> elec > misc > how-to-remove-power-supply-ripple-capacitor-multiplier-eevblog

EEVBlog #1116 - How to Remove Power Supply Ripple

EEVblog - 2018-08-27

Circuit building block time. The capacitance multiplier and how it gives almost negligible power supply ripple compared to a voltage regulator.
Whiteboard theory and then some bench demonstrations and experiments. Plus a twist at the end that proves that the "Capacitance multiplier" is perhaps one of the most mis-named circuits of all time.

Forum: http://www.eevblog.com/forum/blog/eevblog-1116-the-capacitance-multiplier/

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Trent - 2018-08-28

Brilliant, Dave. Wish you did more of this content. Bring back fundamentals Friday!

Bill A - 2018-10-28

This was a really great explanation, learned a lot :)
I've just realized that Dave looks a bit like a slightly older version of the LinusTech Tips guy (lol), here: https://www.youtube.com/watch?v=rmW1evMMMMY

Vishal Dhayalan - 2018-12-28

Yes! Please do those again

Dave Micolichek - 2019-01-13

Great tip for smaller audio circuits!, when you dont have space on the board, or enclosure, for large electrolytic caps. I totally forgot all about capacitance multipliers through the years. I can think of a few times when this would have been beneficial. Now I'll have this in my brain for a few more years, before i forget about it again.

Sjoert de Boer - 2019-05-19

Agreed! Circuits like this, make our lives a little bit better :)

Moshé Levy - 2020-01-10

quand on est brillant, on ne dit pas qu'en sortie d'un redressement double alternance avec un chimique de filtrage, la tension a cette allure mais plutôt une droite horizontale!

FotatoPotato - 2018-08-28

Wow, this video just saved my ass... I had 350mv ripple on a crappy power supply at work and it was messing with some of my testing and this circuit got rid of it all! Thanks so much Dave! You're a real life saver!

The Reynolds - 2018-09-18

@YiZhou Wang You still depend on a voltage regulator. The V+ going into the capacitance multiplier comes from a regulated source.

YiZhou Wang - 2018-09-18

yeah! the drop voltage induced by the capacitance multiplier can not be ignored when output current is large. or we can say that the capacitance is not included in the closed loop.

Revi M Fadli - 2019-01-02

@YiZhou Wang if the power supply's hackablr enough, one could replace its feedback circuit(probably a voltage divider) with one placed after the capacitance multiplier

Juan Carlos Rodriguez - 2019-08-09

YiZhou Wang yo, he specifically states that should one require stable output voltage, then one should use a Voltage regulator

Frank Reiser M.S. - 2019-08-22

Most inexpensive ( yea right!) power supplies simply use a parallel electrolytic capacitor as a filter to reduce ripple. The best filter is an inductive pi filter. It consists of, in parallel, of a capacitor, a series inductor, and another parallel capacitor. It looks on a schematic like the Greek letter pi. This takes of capacitive and inductive ripple. It precedes a linear or switching power supply.

tocsa120ls - 2018-08-28

s. 578 in The Art of Electronics :)
And if the ripple is in the MHz range, you can add a ferrite bead to the base lead of the transistor.

DGerber - 2018-08-30

Chapter 8.15.1! :D

Oğuzhan Yılmazer - 2018-11-09

he could be turkish too

Jim Mitchell - 2019-05-05

Low R's in seriously 1 penny I can't remember how many boxes of 39 ohm chips (that was series and I cent vs 4)

King Solomon - 2019-12-01

Why the base? Why not the emitter? The current coming from the base is very minimal compared to that moving from the collector to the emitter? Can you please explain this maybe im missing something here thanks

Jimmeh B - 2019-12-13

@King Solomon Transistor gain (hfe), and in this configuration It's a "voltage follower".
I've never seen or thought of this before. It's a very clever little circuit.

