TIA Photodiode Bootstrap at 10MHz

I need to bootstrap a photodiode in a TIA circuit similar to the way it is done as shown on page 18 of:

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This example is much too limited in bandwidth. I need a 10MHz bandwidth.

The bootstrapping is needed because of the low impedance of the photodiode. This is 150pF in parallel with 1 Kohm. The problem is one of designing a 10MHz unity gain amplifier with high impedance input, low noise, negligible phase change, and unity gain.

Does anyone have any ideas? I am not sure it can be done.

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Do you have Phil Hobbs' book? That is Step One for issues like this.

That opamp is a little noisy.

John

Reply to
John Larkin

That's funny. Brings back memories of years ago trying to make a photo receiver for a specialized light wall. It worked for what I had to do at the time how ever, the next version I made was with a cluster of 4 small body photo diodes into a 4 channel op-amp. I then summed the results. That generated a cleaner output..

P.S. I was only doing 500 khz and it was a digital stream with a little hysteresis in the circuit.

Reply to
Jamie

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Cool circuit, thanks for the link. (I don't quite understand bootstrapping.... something for me to work on.)

The gain is due, in part, to the changing of input C from 3nF of the PD to 10pF of the JFET+opamp+stray. Bootstrapping a 150pF PD will give you less improvement. But still perhaps enough. Do you have enough photocurrent to reduce the TIA resistor from 1 Meg to 1k? I would then 'naively' expect a bandwidth improvement of sqrt(R) so a factor of 30... X 350kHz... something near 10MHz may not be out of the question. I've never built PD circuits this fast though......

George H.

Reply to
George Herold

One method is to connect the PD directly to the input of a nice quiet

50-ohm amplifier. If you have at least 200 uA of photocurrent, this will work very well--you can get to the shot noise limit that way.

At lower photocurrents, life gets a bit harder. Your particular problem gets quite difficult below about 20 uA--at that point you have to start trading away SNR or reducing that capacitance. The best Si PIN diodes have a capacitance of 40-100 pF/cm**2 when reverse biased, so if your PD isn't at least a half inch square, you can reduce the capacitance by choosing a different PD and/or reverse biasing.

So how big a photocurrent are you expecting, and what's your SNR target?

Cheers

Phil Hobbs

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Phil Hobbs

I don't. A book I do have is by Jerald Graeme, "Photodiode Amplifiers, Opamp Solutions"

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There is a topology for a boot strap amplifier in it. He begins his treatment of them in Chapter 4.3 page 71. He does not give any component values or part numbers for the topology (fig 4.12 page 81). It would take a lot of research for me to figure out if there are any transistors and resistors that can make this topology go up to 10MHz.

Is this the book you are referring to?

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I will get back all of you on the expected photodiode current.

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This one:

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John

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John Larkin

The peak current is expected to be 1 uA.

The latest value for the capacitance I have is now 30pF.

I do not have a choice on photodiodes. The detector I have been assigned to make work for this project is not actually a photodiode in the conventional sense. It is a custom made photoelectromotive force detector for use in a laser ultrasonics application. This device cannot be reverse biased like a PIN diode.

A major concern about the low series resistance is that it will create a high gain noninverting amplifier with the feedback resistor for the equivalent input noise on the inverting input. This gain will also reduce the bandwidth of the opamp circuit.

The zero the capacitance will make is another reason I am looking to bootstrap this.

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If there's a way to make that 10 uA, your life will be much easier.

Bootstraps have the same noise multiplication problem as TIAs, for the same reason: they put their own noise voltage across the PD capacitance. With equivalent devices, you can get a 3 dB improvement by using both, but bootstrapping is not a slam dunk. One good thing about it is that you can AC-couple the bootstrap, which means it can be single-ended rather than differential.

You can get the same 3 dB improvement by putting a TIA on each end of the PD.

If it's a photoacoustic measurement, you may not need DC-10 MHz. What's the actual measurement bandwidth?

