Good evening! The other day, I decided to share my post on my power supply design to r/ECE- a subreddit dedicated to electronic and computer engineering full of some very kind folks who have similar interests to my own. In said post, I also included what was my full design schematic at the time as well as the design for the power supply. Now with me being the naive fool that I can be s̶c̶a̶r̶i̶l̶y̶ ̶o̶f̶t̶e̶n̶ occasionally, there was a couple of things about my design that I missed at first (although would have come to realise later on when I went to focusing on that part), with one being that I had my N and P channel MOSFETs wired backwards for the bridged push-pull configuration I planned to use them in, which would have led to what was in essence a short between the positive and negative rails and many bad things would have followed that. I have since switched them to the orientation they should be in and everything is now looking more okay on that front. One thing I did find however was that nobody made any comment on my 12AX7 pre-amp stage, which either means that a) it’s all absolutely fine with no design issues, or b) nobody who saw the post knows much if anything about valve pre-amp design and I am in for a wild ride! Either way, I like to go with the thought that no news is good news.

Anyhow, the main talking point of the post was my supply design, and people definitely had things to say about that…

Two main points were raised about my design, the first being temperatures. As u/TesticlesTheElder put it ever so elegantly:

This isn’t an audio amp, it’s a heater that happens to amplify audio.

Thank you u/TesticlesTheElder, very cool!

My main reasoning for using linear voltage regulators were initially as follow:

I was under the impression that linear voltage regulators both dropped down the voltage to give a constant, stable output and removed any noise and ripple from the lines. Reducing the voltage down to something significantly lower than the input allowed me to keep my maths relatively simple for calculating power outputs and saved me from having to work to find voltage sag, be that experimentally or going into what I thought at the time would have been quite complex mathematics involving vectors and argand diagrams to calculate output impedance at different frequencies and so on and so forth.

In short, I was dumb and lazy.

However on the same night I shared said blog post and at practically the same time most of the comments were coming in (about 1AM GMT, which is 5PM PT, or roughly when all the Californian and other west coast electronic engineers were getting home and booting up to the front page of the internet) I was watching a video by EEVBlog which had popped up in my YouTube recommendations titled “How to Remove Power Supply Ripple” which opened up talking about different levels of ripple and why you’d want rid of it and that, actually, linear voltage regulators do quite a poor job of removing ripple (which is the main reason I chose to use one in the first place). Finding this video (if you can call an AI algorithm plopping it in front of your wandering bored eyes “finding”) was almost as significant of a source as finding the page on the Valve Wizard that not only pointed out to me that I had my view of valves incorrect and that the anode connection is actually the inverting output, but that also shows me how to incorporate local negative feedback into the loop to control the gain! Mind you, once I’d realised that I just had my ideas a bit backwards I felt like an idiot for not realising it soon. It’s practically like an inverting op amp configuration, except you feed back from an inverting output to the input rather than from an output to an inverting input. Anyway so of course I decided that as soon as I woke up the next morning (12PM) that I would get on modifying my design to incorporate such a technique.

As is the title of the video, not only does Dave discuss the poor ripple rejection of linear voltage regulators but he also does offer a very powerful solution; this solution being the questionably titled “capacitance multiplier” (which I now read in my head with his voice and accent every time I read it because there seems to be something about watching something quite late at night right before you go to bed that means that certain details of what you’ve just learnt are burnt into your memory). I must give it to Dave though, the video is very well put together with regards to going into important information, as well as him addressing himself the issues with the name.

What is a capacitance multiplier?

A capacitance multiplier is a circuit used to remove unwanted ripple from supply rails (if I hadn’t already made that apparent). It is quite a simple circuit that uses the properties of passive low-pass filters and transistors in some nifty ways to achieve some quite remarkable results with regards to ripple rejection. Below is a circuit diagram of one such multiplier.

