C-multiplier again
in [Design]
|
Prev: Futuristic weapon question: Anti-matter-gun, would it have a signature ?
Next: EEVblog Live Event
From: dagmargoodboat on 23 May 2010 21:20 On May 23, 6:33 pm, David Eather <eat...(a)tpg.com.au> wrote: > On 24/05/2010 8:45 AM, John Larkin wrote: > > > On Mon, 24 May 2010 08:28:03 +1000, David Eather<eat...(a)tpg.com.au> > > wrote: > > >> On 24/05/2010 8:07 AM, John Larkin wrote: > >>> On Sun, 23 May 2010 13:26:26 -0700 (PDT), dagmargoodb...(a)yahoo.com > >>> wrote: > > >>>> On May 23, 11:29 am, John Larkin wrote: > >>>>> On 23 May 2010 04:28:01 -0700, Winfield Hill > > > >>>> Win's idea looks pretty decent to me, IIUIC: > > >>>> FIG. 1 (View in fixed font) > >>>> ====== > > >>>> Q1 > >>>> 2n3906 > >>>> Vin>--+----. .-------+---+------+--> +13.3v > >>>> | V / | | | > >>>> R1 ------ | R2 --- C1 > >>>> 470 | Q2 | 1k --- 15uF > >>>> | | 2n3904 | | | > >>>> '------+---. / === === > >>>> \ ^ > >>>> ----- > >>>> | > >>>> 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) = 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 > >>>> ====== > >>>> R1 > >>>> +15V>--+------------------/\/\/\--------+--> Vout = 14.8v > >>>> | 5 | > >>>> | | > >>>> | | > >>>> | | > >>>> | .-------+------+--------+ > >>>> | | | | | > >>>> | | | R6 | > >>>> | | | 1k | > >>>> | R3 R5 | |<' Q3 > >>>> | 2.7M 10K +------| 2n3906 > >>>> | | | | |\ > >>>> | | | |/ Q2 | > >>>> | | +----| 2n3904 | > >>>> | | | |>. | > >>>> | C1 | |<' | | > >>>> '---||---+----| Q1 '--------+ > >>>> 10uF |\ 2n3906 | > >>>> | R4 > >>>> | 4.7R > >>>> | | > >>>> === === > > >>>> LT Spice says 20dBV rejection @ 1KHz, zero @ d.c., natch. > > >>> Only 100 dB to go! But I don't understand Q1s biasing. > > >> 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. > > > 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. It works a lot better with the missing resistor (R2 = 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
From: dagmargoodboat on 23 May 2010 22:00 On May 23, 6:54 pm, Winfield Hill <Winfield_mem...(a)newsguy.com> wrote: > dagmargoodb...(a)yahoo.com wrote... > > > This shunt filter only needs 200mV headroom: > > > FIG. 2 > > R1 > >+15V >--+------------------/\/\/\--------+--> Vout 14.8v > > | 5 | > > | | > > | .-------+------+--------+ > > | | | | | > > | | | R6 | > > | | | 1k | > > | R3 R5 | |<' Q3 > > | 2.7M 10K +------| 2n3906 > > | | | | |\ > > | | | |/ Q2 | > > | | +----| 2n3904 | > > | | | |>. | > > | C1 | |<' | | > > '---||---+----| Q1 '--------+ > > 10uF |\ 2n3906 | > > | R4 > > | 4.7R > > | | > > ------+----------------+---- > > 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 | |<' Q3 > | 2.7M 10K +------| 2n4403 > | | | | |\ > | | | |/ Q2 | > | C1 | +----| 2n3904 | > '---||---+ | |>. | > 10uF | |<' | | > +----| Q1 '--------+ > | |\ 2n3906 | > R7 | R4 > TBD 27k | 4..7R > | | | > --+------+---------------+---- 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
