C-multiplier again

That's not simpler! At least over frequency.

John

I just used the default cap, zero ESR and maybe some Spice thing lurking in the background. The bad PSRR is at low frequencies when the cap isn't doing anything.

John

Only 100 dB to go! But I don't understand Q1s biasing.

But I can replace all that stuff after R1 with a big polymer aluminum cap and get about the same rolloff, probably better at high frequencies.

*I'm* not having much fun. I've got a circuit that needs nV/fA noise levels and it's beseiged from all directions. Johnson noise. Shot noise. Power supplies coupling in through diode junctions. Switchers inches away. And I'm supposed to Gerber it tomorrow.

Wish you were here.

John

Improved ripple response (but I think a little defective - it only works when Vin drops).

When Vin drops Q1 turns on via base current drawn out through C1. Q1 robs base current from Q2 turning it off, which in turn turns off Q3 and reduces the current flow and hence voltage loss through R1.

Since you have only a volt to play with and the noise requirement you have I think a shunt regulator of some kind is your only chance.

You can use 2 op-amps with the first low noise and the second R-R

If you were at lower voltages, I'd suggest the LT6200 or the like.

With a dual op-amp, you could use the second section to servo a PNP transistor's current to let the first one always run with a low output current.

So what's the quiescent current of Q1? Of Q3?

John

IQ q1 = 0 IQ Q3 is intended to be dependent on the load 0 ma to 40 ma.

I didn't say I liked the circuit - I don't think it will even work very well or at all.

As a suggestion how about using a shunt regulator made of 2 x NSI50010 constant current sources feeding into a TL431?

My back of the envelope suggests -70db with no bypass caps. Low ESR / ceramic caps on the output should get you very close to what you want.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

Nice ASCII art. Is fig 2 from your feverish brain?

I see your idea, invert the ripple and subtract it out. Good. To do that the cancellation amplifier needs to be biased class A, so it can work over the entire ripple range. It should continuously draw current from the supply through R1, and superimpose the inverted ripple signal on top of that. R4 can be trimmed to optimize. The new R7 should be sized to handle the p-p ripple.

Then John's delicate C-multiplier filter can follower, with all the heavy lifting having been done.

+15V >--+-----------------/\/\/\--------+--> Vout 14.8v | 5 | | | | .------+------+--------+ | | | | | | | | R6 | | | | 1k | | R3 R5 | |. | 10uF | |

R2 is missing - from the base of Q1 to GND - I suggest a value of 18k but it is a weird circuit I think a ripple reduction of no more than 46db

Better would be 2 x LM334

Since you don't need much current, how about a cheap low-noise non-RR opamp such as the LM833 followed by a JFET with its drain tied to the input rail? Ok, then you have your transistor back :-)

Has anyone ever checked the noise performance of logic inverters used as analog amplifiers?

--
Regards, Joerg

http://www.analogconsultants.com/

"gmail" domain blocked because of excessive spam.
Use another domain or send PM.

Being that circuit is acting more like a current source with a max voltage of 13.5- 0.6 = ~12.9. I don't think you are going to get the desire affects.

Yup. A precision feed back is in order here.

But then again, who knows.

R.

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That's 20dBV, e.g., (ripple in)/(ripple out) =3D 100. Pretty good for standard parts with no trimming, I thought. Better cancellation needs more accurate parts, e.g. op amp + precision resistors.

Do you really need that ratio >=3D 10^12 ? Or do you mean dB(power), i.e. (ripple in)/(ripple out) >=3D 10^6?

Nope, you understand it fine! R2 got overlooked in the ASCII-art conversion--5k from Q1(b) to GND.

If you've got the room that sounds fine.

It sounds like a lovely challenge, with a few shields tossed in the mix. Not unlike RF receivers, methinks--those might inspire.

-- Best,

James Arthur

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ks

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It works a lot better with the missing resistor (R2 =3D 5k, Q1(b) to GND), and not at all without it.

Overall it's a an inverting voltage-controlled current sink that sucks more current as the input voltage rises, thus increasing the drop across R1 and sucking down the output exactly(tm) enough to cancel the input ripple. Vice versa for falling inputs.

The R2-R3 divider biases the current sink class A so that both positive and negative input ripple swings can be handled. Q1 roughly cancels Q2's Vbe drift, and buffers Q1(b), the feed forward input node. Q2-3 is a complementary darlington--a Sziklai.

Rejection is 1rst-order limited by the transconductance accuracy, so op-amps could do better.

This gadget is fast, wide-band, and decently low-noise. And because of its high input impedance, it can cover the low end of the frequency scale >10^3 better than R-C filters on the output.

-- Cheers, James Arthur

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Exactly right. I overlooked a biasing resistor--your R7--in the ASCII- art conversion. I chose a bias point to yield ~60-80mV drop across R1, so that that drop could be reduced enough to counteract the +50mV ripple swings in John's simulation.

I intended to produce a bias voltage @ Q1(b) of ~70-100mV. 5k gives

27mV d.c., plus the 7uA bias current from Q1 into 5k makes 35mV, or 62mV total. Yeah, that's a bit cheesy...needs adjusting.

This shunt cancellation approach has the advantage of low drop-out voltage, but super-ripple (and noise) cancellation depends on precision gain setting.

John? He'll probably do better cascading a couple of his simple stages--dirt simple, no precision needed. But he'll need more voltage to run it--there's no avoiding that. Hmmm, maybe I'll sim that...

-- Cheers, James Arthur

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If by 46dB you mean power, i.e. 20log(Vin/Vout) =3D 46, yes, that's easily possible--that implies 0.5% gain accuracy.

If you mean 46dBv, i.e. 10log(Vin/Vout) =3D 46, i.e. Vout / Vin =3D 25ppm, no, that ain't happening, not unless you use op amps and some mighty fine resistors.

-- Cheers, James Arthur

Or just cap couple its output to the output rail.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

The Kanner Kap uses an audio power amp to do this, applying a small amount of positive feedback to multiply the value of a BFC. Works OK, but it isn't worth paying royalties on.

Cap multipliers are magic--especially two-pole ones. It's 0.7 volts well spent IMO. If Early is a worry, use a slower transistor--the ripple rejection is basically C_CE/C_BFC, with some degradation due to Early voltage and capacitor ESR.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net