Cheers

Phil Hobbs

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Phil Hobbs

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"> You can get the same 3 dB improvement by putting a TIA on each end of

Hey, I remember that idea! No one=92s ever tried it though, have they?

Say, will your cascode circuit work for photo detectors other than photodiodes? I have this =91silly=92 idea that electrons from photodiodes are born at half a volt or so, and are thus able to do a bit of work before they are detected.

George H.

Reply to
George Herold

How can a TIA on each end accomplish the same thing as bootstrapping? The idea of bootstrapping is to increase the effective impedance of the photodetector by making the same virtual ground voltage appear on both ends of it. In a dual TIA arrangement the current going into the virtual ground of one TIA comes out of the virtual ground of the other. That means the outputs of the TIA will be equal and opposite with the consequence that the voltages on the virtual grounds will also be equal and opposite (neglecting part tolerances.) This would halve the effective photodiode impedance rather than increase it.

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Sure, but it helps the most with PDs. Most other detectors are either intrinsically slow and noisy, like many photoconductors, or else have low capacitance anyway, like PMTs. Of course there are lots of non-optical transducers, but not too many of them have the difficult feature of PDs, namely high speed with predominantly capacitive impedance and wide dynamic range.

Cheers

Phil

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Reply to
Phil Hobbs

I didn't say it did the same thing, only that it gets the same SNR improvement. Putting a TIA on one end of a PD requires bypassing the other end to ground, so the bypassing might as well be done by the summing junction of another TIA!

Cheers

Phil Hobbs

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Phil Hobbs

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I see there are papers and patents on fully differential TIAs, but no real products.

But isn't this all a wash since you added another noise source (2nd TIA)?

Reply to
miso

snipped-for-privacy@sushi.com a écrit :

Translate one TIA input noise to the other TIA. Being uncorrelated their power add up, so it's +3db noise. The signal is +6dB level for a net +3db S/N ratio...

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Fred Bartoli

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Exactly, I came to think of differential TIA's from correlation noise measurement circuits. (You look at the same noise signal (say a resistor) with two identical amps. You then look for the correlations in the noise in each channel.) They do this by multiplying the two signals. The amplifer noise averages to zero and you are left with the 'signal' from the noise source. So if you have a differential TIA would you be better to multiply the two signals rather than summing them?

George H.

Reply to
George Herold

No, because the noise adds in power and the signal adds in amplitude. That's why it's 3 dB not 6 dB improvement.

Cheers

Phil Hobbs

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Phil Hobbs

Using correlation will help suppress the amplifier voltage noise, but won't affect the amplifier current noise or the shot noise, because both of those are real currents that flow into both amplifier inputs. (It also won't help the pickup and power supply feedthrough much.)

Cheers

Phil Hobbs

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Phil Hobbs

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Sure, But it's often the case that the amp voltage noise is the dominant noise source. Mind you I=92ve never had an occasion to worry that much about 3dB of improvement. It=92s usually a lot easier to increase the signal upstream somewhere... more light, better focus, etc..

George H.

Reply to
George Herold

The low end of the bandwidth is 100kHz.

A bootstrap does not necessarily add noise to the circuit. If the bootstrap amplifier has less noise than the op amp its noise is swapped for that of the op amp. This measured reduction in noise is documented in figures 4b, 5b, and 6b on pages 17 and 18 of:

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Suppose I returned the photoelectromotive detector to ground. Since the expected detector current is 1uA, to get some reasonable output from this first TIA stage I would need a feedback resistor on the order of

1Mohm. Since the series resistance of the detector is 1kOhm there is a noise gain of 1001 before the zero from the 30pF capacitance kicks in at 5.3MHz. Taking into account the DC noise gain, the capacitance on the virtual ground (dominated by detector capacitance) and the needed bandwidth of the TIA (10MHz) I would need an opamp with an 18.8THz GBW. If it weren't for the noise gain I would have needed only 18.8MHz. The only way this is going to be done is by means of bootstrapping to reduce the effective impedance of the detector.
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