into your 50 ohm load.

In reality, the diode has some real (resistive) impedance. You need to match this to 50 ohms, after resonating out the capacitance.

Answer 4

Ideally to achieve MPT the load must be a conjugate match of the source and the transmission line may be the sqrt of the product terms or matched to either end (?) I think

Therefore a current source goes into a “voltage source” load. (TIA inverting input impedance from NFB) and a series cap source is could be matched to a series L shunted by TIA differential short cct.

Therefore the PCB trace if Zs=6pF + tbd R = -Z(f)6pH + tbd R.

Given short trace ESL inductance ~ 1nH/mm what is 6 pH in length ? answer = 6um ( smaller than the diode)

Can you expect some RF attenuation from mismatched conjugate impedance at this frequency into a low f TIA?