Huh, cap multiplier needs an idle resistor...

So testing a photodiode TIA, I put a cap multiplier on the bias supply. Weird behavior, mostly not fast enough. I stuck 10 k ohm resistor to ground from the emitter end, and that cured it. So that's ~1 mA of idle collector current. Is that a good number? Max collector current is ~10 -20 mA. (And you are correct, it's the first time I used a cap. multiplier directly to a photodiode.)

George H.

Reply to
George Herold
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You're setting r_e (effective emitter resistance) with bias current. If bias goes to ~zero, r_e goes up and so does the output voltage.

That should have more to do with output voltage fluctuation than filtering performance, but sure, why not.

Tim

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

How much voltage do you have across the transistor?

If the current is too low the base current may not be enough to give the required headroom across the transistor compared to the noise/ripple at the input.

When I last used a capacitance multiplier I put an additional resistor from base to ground to ensure that there was a couple of volts across the transisto

Can you show the circuit?

kevin

Reply to
kevin93

Do you have a cap to ground at the output of the c-mult? If not, the pd hf response will droop when the transistor has low current/high emitter resistance. You really need an output C.

Unless you expect a lot of pd current, usually a passive RC filter will be as good as a c-multiplier, and doesn't lose as much voltage.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

Sure here's a scribble.

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Vce = 600 mV (with 10 k ohm R on output.)

George H.

Reply to
George Herold

On Tuesday, July 25, 2017 at 1:33:42 PM UTC-7, George Herold wrote: ...

I suspect that there is not enough voltage dropped across the 500 ohm base resistor to provide the working voltage for the transistor - adding emitter current increases the voltage or the way I did it was to add a resistor to ground form the base.

Actually 500 ohm seems a bit low - you could use much larger one there to either get better ripple rejection with the 100uF cap or reduce the cap somewhat.

kevin

Reply to
kevin93

Also of course the bias current will increase the gm of the transistor, which may be useful.

kevin

Reply to
kevin93

Yeah I had just an RC low pass before.. trying to be fancy. (KISS) (Hanging C on circuit.... not much change)

Say I've got another question. How fast can I turn on/ off a 'typical' red LED, through hole, plastic case. Or what is an LED capacitance? For this LED I know C is about 20 pF when reverse biased (to ~20 V)

My step response at the lowest gains looks strange. Here's 333 ohm

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and 1 k

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Chan. 1 is LED voltage (LED + 250 ohms) Chan 2 is PD response.

George H.

Reply to
George Herold

LED capacitances can be all over the place, but easy to measure. I've seen laser diodes with very strange step responses, but LEDs are usually pretty simple. An LED shining directly into a fast photodiode

  • 50 ohm scope would be a good crosscheck. I tested one CREE white LED, and it was nanoseconds fast, phosphor and all.

Is the blue trace your TIA output? It does look a little strange.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

Add a cap across that 10K!

Do you have Phil's book? He talks a lot about PD amps.

What's the voltage rating on that tantalum cap?

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

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

500ohms B-C is the same as a direct short. Here is what you need(Warning: LTspice file):

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Reply to
Steve Wilson

The led is kinda dim, turn off is slower. I've got a diode laser I can wiggle at (at least) ~100 MHz. But it's more complicated than an LED.

Right I think I need more compensation capacitance. but the led is not fast enough... (maybe..?) I should post the step response with no compensation across the feedback R... It rings, many times, (~5) but not with enough amplitude.

George H.

Reply to
George Herold

I did that, it brought out the ringies a bit. Tomorrow, I'll go back to RC low pass.

Yeah I love Phil's book. (I guess I should read again.)

35V. (I had a 35V (cheap) tant. fail at less than 24 V... that circuit now has a 50V tant. but maybe I should make it a ceramic.)

George H.

Reply to
George Herold

Steve, I'll have to look at that tomorrow. I picked

500 ohms assuming a beta of 100 and a max current of 25 mA, so 250 uA of base current, times 1 k ohm would be 250 mV, I wanted something less... maybe more R would help..?

George H.

Reply to
George Herold

You mentioned ringing, so I took a look at it. The PNP does have a bit of overshoot like a second order system with a damping factor of about 0.7

I did the same with your emitter follower, modified with a much larger base resistor and a cap on the output. Surprisingly, it gave deeper attentuation than the PNP. It also had no ringing, but sinply decayed to a slightly lower output voltage when current drain was applied to the output.

I'd upload the files, but Google Drive is giving me serious problems with my browser and logon issues. I'm looking for a better file host with fewer hassles.

Reply to
Steve Wilson

maybe it's saturating..

vr1=ic/beta*r1 vr2=ic*r2

vr1+vr2+0.6=12

ic/beta*r1+ic*r2+0.6=12 ic*(1/beta*r1+r2)=11.4 |beta=100,r1=500,r2=10k ic=11.4/(5+10k)=1.14mA

vr1=0.0057 maybe too small

2V here?

either r1 to 175438Ohm

or change r2

2=ic*5 ic=0.4 a bit too hi pick: ic=0.01 r1=20k

2+0.01*r2+0.6=12 r2=9.4/0.01=940

Reply to
Johann Klammer

That's important. The transistor hasn't much gain until you bias it, and the stray capacitance (base stored charge) is rather high in saturation. The C-B junction, like any other diode, has lower capacitance when it's reverse biased. And, that C-B diode capacitance is a sneak path from the raw DC source of ripple. At some (high) frequency, Rbb will decouple the base filter capacitor... but not Ccb.

Reply to
whit3rd

It follows the concept of a constant current source driving a capacitor and is intended to filter noise from the incoming supply. What was surprising is the heavy voltage drop on the output with a very light load. No wonder Jeorg doesn't like PNP LDOs

Reply to
Steve Wilson

My fave for low current cap multipliers is the 2SK2704K superbeta low V_CEsat NPN. The main sneak path for a high-isolation CM is Early voltage, so an RC in the collector circuit helps a lot.

A two-pole filter with 2k and 10 uF in the collector and a couple of

20-us sections in the base gets you about 130 dB suppression in the 10-300 kHz region. (You have to manage the bias conditions of course--this one is making a quiet 3V @ 4mA from a noisy 15V, for the drain of a JFET bootstrap in a 10-Gohm TIA.

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

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