If you aren’t aware of what a low-pass filter is, it involves a resistor in the path of an incoming signal with a capacitor coupling it to ground. Because of the nature of capacitors, they have an impedance that varies with the frequency of a signal and as such can be used to make a voltage divider to ground with the resistor. What this does then is that at higher frequencies the capacitor’s impedance decreases towards zero, meaning that most of the voltage at higher frequencies is dissipated across the resistor, whilst lower frequencies the capacitor’s impedance goes towards infinity, meaning a smaller fraction of the voltage signal is dissipated across the resistor and in turn it can “pass” through without issue. When used with some mathematical trickery, the resistor and capacitor can be valued as such so that frequencies below a certain frequency pass without issue, whilst signal above that frequency are attenuated by 3dB for every octave above that cut-off frequency you go.

These passive filters are great for things such as bookshelf speakers, where you have a smaller tweeter that is meant for high frequencies, and a larger mid range and low frequency driver for lower frequencies. If you have a tweeter designed for high frequencies (that are usually much less abundant in audio and as such are generally of a lower voltage) then you don’t want your large low frequency bass running through it and blowing your speaker. However, for things such as power lines passive filters on there own can pose issues. If you were to use it on it’s own, the resistor would be effectively in series with your load, which limits the voltage out to your device as well as being wildly inefficient as larger loads would draw more current which would mean much more power lost to the resistor and heat generated. Instead, what we’ve done with a capacitance multiplier is connected the base of transistor Q1 to the output of the low-pass filter made by resistor R5 and capacitor C11. Given that Q1 is in a voltage follower configuration, the output voltage will ideally be the same as the input voltage but with a large current gain, which means we get the smoothed DC signal but with a much greater current than the passive filter would be able to supply otherwise.

If you look at the top graph of voltage in against time, you will see the talked about ripple on the rail. The dashed line represents the DC portion of the voltage rail at an unspecified voltage, whilst the solid line represents the actual output with the AC portion added on top of the DC. The AC is the ripple, and we want rid of that.

Looking at the bottom graph of voltage out against time, we can see that the capacitance multiplier has in fact removed the ripple. However, you will notice that the actual voltage out is lower than our dashed line, which I’ve left in the same place on the bottom graph as I have on the top. This is due to the imperfect nature of transistors, and represents a voltage drop of about 0.65V across the transistor. In this case, the dashed line represents the theoretical maximum DC voltage out.

Why would you need to get rid of ripple?

For the most part, ripple is bad. Whilst in some cases the ripple rejection of a simple voltage regulator- whilst poor by comparison- is probably enough, in my case it is very much necessary to remove as much ripple as possible. This is because the ripple in the supply rails can carry down to the other parts of the amplifier and be amplified to the point where they’re audible in the output signal, which can give unwanted hum and distortion. As well as this, a large enough ripple can cause the degradation of some capacitors as it causes unwanted power losses and heating in them which will shorten the life span. It can also cause issues with sensitive digital electronics that require very smooth, regulated supplies.

Why are they called capacitance multipliers?

The main reasoning behind calling this circuit design a capacitance multiplier is that it, in a way, multiplies the capacitance of the capacitor in the low-pass filter. However by using that naming process, they could as easily be called resistance dividers. If you go to the Wikipedia page for capacitance multipliers, it will tell you:

… the capacitance of capacitor C1 is multiplied by approximately the transistor’s current gain (β).

While this stands relatively true for how they work, it’s also not the theory in it’s entirety.

For any passive filter, be it high-pass or low-pass, the equation for the cut-off frequency is as follows:

f=(2πRC)⁻¹

With R being the resistance of the resistor in Ohms, and C being the capacitance of the capacitor in Farads.