Herby Bey - 2018-08-28

Nice explanation, Dave. There is a trap for young players here. If the input voltage ripple amplitude exceeds Vbe-Vce~=0.6V, this circuit doesn't work properly anymore, because the collector voltage drops too close or even below the emitter voltage. At that point, only the base current makes it to the output and there is no current gain anymore. An additional resistor across the capacitor can be used as a voltage divider to solve this issue. This will lower the base voltage, which lowers the emitter voltage, and allows the input voltage to drop lower before there is no current flowing from the collector to the emitter. I ran into this problem with a low power, high voltage boost converter that had more than a volt of ripple.

Zarcondeegrissom - 2018-08-29

yeah, I was kind of thinking about that, and wondering if putting a cap on the output would even make any diff at all other than smoothing out the current draw of the load through the bypass transistor. and, at that, what is preventing that second resistor in the divider, from being a simple zener, to further stabilize the voltage of the output. I guess it mostly depends on what the load is, and what the supply is doing. lol.

I also found, for many low power things, a simple in-line diode with a cap after it was sufficient for a single device without trying to stabilize an entire power buss just for a single chip to be happy.

Nine thirty one - 2019-03-27

Does this still occur if I were to use a darlington pair?

Jonas Daverio - 2019-05-11

@Nine thirty one I think this is getting even worse, because the voltage drop is even bigger, about two times the voltage drop of a normal BJT.

Paul Max Avalos Aguilar - 2019-07-28

Does this happen with a mosfet?

Jim Viau - 2019-08-16

Paul Max Avalos Aguilar : Actually, with MOSFET, you have the threshold gate voltage lost necessary to turn the MFT on. Which is typically in the range of 2 to 6 Volts. This value would be the "room" available for ripple elimination. The hick with MFT is that this threshold gate V significantly vary with Source current, so this would create a situation where you would loose even more voltage than with a BJT. The advantage though is a better ripple filtering than with BJT, as long as your initial ripple is lower than V gate.

Lasse Knudsen - 2018-09-01

This type of video, where one "building block" is explained in detail is my favorite.

Fran Blanche - 2018-08-28

NPN transistors for the win!

SJD - 2018-09-22

That was interesting Dave hope u keep posting such fundamental videos.

Philip Akpojiyovwi - 2018-10-22

A mosfet or igbt depending on the amount of current/voltage u need then clamping the gate voltage with a zener diode then another low pass filter will do well

Jimmy O. - 2019-01-04

@Philip Akpojiyovwi i would be worried the Zener diode would interfear with the function. wouldnt it stop some of the ripple being seen by the gate? thus allowing the ripple to simply pass through the fet?

Philip Akpojiyovwi - 2019-01-04

@Jimmy O. Remember the gate signal is a replica of what must pass through the MOSFET /IGBT in an amplified form, a ripple free gate signal means a ripple free amplified signal

Frank Reiser M.S. - 2019-08-22

They are great for amplifiers.

David Perkins - 2018-08-29

12:30 - This circuit also provides a soft-start as C charges on power-up

Dave Micolichek - 2019-01-13

Yup. No NTC thermistor needed.

Josias von Leiswolf - 2019-02-06

If one wanted to avoid using an op-amp, could a „capacitance multiplier“ circuit be used as an active low-pass filter for audio signals?

DJ Fix - DJ and Synth Repair - 2019-07-21

we usually just hire an intern to watch the scope with one hand on the regulator's Vout control. They've gotta have good reflexes though. And coffee.

TrickyNekro - 2018-08-27

I really missed these videos! And well... that was (maybe?) new to me, naming is indeed as you said tricky... It makes sense though, plus the transistor as a primarily a current source has a really high noise rejection by itself... Yeah! Cool stuff! BTW... I really don´t get why it would work with a MOSFET, you have to run it in the non linear region and it would be rather messy.

