transimpeadance vs voltage amp for photodiode

I am working on an amplifier circuit for a avalanche photodiode that is looking at very high speed 2 nsec rise 10nsec wide pulses. Previous work had been done on a circuit that uses 2 TI OPA695 X8 amps looking at the photodiode signal across 50 ohms. I am pressuring a transimpeadance amplifier using the OPA695 with a 1k Rf. The photodiode current is ~300ua. On a general basis what are the pluses and minuses considering noise, bandwidth, etc. of using a voltage vs. a transimpeadance amp in this application.

Reply to
Bill
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Dumping your signal into 50 ohms is a waste, noise-wise. The place for signal to go is into the silicon.

I just did a PIN diode amp using two AD8014's, current-mode opamps. The first stage was a TIA with (I recall) 680 ohms feedback || 1 pF, and the second stage was an inverting gain-of-3 or so. Risetime (from a vcsel laser pulse) was just under 2 ns overall with about a 1.5 pF silicon pin diode. I had a similar amount of signal as you do, a few hundred uA, so s/n was pretty good and edge jitter was around 5 ps RMS, not bad. I think (data's at work, I'm not) I'm getting just about

1 volt out per optical milliwatt in at 850 nm.

Phil, avowed enemy of bad TIA designs, will probably rip me up for this one.

John

Reply to
John Larkin

If you have to put a load resistance on the photodiode to make the frequency responce flat, the transimpedance amplifier will produce a lower noise The "noise power" of a resistor at a given temperature is a constant. As R goes down noise current goes up. The transimpedance design gets a low impedance with a higher value resistor and hence les noise current.

If you use the capacitance of the photocell to integrate a small signal, the transimpedance and the voltage amplifier method end up with the same noise.

Until you get to frequencies that require silly values of resistance, the transimpedance amplifier is the best option from a bandwidth point of view. At really high frequencies, you want to match the amplifier transistor to the photodiode's impedance.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

He won't be happy unless we're using a biased common-base input stage.

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 Thanks,
    - Win
Reply to
Winfield Hill

Well, that lets me off the hook. The inverting input of a current-mode opamp *is* a biased common-base stage.

John

Reply to
John Larkin

OK, given all that, a standard MMIC should be a great, simple, superfast amp for an APD, unless DC performance matters. MMICS are crap dc-wise. But cheap to replace every time the APD blows one out.

John

Reply to
John Larkin

I am of course The TIA Avenger, but that's because I have this deep committed long-term faithful monogamous relationship with my signal-to-noise ratio.

Circuit details are much less important, and in this case 50 ohms is probably fine. This is an APD we're talking about, after all, so assuming that it is running at a gain of 100 (pretty typical), the noise current is sqrt(100)=10 times larger than that of a PIN diode running at the same output current.

With a current having full shot noise, such as a photocurrent from a PIN diode, shot noise will dominate Johnson noise if the current drops at least

50 mV across the load resistor.

That means that the APD's output current, running at 20 dB above full shot noise, will be quantum limited if it drops at least 500 uV across the load.

M sqrt(2q I_primary) = sqrt(4 kT/R_L)

makes amplified shot noise equal to Johnson noise, where I_primary is the photocurrent _before_ multiplication, M is the APD gain, and everything else is as usual.

Rearranging this a bit, the voltage drop required

M**2 I_primary R_L = 2 kT/q (50 mV at room temp)

so since amplified photocurrent I_photo=M I_primary,

I_photo R_L = (50 mV)/M

puts you in the quantum limit.

Thus for a 50-ohm R_L, you need only (50 mV)/((50 ohms)*100) or 10 uA of amplified photocurrent to get into the quantum limit.

With a 300-uA pulse, 50 ohms is better than good enough.

Cheers,

Phil "TIA Avenger--I like that" Hobbs

Reply to
Phil Hobbs

Apologies for the faux pas of replying to my own post:

The noise figure of the amplifier comes in here as well. Since we're not conjugate-matching the input of the amp, we get the full benefit of the low noise temperature of the amplifier. A good amp can have a 35K noise temperature, and that's the number to put in to the sqrt(4kT/R) Johnson noise formula. Therefore with a really good amp, you could be in the quantum limit with an 8x smaller voltage drop--only 60 uV or thereabouts. It's the dc current times the ac resistance that matters.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

OK, check me on this if you will:

If the APD is running at 300 uA out at a gain of 100, we can pretend it has a photocurrent of 3 uA followed by an (optimistically) noise-free gain of 100. Assume a bw of 180 MHz to get about the 2ns risetime specified.

