Methods to increase bandwidth of a transimpedence amplifier?

[Topher925]

Long story short, I'm trying to "easily" increase the bandwidth of a transimpedance amplifier which is limited by the intrinsic capacitance of the feedback resistor, R. The knee frequency is of course given by f = 1/(2*pi*R*Cr), where Cr is the intrinsic capacitance of the resistor.

As of right now, the only thing I can think of is putting two (or more) resistors in series equal to the value of R to reduce the total capacitance, Cr. However, I don't know if this will actually reduce the capacitance or increase it since some capacitance will be created where the resistors are soldered together.

I currently have the PCB designed so the resistors are tombstoned or facing up with one soldered on top of the other. Is this going to work? I thought about actually trying it but I don't have a LCR meter that can measure in the tenths of picofarads. Are there other more conventional methods I should look at?

 

[vk6kro]

Could you put two resistors in series and a small bypass C to ground between them?

This would give a bit of top boost and may extend a bandwidth a bit.

 

[Topher925]

Could you put two resistors in series and a small bypass C to ground between them?

This would give a bit of top boost and may extend a bandwidth a bit.

Nope, I can't. One important aspect I should have mentioned is that the feedback resistor is in the gigaohm range and therefore leakage current is extremely crucial. Adding a decoupling cap could completely null out the feedback resistor.

 

[vk6kro]

What if the capacitor was a 3 to 30 pF variable? You are trying to reduce the AC feedback while maintaining DC stability.

What frequency range are you getting and trying to get?

 

[Adjuster]

What about equalising the response by adding a high-boost filter after the transimpedance stage? That might not be so good for your noise performance / dynamic range though.

 

[Bob S]

Adjuster has the right idea. Use the minimum R in the transimpedance amp to get well above the noise of the circuit. Add a high bandwidth booster voltage amplifier. Remember that both resistors and amplifiers have noise. See Nyquist noise in

http://en.wikipedia.org/wiki/Johnson–Nyquist_noise

High meg resistors are especially bad for both noise and distributed capacitance.. Using two amplifiers is better for both noise and bandwidth.

Are you using a solid state photodiode? I believe vacuum photodiodes are better for both sensitive area and noise.

Bob S

 

[Topher925]

What if the capacitor was a 3 to 30 pF variable? You are trying to reduce the AC feedback while maintaining DC stability.

What frequency range are you getting and trying to get?

I'm not really sure how making the cap variable is going to help. There's less than a pA going through that resistor. Adding something like a variable cap between the two resistors will completely wipe out my signal. I need a frequency range of ~250-1000Hz.

Adjuster has the right idea. Use the minimum R in the transimpedance amp to get well above the noise of the circuit. Add a high bandwidth booster voltage amplifier. Remember that both resistors and amplifiers have noise.

Ok, you've totally confused me. You're saying I should replace the feedback resistor with a booster? Won't that be a complete disaster?

Are you using a solid state photodiode? I believe vacuum photodiodes are better for both sensitive area and noise.

Solid state avalanche photodiode. I don't believe vacuum photodiodes work in the light spectrum I am working with.

 

[vk6kro]

Less feedback means more gain.

I modeled this with a normal Opamp, not having details of your setup.

A 50 pF capacitor between two 50 K feedback resistors in series (with a 10 K input resistor) produces a marked peak at about 60 KHz.
If this coincided with an existing falloff in gain then it would produce an extension of bandwidth.

Why do you have such a large feedback resistor? Are you trying to use the photodiode in almost dark conditions?
What area of light spectrum are you using?

 

[Bob S]

One way I have sped up the risetime of picoamp current signals is to first amplify them through a current integrator circuit, through a small series resistor to the summing junction and then a capacitor in the feedback loop, sometimes as small as 100 pF. A fast 100 pC charge pulse would produce a 1 volt step function at the output. There is no problem with distributed capacitance RC risetime limitations. If necessary, the integrator can be followed by a conventional wide bandwidth amplifier.

Bob S

 

[vk6kro]

OK Bob, I think I get what you are describing, but maybe a diagram would make it clearer.

Here is my circuit and frequency response:

[PLAIN]http://dl.dropbox.com/u/4222062/opamp%20response%202.PNG

The yellow line is without the 32 pF capacitor and the blue line is with the capacitor. Makes a worthwhile improvement.

 

[Adjuster]

I'm not really sure how making the cap variable is going to help. There's less than a pA going through that resistor. Adding something like a variable cap between the two resistors will completely wipe out my signal. I need a frequency range of ~250-1000Hz.



Ok, you've totally confused me. You're saying I should replace the feedback resistor with a booster? Won't that be a complete disaster?



Solid state avalanche photodiode. I don't believe vacuum photodiodes work in the light spectrum I am working with.

I did not mean to suggest discarding the transimpedance stage or its feedback resistor. What I meant was to add an equalisation filter later in the signal path, so as to flatten the frequency response over the required range. This may be only worth considering once you have exhausted the possibilities for extending the first stage response.

