some pics

Purple Rock:

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Tek SD43 o/e converter:

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The actual converter is in the gold brick. It has one fiber input, one SMA output, and one power pin going in. So, what is all that electronics for?

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin
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It has DC outputs on pin jacks as well as the SMA one. I expect that the gold brick is an InGaAs photodiode with a 50-ohm resistor on the anode and a bias tee on the cathode, to apply bias and allow them to measure the DC photocurrent. In the process they need to not screw up the settling transient at late times, while not turning the PD to lava if you send in too much light. (There's probably 10V of reverse bias, so once you get into the milliamps, the PD dissipation can be significant.)

So the cathode circuit will need a medium-speed TIA with carefully controlled input impedance, a current-limited bias generator, and a diff amp to ground-reference the TIA's output.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Seems like a lot of parts. Competitive boxes have one battery and a cap. The gain is low, so it's probably not an avalanche diode.

I wish some patriot would bootleg the SD-series schematics. And the

11801.
--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

No, it's a regular photodiode. But it's a really small one, which makes it electrically and thermally delicate.

The 1180x team were really bird-dogging their step responses, at both short and long times.

Ditto.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Can you get close to a mA from a PD that does 8 GHz into 50 ohms? I was blasting a big PD the other day.. measuring laser power And my PD saturated at ~10 mA

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(posted on SEB)

That was with a laser spot smaller than the PD area, so it might be a "local" saturation effect.

George H.

Reply to
George Herold

Reverse bias helps a lot. Other things that help:

Thick epi. Reduces lateral voltage drops and local carrier depletion, at the expense of short wavelength response. (You basically only see the light that makes it through the epi before being absorbed.)

Large spot size. Reduces the local current density and injection level.

Good transimpedance amps. Keeping the voltage swing across the diode under control gets rid of a lot of nonlinear capacitance funnies.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Yeah this is a PD->TIA that I made before reading your book. No bias on the PD. A DC measurement in this case. I'll have to try bashing in a bias.

George H.

Reply to
George Herold

Reverse bias helps, because it makes depletion region wider? I guess I don't know what causes the saturation.

George H.

Reply to
George Herold

Reverse bias drops the junction capacitance radically. And after that effect levels out, more bias still sweeps charges out faster, which continues to speed things up.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Recombination due to high level injection. The recombination rate goes as n_H*n_E.

With enough carriers, you shield out the built-in E field in the junction, so the carriers just diffuse around till they recombine.

A big reverse bias pulls carriers out faster, so it takes more light to make that happen.

Speed in PDs is limited by RC delays (R is mostly in the epi) and by diffusion in the undepleted regions. Some PDs speed up amazingly when you put on enough bias to deplete them right up to the contact.

There are two types of ohmic contact: (a) a heavily-doped region and (b) a Schottky barrier of negative height. I don't know if anyone uses (b) in PDs, but it ought to make them faster.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

I should imagine the contact itself should have an interesting shape; a fractal of the right type ought to have minimum ESR and maximum transparency (small optical cross section, and if extending to very small scales, thin enough wires or short enough segments to be transparent to relevant wavelengths too).

Same applies to transistors, which always bothered me; the only conclusion I can think of is, they /want/ power transistors to perform poorly. Perhaps so designers don't accidentally get them singing at 200MHz+. In principle, there should be very little difference between, say, a 2N7002, or an IRF540, and comparably rated RF parts.

Tim

--
Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Doesn't work that way with metal, which is why PDs use the top epi to conduct photocurrent to contacts at the edge. You can use interdigitated Schottky contacts, but you have to really bias it up in order to get the opposed diodes to conduct, and you lose sensitivity due to the metal area.

(I'm doing optical antenna design right at the moment.)

The doping profiles are really different though. Getting high breakdown voltage slows down the diffusion-limited speed.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Great, that makes sense. By the epi layer you mean the top, p-doped layer of the Photodiode. (just checking) So here's another question. You've often said that you speed things up, by cranking up the reverse bias such the epi layer starts to deplete. Now if that happens shouldn't that increase the resistance of the epi layer? (or do you have to deplete right to the layer... but no further.)

George H.

Reply to
George Herold

....

...

I'm always impressed by the quantum efficiency of InGaAs photodiodes.

At 1550nm the resulting responsively usually works out to close to 1A/W. Which is convenient for back of the envelope calculations.

In this case the sensitivity shown on page 11 is 25mV/mW this is consistent with a diode terminated directly with a 50ohm resistor at the diode and a 50ohm termination at the other end of the cable with no amplification, as Phil said.

They state the sensitivity as 30mV/mW at 1330nm - I'm a bit surprised why it is more than at 1550nm, usually I have seen a rising response up the cutoff at ~1700nm. Page 12 shows a peak at ~1350nm.

I agree it seems a lot of bits to do the biasing and low frequency optical power measurement.

kevin

Reply to
kevin93

The specs say that the output impedance is 1K nom, so they don't terminate internally. It ohms just about 1K with power off.

The only user-accessible output is the SMA connector. Mysterious.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin
[on low-resistance ohmic contact to silicon]

It's amusing to look at microphotos of transistor dice, the frequency DOES affect the emitter and base connection geometry. Most important, though, is the effect in SCRs (and LASCRs). Speed matters a LOT when you're putting kiloamps into the wafer, and turn-on time is dominated by a dI/dt limit (comply or things melt). The gates look like christmas-tree fractals sometimes. For LASCRs, the 'gate' is a set of holes through the metallization, and the dI/dt limit basically goes away if you can flash-lamp the turnon signal through all of 'em.

Reply to
whit3rd

I'm not really a semiconductor physics guy, but I don't think so. You have metal contacts at both ends, so there's an effectively infinite reservoir of both electrons and holes, and a fully depleted epi has a whole bunch of available travelling-wave states for them to move in and out of.

Fully depleted is quite different from intrinsic.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

If it had a QE of 100%, it would reach 1 A/W at 1.23 um. It has a lot to do with the epi absorption, junction thickness, and AR coating (if any).

Keeping the junction thin makes the device faster, but hurts the long-wavelength efficiency because a lot of the light doesn't get absorbed within the depletion zone (where the carriers can be collected).

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Nah, there are one or two pin jacks for the DC stuff too, aren't there? There are on my SD-46es.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

This is an SD43. There are lugs on the board but they don't come out. Maybe they used the same PCB but didn't bring out the slow signals for some reason.

Still mysterious.

The '43 is 8 GHz and works at 850 nm, where many other o/e converters are blind.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

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