[time-nuts] Photodiodes for high frequency OPLL
lists at lazygranch.com
lists at lazygranch.com
Sat Mar 30 17:40:47 EDT 2013
The circuit is something like the instrumentation amplifier. The description starts on page 207 with a schematic on page 208. I can scan it later, but the circuit is easy to describe. Think of two op amps in the classic current multiplication (I to V) circuit, that is positive input to ground and a resistor from output to negative input. Now place the diode between the two negative inputs. The current flow will cause the outputs of the op amp to move differentially, which can then be made single ended with an op amp circuit. But the thing to keep in mind is that the voltage potential across the photodiode has been kept to zero, so the capacitance doesn't matter. But it is only held to zero as long as the amplifier has loop gain, that is create the virtual ground, hence you need a high GBP for the capacitance to be neutralized.
Now I haven't seen this published, but it seems to me if you wanted a negative bias across the diode, you could just use a transformer. I ran into a patent on transformer coupled photodiode circuitry. It was for high speed flash detection. (Amazing was common sense obvious design technique can be patented.)
The bootstrap technique basically takes the virtual ground signal of the current multiplier circuit and replicates the AC portion of the virtual ground on the other side of the photodiode, keeping the AC signal across the diode at zero volts, but letting the current flow into the circuit to be multiplied by the op amp&resistor.
Since the bootstrap needs to "sample" the virtual ground, it itself can't steal any current from that point, so it usually employs a JFET.
These amplifier circuits usually go down to DC, so they will also have response to ambient light, which could eat into the dynamic range of the circuit.
Linear Technology app notes use the bootstrap often. They use a common low noise JFET from NXP. Noise from the bootstrap adds right to the diode, so the bootstrap components need to be low noise. The fully differential circuit doesn't have the bootstrap noise source, but it has two amplifiers on the front end, hence two uncorrelated noise sources.
Going back to the fully differential circuit, you could bias the diode by placing the positive inputs of the I to V circuits at different potentials, then reject the DC components at the double ended to single ended converter by capacitive coupling. You would probably want to meditate on start up issues if the DC bias is large, that is use clamping diodes on the double ended to single ended converter if it looks like something will be stressed on start up.
When you read the photodiode literature, bandwidth is stated into an impedance, so I think they just treat the capacitance as the limiting factor. Maybe that is real life, or maybe it is oversimplified. At that point this is a solid state physics problem and not a circuit design issue. (I'd have to crack a book on the physics.)
From: Attila Kinali
To: lists at lazygranch.com
To: Discussion of precise time and frequency measurement
Subject: Re: [time-nuts] Photodiodes for high frequency OPLL
Sent: Mar 30, 2013 12:03 PM
On Sat, 30 Mar 2013 17:29:45 +0000
lists at lazygranch.com wrote:
> You lose me at damping per decade? Is damping the right word?
> Do you mean high frequency rolloff?
Er.. yes. Frequency rolloff... Sorry, my native language got the better of me.
> Most texts on photodiodes go into bootstrapping them to reduce the
> effect of capacitance. But if you design fully differential amplifier
> circuits, they have the same effect as bootstrapping. Jerald Graeme's
> "Photodiode Amplifiers" goes into this. I'm not a fan of Gain technology
> as a company, but Graeme's book are good texts.
Thanks! I just ordered this book.
Although i will have to build a discrete amplifier for the first stage
until i can mix the signal down to something that can be handled easier.
> The bootstrapping circuits use simpler buffers to keep the voltage
> potential across the diode low, while the fully differential circuits
> depend on gain bandwidth product to do the same.
Well, as i currently lean torwards using an avalanche photodiode,
which needs an operating voltage in the range of 100V, keeping the
potential accross the diode low is not really an issue :-)
How do the fully differential circuits get to keep the potential low?
As far as i can tell, from the circuits i've seen so far, the differential
circuits just use the single ended signal from the diode take the difference
to a bias voltage.
The people on 4chan are like brilliant psychologists
who also happen to be insane and gross.
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