As the theory claims, the capacitance of the capacitor C11 in the diagram above is multiplied by the gain of transistor Q1. In this case, let’s say the capacitance of C11 is 31.8 microfarads (31.8uF; 31.8*10⁻⁶F). If we wanted a cut-off frequency of 5Hz, we’d need the value of R5 to be 1 kiloohm (1kΩ, 1000Ω) for the signal to be attenuated by 3dB at the output of the filter. The reason I claim that the theory isn’t stated in it’s entirety is that for a transistor of gain 100, the capacitance would be multiplied by 100 to give a value of 3.18mF, and in turn give a cut-off frequency of 0.05Hz. Experimentally, this isn’t the case. The cut-off frequency stays the same.

As I stated earlier, the circuit could just as easily be called a resistance divider as it also does this.

For our filter circuit, if the capacitance of C11 was to increase, the resistance of R5 would have to decrease by an equal multiplier to give the same cut-off frequency. Because we have the same cut-off frequency out of our capacitance multiplier and a claimed “multiplied capacitance”, our input resistance must have also decreased by one over the multiplier in order to give the same value for the cut-off frequency.

So what is this multiplier that the capacitor is increasing by but the resistor is decreasing by?

β.

The capacitance increases in value by the product of it’s capacitance and the gain of the transistor, whilst the resistance has been divided by that gain.

To put it into numbers, our theoretical gain is 100, our capacitance is 31.8uF and our resistance is 1kΩ. If the capacitance is increased by a factor of 100, our new capacitance is now 3.18mF. However, the cut-off frequency of our filter is the same, and as such the only other variable in the equation is resistance and in turn that must mean that the resistance of R5 must have decreased by a factor of 100. This gives out new value to be 10Ω. If we plug these new values into our equation for cut-off frequency, we’ll find that we get the same value of 5Hz.

Or you could simplify it down to the fact that the transistor is giving unity gain on the voltage so it only passes the constant DC with an amplified current and that neither the capacitor nor resistor are actually changing in value.

My new design

So far I’ve written 2105 words in this post and haven’t even started to talk about the actual modifications I’ve made to my power supply design, so I think it’s about time I get to the moment everyone has been waiting for.

So probably the first notable difference is the complete lack of regulators. I decided to do away with them entirely after incorporating the capacitance multipliers into the circuit to smooth the rails, as well as taking into account everyone’s comments on how toasty it would all run (and a few people who made comments that having one running at the negative rail to supply a negative voltage with a positive regulator would be very difficult if not impossible, and considering the time frame I have left to produce this piece I decided against researching methods of regulation for the negative rail). I said earlier on that two main points were raised. This point about negative voltage regulation using a positive voltage regulator was that point.

Doing away with the voltage regulators now meant that I had to work with the maths involved in voltage sag, which I at first thought might be somewhat difficult.

When I was studying for my A-level physics last year, one of the topics that fell under the electricity category was that of internal resistance of batteries. We learnt that the voltage out of a battery drops relative to the load on it, as they have an internal resistance that dissipates a larger voltage as the load increases as it is in series with the load and acts as a voltage divider, with the larger resistance dissipating the larger voltage, so as the resistance of the load decreases and the current increases, the voltage lost to the internal resistance also increases as it makes up a larger fraction of the overall resistance of the circuit. However, one portion of the exam (which we weren’t meant to do in the exam but I did on a whim because I had a better feeling about it than the medical physics stuff we had actually covered in class) was alternating current. In said portion, one part we were required to be able to calculate was the total impedance as well as the phase angle of a circuit consisting of a resistor in series with an inductor or a capacitor. In the case of the total impedance, it is not as simple as just adding the resistance of the resistor to the reactance of the inductor or the capacitor but instead requires the use of the pythagorean theorem involving the resistance and reactance to calculate the magnitude of the impedance.

Because of this I was under the impression that to calculate the voltage sag of the transformer under load I had to account for things such as it’s inductance and in turn it’s reactance at mains frequency, as well as series resistance of the transformer’s secondary with the load itself. Unfortunately, no matter what I searched I struggled to find any answers to my questions and as such ended up turning to r/AskElectronics as a last resort in the hopes of finding an answer. Fortunately, I found an answer. Very quickly. And it was very simple. My initial question was:

And I received a response from u/1Davide who said, quite simply,

“Yes, it’s the same.”

