Transimpedance amplifier (TIA) oscillating at above 1 GHz. Need Help

resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

up to 10pf.

oscillation is getting worth.

Wow, that's one ugly schematic. Conventionally you use a triangle for an op amp, and put the input on the left and the output on the right.

With no photodiode and no feedback cap, you're trying to run the LMH6629 at a noise gain of 2, whereas it's only barely stable at a noise gain of

1. It's also a really hot amplifier, with a GBW of nearly 10 GHz, so the layout is going to be very critical.

You overestimate the capacitance of the feedback resistor itself--it'll generally be between 0.05 and 0.15 pF, depending on the type.

The photodiode capacitance is key to making this thing stable, because at high frequency the noise gain is set by the ratio of the input and feedback capacitances--you need a PD whose capacitance is at least 10 times the feedback cap to make it even marginally stable.

What sort of photodiode are you planning to use?

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 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net

Am Dienstag, 14. Mai 2013 10:25:29 UTC+2 schrieb snipped-for-privacy@gmail.com:

resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

up to 10pf.

oscillation is getting worth.

Hey,

Actually it is so ugly because it was my first approach to add a custom part to my layout editor ;)

How do you calculate the noise gain of 2? Without the feedback cap and the diode it should be 1, shouldn`t it?

I am using a pin diode with 1.2pf of capacitance at 3.3V reverse bias. I thought the more input capacitance the more unstable the circuit is so I tried to keep it small..

Cheers, Julian

(Context restored)

The noise gain is the closed-loop noninverting gain of the stage. With no PD and no cap, you've got 1.2k input and 1.2k feedback resistors, for a noninverting gain of 2, which is guaranteed to oscillate with this part.

If your PD is only 1.2 pF (which is a bit hard to believe at zero bias), then if the pads are small and the traces short, you should have ~ 1 pF from the board, 1.2 pF from the PD, plus the differential input capacitance of the amp. This number is suspiciously absent from the datasheet, but I'm guessing that it's about 2 pF. Thus your total C_in is about 4 pF.

To maintain stability, your feedback capacitance needs to be less than

0.4 pF. I'd try replacing your PD with a 1-pF capacitor and running with no feedback cap, first. If it still oscillates, try pressing your finger down so that R_f squashes right into the pad of your finger, and see if you can make it stable. If so, choose the right C_f and you're probably done. If not, try reducing R_f.

With too high an R_f, the phase shift of R_f*C_in can destabilize the loop even if the amp were unity-gain stable. The corner frequency of

1.2k and 2 pF is only 66 MHz, so you could also have an oscillation in the low hundreds of megahertz if the 1.2 GHz one didn't take over.

Do you really need such a hot amplifier?

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

resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

to 10pf.

oscillation is getting worth.

Is the pd connected to pad1-pad2?

Why is there a resistor across the photodiode? To Vcc? That will make troubles.

What is Vcc?

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation 
Picosecond-resolution Digital Delay and Pulse generators 
Custom timing and laser controllers 
Photonics and fiberoptic TTL data links 
VME  analog, thermocouple, LVDT, synchro, tachometer 
Multichannel arbitrary waveform generators

Am Dienstag, 14. Mai 2013 10:25:29 UTC+2 schrieb Julian Arnold:

dback resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

mething above 20pf. In my opinion it should be stable for higher caps and o nly start oscillating if the cap is to small.

a perfect sine, even without a signal applied!

1pf up to 10pf.

scillation is getting worth.

ue.

Thanks a lot for the responses!

Unfortunately I need to build up the circuit for a university project where I need to transmit an OOK modulated signal at around 100 MHz...

The resistor across the diode is not soldered to the board. I think I should remove it from the schematic. It was placed there to remove biasing voltage if the diode would not be con nected to VCC, which is at 5V.

So if the stability of the circuit depends on the noise gain, which needs t o be above 10, than wouldn´t it be impossible to get it stable without a signal applied to the diode? Because without the resistor across the diode the noise gain should always be 1 without applying a signal.

Cheers, Julian

Am Dienstag, 14. Mai 2013 10:25:29 UTC+2 schrieb Julian Arnold:

resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

up to 10pf.

oscillation is getting worth.