TrickyNekro - 2018-08-28

You know, I was giving it some thought and what confused me was the DaveCad layout. Yeah that´s practically an emitter follower but you are saturating the transistor. Saturated stuff generally not behave well especially with AC. What if you replace that resistor with a divider, depending on the transistor the drop you aim for isn´t that big (0.2 - 0.5 of a volt for a BJT) and you anyhow accept that you´ll have some voltage drop there. You can still calculate everything easily with Thevenin equivalent resistor for the divider. And you´ll have the transistor operating in the linear region which makes it a current source which in AC is an open circuit (of course you´ll still have noise coupled from the capacitances of the transistor but that might be more manageable). And yeah you´ll be burning power on that transistor a bit more from your suggested citcuit, but when noise is the problem and you only power an op-amp or two. Well, that might be a better solution. I haven´t tried it though to see the difference, but I know why it reminded me something. I hope you see it and give a reply, or why not make a follow up video! Cheers, mate!

Vedran Alajbegovic - 2018-08-28

So nice Dave...thnx

Dan Douglas - 2018-08-29

|How about an op-amp compensated mosfet?

TrickyNekro - 2018-08-29

Eh? Well that sounds awfully a lot like a regulator :-P

Evil Gremlin - 2019-01-06

Yeah, that's why people invented IGBT's :) You could use MOSFETs in such cases, no problem, really. But IGBTs at least will heat up twice less :)

Xmod - 2019-10-04

"Active low pass filter" sounds better than "capacitance multiplier"

Richard Smith - 2019-11-19

Inductor + capacitor instead of resistor + capacitor. Really effective.

kbruin79 - 2019-07-08

I can’t thank you enough! Sharing your knowledge is such a big a gift to the EE community.

Tony Bell - 2018-08-28

Brilliant, thanks Dave..... Love these fundamentals..... Useful circuit.....

Cynthia Cantrell - 2018-12-28

In telephony, this circuit ( Rb, C, Darlington Q, and Re) are used as an "electronic inductor." (Re is the load resistor - RL in Dave's circuit). If you measure AC impedance from V+ to ground, (or do the math for this circuit) you'll find that at low frequencies, this impedance is Re - which is typically small. At high frequencies, the impedance is Rb, which is larger. In-between, the impedance goes up at 20db/decade - the same slope as an inductor.

This assumes you are below the frequency where Miller capacitance is a problem - above that frequency, the impedance will start dropping again.

This circuit was used to replace the inductance of the ringer coil when newer phones came out. The phone lines were specified to require a certain amount of inductance across the leads when on-hook. When big mechanical ringers were replaced with electronic ones, they still had to keep that inductance in the circuit - hence the "electronic inductor" circuit came to be. The effective inductance of this circuit only needed to appear in the audio band.

When using this circuit for ripple reduction, I've heard it referred to as a "rip-red," i.e. short for "ripple reducer." In that case, Re is replaced with the load you're reducing the ripple to.

TheTruthSentMe - 2018-08-28

Glad you're still doing these. They're fantastic.

toast_recon - 2018-08-27

Thanks Dave! Even after getting my BS, your fundamentals videos continue to educate. School can't cover everything :)

Mdmchannel - 2018-08-28

toast_recon You’ll learn more by designing and building circuits.

toast_recon - 2018-08-28

I've found myself kind of accidentally in communications research at the moment, so at this point I'm not really advancing my circuits knowledge outside my free time! I need to dedicate more time to DIY electronics fiddling.

Vaes Joren - 2018-08-28

A good education does not teach you all the circuits, because it can't and there are too many. Instead, it focuses on the concepts, rules, and principles we use to build up those circuits, so you can more effectively compare and learn new circuit topologies as you progress through your engineering carreer.

Mdmchannel - 2018-08-28

toast_recon I was lucky as an engineering tech in the early 80’s prototyping circuits and testing them before the simulation days.

Robert - 2018-08-28

Learned something again, thanks Dave. Great stuff.

Kevin - 2018-12-25

For one reason or another, I forgot this circuit! Genius! Thank you!

elektrinis - 2018-08-28

its a really smart way of making your SMPS "linear". Probably a nice solution for compact audio amps and lab psus.

Shim267 - 2018-08-31

I fell for the voltage regulator meme...
I just recently learned this lesson the hard way and I was looking for a better solution.

Bless you, Based Aussie.

John Overstreet - 2019-01-04

Great description Dave! In the "old days" we called that a ripple stripper, not a capacitor multiplier.

