On Thursday, May 14, 2015 at 9:15:16 AM UTC-7, Phil Hobbs wrote:
> On 05/14/2015 11:52 AM, Felipe Jimenez wrote:
> > On Thursday, May 14, 2015 at 4:08:04 AM UTC-7, Phil Hobbs wrote:
> >>> Oh and I'm fairly certain the non-inverting amplifier is wired up
> >>> correctly. Between switching back and forth from the
> >>> non-inverting to the inverting configuration several times, I am
> >>> fairly certain that both configurations were wired up properly.
> >> Hmm... "several times" does make it sound like a solderless
> >> breadboard, as John suggested. Those things are slabs of misery,
> >> for sure.
> >> The AD8066 doesn't come in a DIP package, so I assumed it wasn't
> >> one of those things, but having bad layout strays is one of the
> >> ways to not build what you think you're building.
> >> Oh, one other thing--did you get your parts from an authorized
> >> distributor? Otherwise it's possible the op amp is counterfeit.
> >> (And do knock off the 'Dr.' bit--that's in my sig for SEO reasons.
> >> It's all free-men-and-equals round here.)
> >> Cheers
> >> Phil Hobbs
> > Sorry Phil...Still in grad school, so I'm just used to saying "Dr."
> > if I don't know the person and I see those letters near their name.
> > As for all the other replies including your own:
> > The pnp C-multiplier guy was drawn wrong. Drew it last night. I know
> > it was as we measured both +/-10V with a DMM and with a scope to
> > check for oscillations.
> > The whole circuit was built on piece of copper clad board. Made the
> > job tight. Lifting input pins of the op amps off of the copper clad
> > board. Used 0603 caps and resistors and 30AWG for connections. I will
> > tell my co-worker to add another C-multiplier for the photodiode and
> > to add some beefy caps on the supply pins. And to consider using
> > single package opamps instead of dual package.
> > As for his specs: 50nW of light, PD: PC5-6-TO5, BW = 50kHz, light
> > wavelength = 852nm
> Okay, good.
> As a SWAG, I'd say it's still getting in via the negative supply. As
> John says, the fact that inverting the output polarity (and hence the
> current required to drive the cable) changes the bandwidth is pretty
> At 50 kHz, a 6-foot cable (200 pF) will be pulling
> i_out = 2 pi * 50kHz * 200pF = 63 uA per volt of output amplitude, with
> a 90-degree phase lead.
> That volt of output is 1V/45M = 22 nA. To get 22 nA through 10 pF at 50
> kHz requires a supply jiggle of
> v_ripple = 22nA/(2 pi * 50kHz * 10e-12) = 7 mV.
> That's a very believable number for a lightly loaded cap multiplier.
> Using a separate one for the PD, and loading it down so that it 's
> drawing a few milliamps ought to fix it. A 2k resistor from each supply
> pin to ground will stiffen the supplies a lot.
> Meanwhile, try measuring the output with a x10 probe!
> 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
Hey Phil...Just to clarify, you're suggesting a 2kohm resistor to gnd on all c-multiplier outputs (3 of them: +/- 10V for the opamps and -10V for the PD), right? we'll try using a x10 scope probe for the non-inverting configuration.