Just made a few measurements with different caps and no optical signal applied: R_f = 1,2k C_d = 3pf => V_o: Sine wave with 440mVp-p at 418MHz

R_f = 1,2k C_d = 3,9pf => 448mVp-p at 477MHz

R_f = 1,2k C_d = 1pf => 190mVp-p at 1,2GHz

feedback resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

up to 10pf.

oscillation is getting worth.

need to transmit an OOK modulated signal at around 100 MHz...

connected to VCC, which is at 5V.

above 10, than wouldn´t it be impossible to get it stable without a signal applied to the diode? Because without the resistor across the diode the noise gain should always be 1 without applying a signal.

Phil is the expert here, but I'd go for a parallel, 100 MHz tuned LC network driving the gate of a phemt.

If s/n is not critical, just dump the photodiode current into a cheap MMIC.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation 
Picosecond-resolution Digital Delay and Pulse generators 
Custom timing and laser controllers 
Photonics and fiberoptic TTL data links 
VME  analog, thermocouple, LVDT, synchro, tachometer 
Multichannel arbitrary waveform generators

eedback resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

h a perfect sine, even without a signal applied!

m 1pf up to 10pf.

oscillation is getting worth.

ssue.

re I need to transmit an OOK modulated signal at around 100 MHz...

So the PD is across pad 1 and pad 2? And R5 1.2k is not in the circuit? (I was wondering what R5 was doing.)

I must admit I'm also a bit confused by Phils advise that you need more C on the input. But I've never used a opamp in a TIA that wasn't unity gain stable.

Your circuit looks very similar to one in the LMH6629 data sheet... but there they do have a 10 pF PD on the input.

Finally I assume that you've got some nice bypass caps on the power supplies.

Please let us know when (and how) you get this working. I might have learned something today :^)

George H.

onnected to VCC, which is at 5V.

to be above 10, than wouldn´t it be impossible to get it stable without a signal applied to the diode? Because without the resistor across the diod e the noise gain should always be 1 without applying a signal.

Yes, the Resistor across the diode is not uses, sorry for that. The schematic actually is:

What I am confused about is, that the datasheet tells, that at least a gain of 10 is required for a stable operation. What gain does this refer to? If it is the noise gain, than a stable operation could never be possible with out a signal applied because than the noise gain is 0. Furthermore, if I use no feedback cap, than the TIA is nearly oscillating f rom the upper to the lower rail. With the feedback cap it is only oscillati ng with an amplitude of around 70mV. These 70mV even stay the same if I red uce the feedback resistor form 1,2kOhm to 330Ohm. When I press my finger onto the feedback network, the amplitude is only inc reasing slightly.

Maybe Phil can take away some of our confusion tomorrow;)

Cheers, Julian

Am Dienstag, 14. Mai 2013 10:25:29 UTC+2 schrieb Julian Arnold:

dback resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

mething above 20pf. In my opinion it should be stable for higher caps and o nly start oscillating if the cap is to small.

a perfect sine, even without a signal applied!

1pf up to 10pf.

scillation is getting worth.

ue.

What is your major? Because you need to go back and renew your knowledge of (if you're an EE) or learn (if you're not) the basics.

Specifically, the concept of "gain" is one that comes from circuits that have been linearized assuming their response to infinitesimal perturbations. So the noise gain of the circuit remains the same as long as the linearized models of the circuit elements remain the same.

You have the diode zero-biased, which is an error (it should be reverse- biased for speed's sake). Even so, with a small enough signal applied to the diode it's characteristics remain essentially the same -- it's going to act like a current source in parallel with a capacitance. So your gain does not change.

The noise gain, as Phil stated, comes from the capacitance of the diode (which is higher than it ought to be because of the zero bias) and C5. Unless you shine so much light on the diode that its voltage gets over a couple of hundred millivolts, its capacitance and conductance aren't going to change much.

You should have that diode reverse-biased, though. A diode's capacitance varies with applied voltage, decreasing as the diode gets less reverse biased (or, to an extent, as it gets forward biased, until it is conducting so much that "capacitance" ceases to be a meaningful property to assign to it). Reverse-bias the diode and you can get more bandwidth out of the assembly.

--
My liberal friends think I'm a conservative kook. 
My conservative friends think I'm a liberal kook. 
Why am I not happy that they have found common ground? 

