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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 6 pF PIN photodiode which driven by a 2.5 GHz optical signal. At 2.5 GHz the PIN diode has an impedance of -10j Ω.

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

schematic

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 Ω and I want to match this to 50 Ω by using some transformer method (single stub, 1/4 wave, or combination, or whatever).

One conceptual problem I have is that that the load and sources should be complex conjugates of each other. 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, i.e. 50 Ω.

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  • \$\begingroup\$ You have a block marked "Tx?" -- normally in this context "Tx" means transmitter, but you're showing a signal processing chain for a receiver. What is the "Tx?" block? \$\endgroup\$
    – TimWescott
    Oct 6, 2022 at 18:00
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    \$\begingroup\$ Just as a comment -- you're building a receiver. You don't directly care for the best power coupling between the incredibly wimpy power available at the diode and the following stage. You care about the best balance of linearity and SNR for your diode and its following stage. Not only is there no guarantee that getting the best possible conjugate match also gets you the best possible signal out of your device, it is very common that a device with the best conjugate match to the sensor has no chance of being a suitable preamplifier. So -- consider the road you're on, and if it's right. \$\endgroup\$
    – TimWescott
    Oct 6, 2022 at 18:04

5 Answers 5

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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): -

enter image description here

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.

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  • \$\begingroup\$ Thanks Andy, I was trying to avoid using a TIA, and feed the 2.5 Ghz straight into a cheap wifi amplifier because they are so cheap, and sensitive and so I only need a relatively narrow bandwidth around 2.5 Ghz. \$\endgroup\$ Jan 16, 2017 at 10:50
  • \$\begingroup\$ I would suggest using a sim tool for this. Only you know the "shape" of the data and you can experiment with inductors etc.. \$\endgroup\$
    – Andy aka
    Jan 16, 2017 at 10:53
  • \$\begingroup\$ This is not conjugate matched impedance for 2.5GHz from diode into circuit board, but can be. \$\endgroup\$ Oct 18, 2020 at 21:07
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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.

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  • \$\begingroup\$ Thanks for your suggestion, Ill keep it in mind. I didn't know about those transistors, their bandwidth is really quite impressive. \$\endgroup\$ Feb 5, 2017 at 8:56
  • \$\begingroup\$ @analogsystemrf can you please add a schematic? \$\endgroup\$
    – Mike
    Feb 16, 2019 at 9:21
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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.

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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?

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As noted by others, the impedance is quite low so an active solution will be challenging, and is unsuitable if a passive receiver is required for whatever reason.

To match it to say 50Ω, a matching network is required, which effectively either transforms or resonates the capacitance such that it's well damped (or whatever filter spectrum is desired or tolerable) by the transmission line impedance (or internal loss if reduced SNR is acceptable). The far end of which must be terminated to validate that assumption. We cannot rely on device resistance to provide damping (at least, not without a more detailed equivalent circuit -- s params), so this will be a one-port-terminated network, considering the loop between device (photodiode), matching network, cable, and receiver input equivalent (termination).

If this is a baseband/wideband application, then LF or DC, up to some lowpass cutoff, is required. The basic idea is to resonate the capacitance with some inductance, just slightly enough to extend bandwidth, without peaking the response much. We can direct connect it (photodiode straight into transmission line), which gives a cutoff of 530MHz. We can use a series inductor to peak it, extending about 30% or 690MHz. We can use a more complex network to go further, by choosing different prototypes, and increasing filter order (higher order filters extend the band edge slightly, in exchange for steeper stopband attenuation; but returns diminish steeply past 2nd or 3rd order).

Possibly, the capacitive peaking circuit (using T-coils) can be used here, since the source is [largely?] non-resistive: this network is particularly famous for its use by Tektronix engineers in the vacuum tube days, where amplifier inputs and outputs, and CRT deflection inputs, are largely capacitive, and the voltage response is the only thing that matters. Bandwidth can be nearly tripled -- but keep in mind it's purely voltage gain, not power or current gain. As far as I know, power gain can not be extended more than about 40%. I'm not sure about current.

This is still not enough though, so we must compromise by reducing impedance. Matching can then be restored with a narrowband L network, wideband transformer, or baseband resistor (minimum-loss matching attenuator). Evidently, 2.5GHz needs at most 32Ω impedance, but more likely 14Ω or less, at the diode. The SNR budget will be relevant to deciding whether this is acceptable.

Curiously, the goal here is the inverse of traditional peaking (current vs. voltage!); perhaps an inversion of the circuit is possible -- I don't know, I haven't played with that goal. Perhaps not, because the source itself (diode capacitance) cannot be transformed.

References: consider Starič and Margan, Wideband Amplifiers (Springer), especially Part 2: Inductive Peaking Circuits. Other filter design references may be helpful, like Zverev, Handbook of Filter Synthesis, or Williams and Taylor, Electronic Filter Design Handbook (McGraw-Hill). There ought to be papers or books targeting photodiode networks specifically, as well.

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