Could you suggest if it's possible to effectively use SiGe/GaAs RF MMIC (which have 50 Ohm input impedance) as first stage of photodiode amplifier?

The idea is to reverse bias diode from some +50V, and connect it directly to MMIC input, so that photodiode is loaded to MMIC's internal 50 Ohm termination. 50 Ohm might be quite low value though.

I am mostly worried about noise figure, frequency range is 1-100Mhz (maybe even 1-4Ghz???). Final target is to detect individual photons (with some probability) or dozens of photons (photo-diode QE is 30-70%).

I understand that typically one would need transimpedance amplifier for this job, but using common off-the-shelf SiGe/GaAs part in my dreams might yield lower noise...

PS. I understand that silicon photomultipliers / APD might be more useful here, but they are prohibitively expensive.

PPS: Parts I look at are :
Mini-circuits MAR-6 2GHz 3dB NF
Mini-circuits PSA4-5043+ 4GHz <1dB NF

  • \$\begingroup\$ Please provide a link to the part you are considering. \$\endgroup\$
    – Andy aka
    Nov 29, 2015 at 16:24
  • \$\begingroup\$ Individual photons imply individual electrons. These are probably better detected with impedances higher than 50 ohms. Much higher. Much much higher. Though as the aptly named @Photon suggests, an avalanche diode is probably the answer. \$\endgroup\$
    – user16324
    Nov 29, 2015 at 17:07
  • \$\begingroup\$ @Andyaka Added links to the parts. \$\endgroup\$ Nov 29, 2015 at 19:31
  • \$\begingroup\$ @BrianDrummond 10pF capacitance with 50 Ohm gets us around 0.5ns RC constant... So if we bump up the frequency to 1-2-4Ghz we should be fine with 50 Ohm? \$\endgroup\$ Nov 29, 2015 at 19:32

2 Answers 2


I'm afraid the simple answer is no. Even assuming 100% QE, one photon produces one electron. What sort of current does this imply?

Well, one ampere is one coulomb per second, and $$k_e = 1.6 \times 10^{-19} C$$ Applying Ohm's Law is not straightforward here, since detecting a single photon is clearly a short-term event. But let's say that, since the amplifier bandwidth is 100 MHz, the arrival of an electron can be considered a current pulse with a width of 100 nsec. Then $$V = iR = \frac{\Delta Q}{\Delta t}\times R = \frac{1.6 \times 10^{-19}}{10^{-8}}\times 50 = 8\times 10^{-10} \text{ volts}$$ While this is obviously pretty small, you need to compare it to the amplifier noise, and assuming the amp is Johnson noise limited with an effective temperature of 300 C, $$V_n = \sqrt{4k_{B}RT\times BW} = \sqrt{4\times1.37\times10^{-23}\times300\times50\times10^{8}} = \sqrt{82.2\times 10^{-12}} = 9\times 10^{-6}\text{ volts}$$

In this calculation, the expected signal is about 10,000 times less than the rms noise voltage, and detection would be, shall we say, challenging.

This ignores shot noise, which makes things worse, but this seems perfectly reasonable in view of the already-unfavorable situation.

  • \$\begingroup\$ I see... Your formula literally asks to use faster amp, right where 50 Ohm termination with intrinsic diode+amp capacitance will give matching RC constant :-) \$\endgroup\$ Nov 29, 2015 at 19:34
  • \$\begingroup\$ @BarsMonster - No, look up the impulse response of a low-pass filter. \$\endgroup\$ Nov 30, 2015 at 21:55
  • \$\begingroup\$ 50ohm at 5pf gives around 0.5ns risetime and ~600Mhz cutoff frequency... So we can get ~10x improvement by using ~1Ghz amp? When looking at this from another side: given that photodiode+amp capacitance is 5pf, 1 photon will give us 3.2*10^-8 V peak voltage which is quite a bit closer to the detection limit. This peak voltage will then decrease according to RC impulse response. \$\endgroup\$ Dec 9, 2015 at 9:43
  • \$\begingroup\$ @BarsMonster - Also consider that increasing the bandwidth increases the noise. \$\endgroup\$ Dec 9, 2015 at 14:31
  • \$\begingroup\$ 1 period of 100 MHz is 10 ns, not 100 ns. And for a single pulse you might consider one half-cycle instead of a whole cycle. So that improves your "V" calculation by a factor of 20 without changing \$V_n\$. Doesn't change the final result, of course. \$\endgroup\$
    – The Photon
    Dec 9, 2015 at 16:53

You can do that. I see it done mostly above 10 GHz. For 100 MHz you do not need SiGe or GaAs, you can use an ordinary CMOS or bipolar silicon amplifier.

The only reason to use a 50-ohm input impedance on your amplifier should be if you can not put the amplifier close enough to the photodiode. For 100 MHz, this means closer than about 30 cm.

If you can put your amplifier closer than 30 cm from the photodiode, you should just use any of the well-known transimpedance amplfier (TIA) designs. At 100 MHz I am not sure if you can build an op-amp TIA or if you will need to look for a purpose-built IC. At 10 MHz, I'm sure you could use a well-chosen op-amp, and I'd guess it's possible at 100 MHz, but you'll need to research this. You'll want an op-amp with a GBW product at least 10x your operating frequency, so about 1 GHz.

Edit I didn't see where you say you are trying to detect individual photons. That is a very challenging requirement. All the solutions I have seen use, as you have said, avalanche photodiodes (APDs) or photomultiplier tubes (PMT). There is a reason why $0.50 electronics are not able to do this. But I think you can find an APD below the $10 range, which will probably make it the most cost-effective solution to your requirement.

  • \$\begingroup\$ There are indeed cheap APD on ebay, but they are for laser rangefinders, and sense area is quite small... 25-100mm2 photodioes are much more accessible... \$\endgroup\$ Dec 9, 2015 at 9:46
  • 1
    \$\begingroup\$ @BarsMonster, nothing in the question says that large sense area is a requirement. \$\endgroup\$
    – The Photon
    Dec 9, 2015 at 16:49

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