From: dagmargoodboat on 23 May 2010 22:11 On May 23, 6:56 pm, David Eather <eat...(a)tpg.com.au> wrote: > On 24/05/2010 8:45 AM, John Larkin wrote: > > > On Mon, 24 May 2010 08:28:03 +1000, David Eather<eat...(a)tpg.com.au> > > wrote: > > >> On 24/05/2010 8:07 AM, John Larkin wrote: > >>> On Sun, 23 May 2010 13:26:26 -0700 (PDT), dagmargoodb...(a)yahoo.com > >>> wrote: > > >>>> On May 23, 11:29 am, John Larkin > >>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > >>>>> On 23 May 2010 04:28:01 -0700, Winfield Hill > > >>>>> <Winfield_mem...(a)newsguy.com> wrote: > >>>>>> John Larkin wrote... > > >>>>>>> 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.. > > >>>>>> 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. > > >>>>> 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 > > >>>> Win's idea looks pretty decent to me, IIUIC: > > >>>> FIG. 1 (View in fixed font) > >>>> ====== > > >>>> Q1 > >>>> 2n3906 > >>>> Vin>--+----. .-------+---+------+--> +13.3v > >>>> | V / | | | > >>>> R1 ------ | R2 --- C1 > >>>> 470 | Q2 | 1k --- 15uF > >>>> | | 2n3904 | | | > >>>> '------+---. / === === > >>>> \ ^ > >>>> ----- > >>>> | > >>>> 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) = 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 > >>>> ====== > >>>> R1 > >>>> +15V>--+------------------/\/\/\--------+--> Vout = 14.8v > >>>> | 5 | > >>>> | | > >>>> | | > >>>> | | > >>>> | .-------+------+--------+ > >>>> | | | | | > >>>> | | | R6 | > >>>> | | | 1k | > >>>> | R3 R5 | |<' Q3 > >>>> | 2.7M 10K +------| 2n3906 > >>>> | | | | |\ > >>>> | | | |/ Q2 | > >>>> | | +----| 2n3904 | > >>>> | | | |>. | > >>>> | C1 | |<' | | > >>>> '---||---+----| Q1 '--------+ > >>>> 10uF | |\ 2n3906 | > >>>> R2 | R4 > >>>> 5k | 4.7R > >>>> | | | > >>>> === === === > > >>>> LT Spice says 20dBV rejection @ 1KHz, zero @ d.c., natch. > > >>> Only 100 dB to go! But I don't understand Q1s biasing. > > >> 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. > > > So what's the quiescent current of Q1? Of Q3? > > > John > > 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 If by 46dB you mean power, i.e. 20log(Vin/Vout) = 46, yes, that's easily possible--that implies 0.5% gain accuracy. If you mean 46dBv, i.e. 10log(Vin/Vout) = 46, i.e. Vout / Vin = 25ppm, no, that ain't happening, not unless you use op amps and some mighty fine resistors. -- Cheers, James Arthur
From: Tim Williams on 23 May 2010 22:28 "Joerg" <invalid(a)invalid.invalid> wrote in message news:85u0ihFmddU1(a)mid.individual.net... > 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 :-) Or just cap couple its output to the output rail. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: Phil Hobbs on 23 May 2010 22:44
Winfield Hill wrote: > dagmargoodboat(a)yahoo.com wrote... >> This shunt filter only needs 200mV headroom: >> >> FIG. 2 >> R1 >> +15V >--+------------------/\/\/\--------+--> Vout 14.8v >> | 5 | >> | | >> | .-------+------+--------+ >> | | | | | >> | | | R6 | >> | | | 1k | >> | R3 R5 | |<' Q3 >> | 2.7M 10K +------| 2n3906 >> | | | | |\ >> | | | |/ Q2 | >> | | +----| 2n3904 | >> | | | |>. | >> | C1 | |<' | | >> '---||---+----| Q1 '--------+ >> 10uF |\ 2n3906 | >> | R4 >> | 4.7R >> | | >> ------+----------------+---- > > 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 | |<' Q3 > | 2.7M 10K +------| 2n4403 > | | | | |\ > | | | |/ Q2 | > | C1 | +----| 2n3904 | > '---||---+ | |>. | > 10uF | |<' | | > +----| Q1 '--------+ > | |\ 2n3906 | > R7 | R4 > TBD 27k | 4.7R > | | | > --+------+---------------+---- > > 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 |