Fried Mule - 2018-12-18

4:45 Am I an idiot (dont answer that:-) I cant find the "linked" video.

TheCreeperkiller - 2019-11-28

26:00, the Op Amp also has its own power supply rejection ratio, if its positive supply is the same ripple injected power supply going into the multiplier, then the ripple rejection will only be as good as the op amp's PSRR.

Vital_net Vital_net - 2019-06-28

I can't find the link of the full demonstration video of regulator voltage you mentioned
I love all your videos !!

Hedley Davidson - 2018-08-28

Thanks - reminds me of the early days of eevblog - great value . “ the best form of marketing is education “

worvtube - 2018-08-28

Nice! Could we use LC filter instead of RC?

zx8401ztv - 2018-08-28

Smashing video dave :-D
I've seen this configuration a lot in basic linear power supplys, but never thought too deep about the filter magnifier effect.
A handy ripple reducer as long as you don't mind a bit of voltage loss :-).

Uwe Quast - 2018-08-28

enjoyed this "found a mental fryed a" :-) episode very much, ... as usual
absolute brilliant, Dave, keep going !

Tom Sparks - 2019-07-19

More of an RC filter effectiveness amplifier. I've seen this before, but it was never explained this well before.

Charles Smith - 2020-01-07

We used to use chokes with capacitors on both sides. Worked pretty well in power hungry circuits. What happened to that?

Fried Mule - 2018-08-28

Perfect video, just as I need it! :-)
These types of videos was why I did subscribe in the first place, so thanks a lot for making more of this kind!!

David Perkins - 2020-02-08

25:50 - Positive feedback via R2 and R1--an accident waiting to happen :)

jarrod1937 - 2018-09-08

Thanks for covering this, first time hearing of this topology, really cool!

Wade Spring - 2019-09-02

This is amazing. This is the first video I've watched about ripple removal that I both understand and actually works!

Denny - 2018-08-28

The fundamental content videos are always nice, thanks for showing!

gamerpaddy - 2018-08-27

What power does a fet or bjt dissipate in this circuit ? Just the ripple voltage x current or are there more factors than internal resistance / wiring.

EEVblog - 2018-08-27

Basically the input-output voltage drop multiplied by the load current

elektrinis - 2018-08-28

0.6V times current. A small price to pay in low power applications (below 10A or so).
You may look for lower dropout device, however you have to make room for the ripple, so your drop has to be higher than ripple Vpp+0.2V or so.

Hu Sun - 2018-12-26

The only thing is that voltage drop is really high, which compromises the efficiency of DCDC converter

pepe6666 - 2019-12-27

this is the ripple solution of my dreams. dave, you are a wizard of the blue smoke.

DriveJapan - 2018-10-04

Every time I hear Dave say "Ripple" I keep thinking of Sanford and Son

Peter Kara - 2019-04-06

Wou, so much enthusiasm and delight. I love this video.

Yaroslav Salnikov - 2018-08-28

This circuit can't filter out ripples higher then voltage drop across transistor.

FangFei Yang - 2019-01-01

"active ripple cancellation"

Lic. Eugenio R. Rodríguez - 2019-03-09

EEVblog one of the best on youtube, excelent !!!

Angel Calzada - 2018-08-28

Nice tip, great demonstrations and awesome explanations! I was also missing these basic electronics videos ;). Excited to try it this weekend :D

Fir3Chi3f - 2018-08-28

Love that frame rate, thanks for the video Dave!

Cristian Bureriu - 2018-12-23

I was gonna say you could use a Darlington...

antigen4 - 2019-07-23

but how does it perform under a LOAD?

Wade Spring - 2019-09-02

17:34 There's a 270-ohm resistor for stimulating the load.

Gibran PantalaLabs - 2019-06-03

Excellent video and explanation !!! Thank you a lot ! I have a question : how to remove the ripple from the negative rail of a symetric PSU?

guillep2k - 2018-08-28

Thank you, Dave!!!

Julian Mallett - 2018-12-29

its like watching Horowitz and Hill the movie, my all time favourite electronics manual in video form. Love this stuff Dave.