The shot noise on 3 uA is 1.32e-8 amps RMS in this bandwidth. Times

100, we actually see 1.32 uA RMS of shot noise at our 50 ohm load, giving 66 uV RMS. The signal is 300 uA, or 15 millivolts into our 50 ohm load.

The Johnson noise of the resistor is 12.2 uV, so the shot+Johnson noise is 69 uV RMS. If the amp has a 3dB noise figure, total equivalent noise voltage is a bit more, about 73 uV RMS. All this is pretty small compared to 15 mV of signal, and the total noise is way dominated by the shot noise, so you may as well dump the apd current into a 50 ohm resistor and follow with an ordinarily good opamp.

Is that sensible?

Since shot noise is dominant, does that automatically mean you can resolve photons?

The only part I don't get is the assumption that you can use an amp with a 35K noise temperature, not that it's needed here. I've never seen a wideband amp with much better than a 3dB noise figure, which puts it close to room temp equivalent. All the 50-ohm cryo-equivalent amps I've seen have a stepup transformer or a high-Q tuned circuit ahead of the semiconductor. In other words, the best semiconductors still have a roughly 1 nV/rthz equiv voltage noise.

John

who should be testing a prototype o/e converter module instead of playing on newsgroups.

Reply to
John Larkin

I would think it means that your signal to noise is being limited by the fact that light comes in photons rather than continuously. However, this in itself is not the same thing as being able to count the photons individually!

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John Devereux
Reply to
John Devereux

A 1-pole rolloff with a 2-ns rise time is 175 MHz at 3 dB, but there's a factor of pi/2 for the noise BW, so that would be 275 MHz. A Gaussian would have the same rise time but somewhat lower noise BW.

The noise voltage of the op amp would need to be reasonably low--not necessarily 1 nV/sqrt(Hz) but not 40 nV either. A current-fb op amp would be a good choice.

You really can get 35-K amplifiers--Miteq sells them. (In fact they sell some that get to 25K at room temperature and 10K at nitrogen temperature. See

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The noise temperature is T_N = p_N/k, where p_N is the noise power spectral density and k is Boltzmann's constant. Tank circuits and peaking coils can transform the optimum resistance to match the actual source impedance, but can't improve p_N. (If the voltage and current noise are uncorrelated, p_N=e_N*i_N.)

As far as resolving photons, wouldn't it be nice. You can do it if you're using a heterodyne interferometer, where the S/N ratio is dominated by the signal-to-shot noise ratio of the *weaker* beam. (I used to have to do the math every time I used one, just to convince myself that that was true. It is.) Otherwise, you're dominated by the sqrt(N) fluctuations of the photocount.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

In article , Phil Hobbs wrote: [...] A few observations based on experience:

It is more likely that the roll-off will be more than a single pole. Chances are the the system will be designed with about a 45 degree phase margin. This reduces the noise BW value a little but slightly raises the bandwidth requirement ro make the 2nS rise spec.

I'll second the point about current feedback amplifiers. Wide bandwidth CFBs seem generally to have better noise specs than the VFB parts. Getting about 10nV/sqrt(Hz) in a current feedback part isn't hard.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Good grief! I wonder what's inside... a distributed amp maybe? Too bad they don't work below 100 MHz.

I used some Hittite analog switches whose data sheet says, in bold letters, "DC-4 GHz". The "equivalent circuit" on the datasheet sure confirms it. Turns out they screw up below about 200 MHz. I talked to the designer, and he thinks 200 MHz *is* DC.

John

Reply to
John Larkin

This is a very nice distributed amp...

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but the nf is still 3 dB. About $185 each in small quantities.

My customer can afford to fund us to use 'em, but not to grow 'em!

John

Reply to
John Larkin

A distributed HEMT amp is probably a good guess--like using a bunch of FETs in parallel, but without the capacitance worries. Nice piece of work, though, as long as you've got a nice fat grant to pay for them. (Of course, *your* customers are the *DOE*.) ;-)

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

I thought DC frequencies were anything below the FM band..

Reply to
Robert Baer

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