You should probably still consider modifying the transimpedance feedback. Splitting the feedback resistor seems reasonable enough - provided that only a fairly small resistance (perhaps tens of kohms) is added between the amplifier output and the split point. The remaining feedback resistor can still be large if this is necessary, for instance to get a suitable DC response, or if noise current matters more than noise voltage, so that paradoxically a huge load is actually preferable.

Finally, the point of using a variable capacitor would be to allow the response to be set up. You could just as well use a fixed capacitor if the optimum value were known.

 

[Bob S]

OK Bob, I think I get what you are describing, but maybe a diagram would make it clearer.

See thumbnail for a fast integrator circuit. A 1 microsecond long, 100 picoCoulomb current pulse is integrated on a 100 picoCoulomb feedback capacitor. Rise and fall times are 100 nanoseconds. The integrator voltage output could be differentiated to regenerate the original input current pulse.

This circuit could be further optimized if I knew more details about the input signal.

Bob S

 

[Adjuster]

I am intrigued to note that you are using an Avalanche Photo-Diode with a load in the gigaohm range. You will therefore expect outputs of the order of some Volts per nanoamp of current from the APD. How much output voltage will the transimpedance amplifier need to deliver to accommodate the expected dark current and photo-current?

 

[Topher925]

Bob, I still don't really see how the fast integration circuit is going to help my situation seeing as your simulation is very generous with the signal parameters. I think the leakage current of C2 and the input bias of the op-amp will just about wipe out my signal. I guess I didn't really provide enough information.

This circuit could be further optimized if I knew more details about the input signal.

The signal going to the transimpedance amplifier is a 20-80 femptoamp sinusoid (minus DC component) with an SNR of around -12dB. The frequency range is 250Hz-1kHz.

You will therefore expect outputs of the order of some Volts per nanoamp of current from the APD. How much output voltage will the transimpedance amplifier need to deliver to accommodate the expected dark current and photo-current?

I'm expecting about a 20VDC offset when setting the reference of the preamp to ground. The signal I'm measuring I'm expecting to be around 0.2-0.6mV.

 

[Bob S]

Hi Topher-

What is the source of your light signal, and what is the wavelength range/distribution? Is it a Planck/IR distribution or something else? How many photons per second?

Bob S

 

[Topher925]

Bob, the light source is from a fluorescent material. The peak wavelength is at about 700nm with a sharp distribution between ~695-705nm, so not an IR distribution, everything is in the visual range. Number of photons per second I estimated to be about 25,000ish with a photo-detector current of about 35fA.

 

[Adjuster]

The signal going to the transimpedance amplifier is a 20-80 femptoamp sinusoid (minus DC component) with an SNR of around -12dB. The frequency range is 250Hz-1kHz.



I'm expecting about a 20VDC offset when setting the reference of the preamp to ground. The signal I'm measuring I'm expecting to be around 0.2-0.6mV.

In that case your greatest dividend may come from minimising the total input capacitance, not just that across the feedback resistance. I don't know if putting two resistors back to back will help - perhaps try it and see. The PCB layout will also need care to minimise track capacitance effects.

Is there any significant length of interconnection between the detector and the amplifier (for instance to allow for cooling the detector)?

 

[Topher925]

In that case your greatest dividend may come from minimising the total input capacitance, not just that across the feedback resistance.

Just about all the input capacitance is intrinsic and comes from the APD. Not much I can do about it but place everything close together but so far I've never been limited in bandwidth from a photodiode for what I'm doing.

I don't know if putting two resistors back to back will help - perhaps try it and see.

I managed to get my hands on a Fluke 6303A LCR meter and measured the capacitance across some high value resistors, although it can only measure down to 0.1pF. The capacitance did get cut in half when I put two resistors in series (0.3pF to ~0.1pF). Those were both large through-hole packages, I'm hoping for lower values with some 2512 chip resistors. I'd like to put three, maybe four, in series but I'm concerned about stability.

The PCB layout will also need care to minimise track capacitance effects.

Yeah, and SMD soldering with gloves on really sucks.

Is there any significant length of interconnection between the detector and the amplifier (for instance to allow for cooling the detector)?

No, feedback resistor(s), amplifier, and detector are all almost right on top of each other in the PCB design. The sensitive nodes are "floating" (not soldered to the PCB).

 

[Adjuster]

Now that you have given a bit more detail, I'm really not sure how much benefit could be had from doing tricks with the transimpedance stage. Provided that the first amplifier gain-bandwidth is sufficient, you could decrease the feedback around 1kHz so as to increase the gain, but you might then have more difficulty with stability.

Whether or not this would be worthwhile might boil down to whether noise from the following stages or from external sources would be significant. Do you know this, given the level of signal expected with the feedback capacitance as is?

 

[tugrulzure]

hey topher, what is the opamp you use? the f3db of the transimpedance amplifier can be estimated as: f3db= sqrt(pole*GBW) so it pretty much depends on your opamp too.

TI has an opamp which has 1.6 ghz gbw with 2 pA input bias current. You might review your design equations for this amp.