C-multiplier again

I need a super-low noise power supply. I have a 15 volt switching wall-wart input and want as close to 15 volts, regulated, as I can get; 14 would be nice, 13.5 is OK.

The LDOs that I can find are all pretty noisy and have mediocre PSRR.

So I thought about using a Phil Hobbs-ian c-multiplier transistor, an R-C lowpass and an emitter follower, with a slow opamp loop wrapped around it for DC regulation. It looks fine on paper, simple loop to stabilize, but I figured I may as well Spice it and be sure.

What I'm seeing is mediocre PSRR. Stripping out the opamp and such, I have...

ftp://jjlarkin.lmi.net/C-multiplier.gif

which has psrr of about 70 dB at low frequencies, improving as the output cap finally kicks in at around 5 KHz. The transistor equivalent seems to look like the expected dynamic Re of about 2 ohms, with a C-E resistor of around 6.6K. Reducing Vb (and Vout) doesn't help much.

I'm using the LT Spice 2N3904 model, which I take to be a sort of generic small-signal NPN. The 33r base resistor value doesn't seem to matter.

There must be a better way, ideally one that doesn't throw away 0.7 perfectly good volts.

John

Reply to
John Larkin
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A SM boost regulator followed by a LM317 (up to 80db PSRR)?

Reply to
David Eather

You're complaining about a 70dB improvement? There is a simple way to use your 0.7 volts, well maybe 0.8 volts, to get even more rejection: change your simple NPN follower into a Sziklai connection (AoE page 95). The base resistor across the added PNP creates a relatively-fixed collector current for your NPN, which means a fixed Vbe, for improved AC ripple rejection.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Ask yourself, does it really make sense to use a unipolar power supply? With 70 dB PSRR and 1 V ripple would be is 0,3 uV.

You would have to be very careful with the ground wire topology, in order to not destroy the PSRR.

A bipolar power supply would be more appropriate and not suffer from ground loop currents.

Alternatively an RC/RC voltage divider could be used to create the virtual ground reference for the op-amps.

Reply to
Paul Keinanen

Ask yourself, did I do that math right?

It's a 4-layer board with a solid ground plane.

I have bipolar power supplies. I need very low noise ones.

The circuit I'm doing has a lot of discretes, and every nanovolt of noise hurts.

John

Reply to
John Larkin

How much current do you need to produce?

At low currents, a fast rail-rail op-amp can make a good clean power supply. They work with as little as 0.3V of head room. If you get the ones that are stable into a capacitive load, like the LT1498, you can bypass the output.

Reply to
MooseFET

15 mA maybe, fairly steady. I'm running photodiodes and discrete jfets and such.

Right, I've been considering that. I have the LM8261 in stock, a rrio C-load amp that has low frequency psrr of about 100 dB. Noise is a little high, 10 nv per, but that's already 15x better than your typical voltage regulator. And I can get a better opamp by applying money.

I'm thinking about an R-C after the opamp, 10 or 20 ohms and a 120 uF polymer aluminum cap. That only costs 150-300 mV and has a corner frequency in the 100 Hz ballpark, so fixes the opamp's PSRR falloff at high frequencies and rolls off the wideband noise. The DC feedback can still be from the output, so regulation stays good. This is practically my existing circuit, without the transistor!

John

Reply to
John Larkin

Since the problem is the Early effect, namely the effective C-E resistance bleeding ripple through, it didn't seem to me like the Sziklai thing would help. The PNP doesn't insulate the NPN from the ripple. So I spiced it. If the LT Spice transistor models are to be trusted, it's actually worse. The optimum value for the PNP's b-e resistor is zero.

John

Reply to
John Larkin

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The topology like this is stable and provides for ~100dB of PSRR.

You can also simulate huge LC filter with gyrators.

Oh, and the trivial solution would be cascading regulators one after the other.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky
[...]

Phil promised 140dB:

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Reply to
Mike

Thanks for checking that. I'd say look at their Early effect values. They'll often throw in something like VA=50 volts, which is really crazy, IMHO. It's also true that subtracting the PNP's Vbe pushes the NPN's Vce down toward 0V, where it'd be saturated and not working well. So to use the PNP, you need a bit more DC operating room (like I said), and you have to lower the NPN base voltage with another resistor to ground. Just for kicks, try another 200mV. Spice isn't on my computer right now, so I can't try it myself.

--
 Thanks,
    - Win
Reply to
Winfield Hill

To take advantage of the output C, the regulating element should have output with high impedance. Voltage controlled current source, ideally.

VLV

Reply to
Vladimir Vassilevsky

What's a more realistic Early voltage for a gumdrop NPN transistor? It would be silly if I were using a bad transistor model. I will be running at low Vce, as low as I can manage, so maybe I won't be in the really flat part. I suppose I should breadboard, grumble.

John

Reply to
John Larkin

That's cute. And complex.

John

Reply to
John Larkin

And I wonder about that wording, "stable". ...Jim Thompson

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| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

PhD design. What do you expect :)))

Here is a simpler idea:

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Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky

Did you set the ESR with C4 to some low value ?

Reply to
Jamie

Win's idea looks pretty decent to me, IIUIC:

FIG. 1 (View in fixed font) =3D=3D=3D=3D=3D=3D

Q1 2n3906 Vin >--+----. .-------+---+------+--> +13.3v | V / | | | R1 ------ | R2 --- C1 470 | Q2 | 1k --- 15uF | | 2n3904 | | | '------+---. / =3D=3D=3D =3D=3D=3D \ ^ ----- | R3 33 | +14v >---'

LT Spice says 31uV of the 50mV 1KHz ripple gets through (32dBv), and the load step is 340uV. That's a lot stiffer than the original, which had a 4.5mV load step (d(i) =3D 2mA for both).

The Sziklai version has the same ripple; I don't quite understand how Early explains that--Early should wreck the load step response too, shouldn't it?

FIG 1's load step is only 60uV if you replace R1 with a 5mA current source, the 1KHz ripple stays the same.

This shunt filter only needs 200mV headroom:

FIG. 2 =3D=3D=3D=3D=3D=3D R1

+15V >--+------------------/\/\/\--------+--> Vout =3D 14.8v | 5 | | | | | | | | .-------+------+--------+ | | | | | | | | R6 | | | | 1k | | R3 R5 | |
Reply to
dagmargoodboat

Unconditionally stable with proper parameters. I.e. stable at any positive load impedance.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky

kind of reminds me of a simple reg I made (age 14) using an incandescent lamp as part of the bias on the Power transistor.

Those were the good old days!.

Reply to
Jamie

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