Tim Wescott, Communications, Control, Circuits & Software 
http://www.wescottdesign.com

(Top posting fixed)

First: feel free to make your posts conform to normal USENET usage, instead of whatever bizarre format Google Groups is imposing today.

Second: What in heck do believe the definition of "gain" is, and why do you think that the noise gain of the amplifier changes with changing signal?

--
My liberal friends think I'm a conservative kook. 
My conservative friends think I'm a liberal kook. 
Why am I not happy that they have found common ground? 

Tim Wescott, Communications, Control, Circuits & Software 
http://www.wescottdesign.com

Don't you have that backwards? The capacitance decreases with increasing reverse bias.

Yes I do -- thanks for catching that.

I think I started to say it one way, and changed in mid-sentence to the confusion of all.

It's easy to remember if you know your semiconductor device physics: the capacitance is formed by two "plates" consisting of the N- and P-regions, with a "dielectric" formed by the depletion region around the junction. The more reverse bias, the wider the depletion region and the LESS (see, I got it straight that time!) capacitance.

--
My liberal friends think I'm a conservative kook. 
My conservative friends think I'm a liberal kook. 
Why am I not happy that they have found common ground? 

Tim Wescott, Communications, Control, Circuits & Software 
http://www.wescottdesign.com

s e

ge

Geesh, don't be too hard on him Tim. I was wondering myself about the circuit operation.

To the OP, So I also don't see why it's enough to keep peak noise gain (~Cin/Cfb) at ten. But that is the place the circuit wants to oscillate. So try a 10pF input C.

The other way to make it happen, is to reverese bias the PD (with a 9V battery perhaps) into a resistor (maybe 50 or 100 ohms) and then use the 'screaming' opamp (non-inverting) at a gain of 10-20 to 'gain up' the resistor voltage.

I've done that with little PD's and mini-circuits amps.

(Tim's certainly correct about reverse biasing the PD to whatever it can take.. if you want speed.)

George H.

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ttdesign.com- Hide quoted text -

Hmm, Hi Tim, I was reading his post differently. (Your reading is more accurate.) But reading between the lines, I thought he was asking about the noise gain as a function of frequency. Which is only one at low frequencies.

George H.

Well, he should know this stuff!

I don't like the term "noise gain" in this usage. What's _important_ is to keep the open-loop gain of the amplifier in its stable region. The open-loop gain is determined by the amplifier characteristics and the feedback network you put onto it.

It _happens_, by coincidence, that the noise gain (or non-inverting gain, your pick) is also wholly determined by the feedback network you put onto the amplifier.

So you can use that coincidence to equate a certain noise gain to stability. But I still don't like it.

Slapping a cap on the input will reduce the signal available from the diode at high frequencies, though. Your best bet is to decrease the value of C5.

(This is so freaking counter-intuitive for me, by the way, or I have my head up my ass -- I'm used to voltage amplifiers where you _increase_ the feedback capacitance for more stability, because too much capacitance on the input destabilizes the thing).

I believe that doing that will get you a bunch of resistor noise in your circuit. But if the signal's strong enough it's certainly a safe thing to do.

I hope so! It's what I do in these circumstances!

--
My liberal friends think I'm a conservative kook. 
My conservative friends think I'm a liberal kook. 
Why am I not happy that they have found common ground? 

Tim Wescott, Communications, Control, Circuits & Software 
http://www.wescottdesign.com

What?

Tim

-- Deep Friar: a very philos>

You're just lucky! :)

Jamie

Am Dienstag, 14. Mai 2013 10:25:29 UTC+2 schrieb Julian Arnold:

resistor of 330Ohm and a feedback cap of 4pf and a signal of 10MHz.

something above 20pf. In my opinion it should be stable for higher caps and only start oscillating if the cap is to small.

perfect sine, even without a signal applied!

up to 10pf.

oscillation is getting worth.

Sorry for the questions about basic stuff. I am still studying and have not heard so much about OpAmps so far, but should not the gain always be a function of frequency?

Furthermore Tim said: "The open-loop gain is determined by the amplifier characteristics and the feedback network you put onto it" but is not the open-loop gain only determined by the amplifier and can be approximated by a single pole response?