So that came as a relief that it would be as simple as what we covered in physics relating to batteries.

But who is u/1Davide? Are they a reliable source?

I can’t say for certain, of course. They’re just a stranger on the internet. However, from looking at their internet presence they are a moderator on the r/Electronics subreddit (who has in fact removed one of my posts on there in the past where I shared my blog for it not being related to electronics enough, which given the nature of it at the time seems reasonable enough) and they specialise is Li-ion battery management systems and connectors. Clicking on their profile provides considerably more information and shows them to be one David Andrea, who is an electrical engineer in Longmont, Colorado, USA, and is also the author of the book “Battery Management Systems for large Lithium-ion battery packs” alongside a number of other achievements. I think the tl;dr of this is that u/1Davide is quite a reliable source.

With this confirmation at hand, I was able to reference the datasheet of my transformer for the information I needed and make required calculations. The transformer is rated at a peak ±12Vrms, and as such I calculated the voltage drop to be around ±0.52V on each rail, giving us an output of ±11.48Vrms.

As for the output after the bridge rectifier there are a few notable changes. The first one to be encountered is that there are now 4 4700uF capacitors instead of the initial two. This is because, as it turns out, it’s quite hard to find capacitors of such a high capacitance that are rated for the ripple current that this amplifier can draw at max load, especially ones within a reasonable price range. As such I’ve chosen to just add an extra capacitor in parallel with the previous ones on each rail as to double the ripple current capability.

Following these new capacitors are the capacitance multipliers I’ll be using for the ripple rejection. You will notice that it looks slightly different to the example I gave above, and this is because I have chosen to go with a Darlington transistor configuration made up of two MJE3055T NPN transistors on the positive rail, giving a minimum current gain of 440 which will be more than enough to drive the outputs. The issue with this however, is that using transistors in a Darlington configuration as I have done above doubles the voltage drop on them, instead causing a drop of 1.3V instead of the original 0.65V. Needless to say, this still leaves us with a higher output voltage on the rails than the initial 9V the regulators were dropping it down to, so there will be no issue with the output clipping. For the negative rail, I have used the complimentary MJE2955T PNP transistors in a Sziklai pair configuration, which functions in a similar way to the Darlington pair except for that it works with PNP transistors and as such negative voltages. All other aspects are the same. The cut-off frequency of the low-pass filters being used comes in at 0.96Hz, which is more than low enough to attenuate any 50 or 100Hz hum that may be coming in on the lines, as well as any other high-frequency signals.

Another slight modification made is the addition of another pair of 4700uF capacitors on the output just to add any extra finalised smoothing required to the output, as well as two 1MΩ resistors connecting the rails to 0V as to allow the capacitors to discharge after use.

Finally, I chose to add ripple rejection to the high voltage rail also as there was no output on their previously. Again, you’ll note this one looks different to the example given earlier, and this is due to the use of a MOSFET in place of a transistor. I chose to use this configuration, as the voltages out from voltage multipliers can decrease drastically when under load, and given that transistors can struggle to function correctly when starved of a minimum current I chose to go with a MOSFET as these function off charge rather than current and as such require no actual input current on their gate, which allows for minimal load on the voltage multiplier outside of that it is already supplying current to.

Other slight modifications I might make in the future include a capacitor coupling the high voltage rail to ground to act as an earlier smoothing as well as doing the heavy lifting when the transformer voltage is at zero and supplying the ripple current, decreasing the values of resistors R64 and R65 and putting diodes across the transistors of the capacitance multipliers. Other than that I think I’m mostly going to be good to go.

Anyway it’s now twenty five to two in the morning and I am having a very hard time not writing teh everytime I type the and anf everytime I write and so I think I should finish this up and go to bed…