Capacitance multiplier for filtering DC

[aion]

The Boss FZ-2 Hyper Fuzz uses a capacitance multiplier for filtering:



This is the only Boss pedal I'm aware of that uses one of these. It's also the only one I know of that uses shielded input wire, so presumably these are both aiming to solve the same problem, which is that the circuit is more sensitive than normal to picking up unwanted noise.

A couple of questions.

1) I've heard mention that the resistor (R46, 2.2k) works in conjunction with the capacitor to form a filter, similar to a resistor in direct series with the power supply. Is this accurate? In other words, it's not just that the capacitance is multiplied, i.e. equivalent to just using a really big capacitor, but that the resistor doesn't affect the output voltage based on current draw. It'll always drop the voltage by one diode, but you get both the larger capacitor and the larger resistor out of the deal, resulting in a much stronger filter (2k2 where I'd normally use 100R maximum).

2) They use a reverse-biased diode for polarity protection. I've moved away from this practice and started using a series Schottky for protection, so I'm considering if I want to use the diode as shown or adapt it. My question: Is the transistor in the capacitance multiplier providing the same series protection since it's acting as a diode? Or can the resistor and capacitor still cause problems since they come before the transistor and can still be affected by the reversed voltage?

The bottom line is, if it's almost as good to use a large capacitor (say 470uf) and avoid the multiplier altogether, then I'd go that route. But I'm trying to better understand how it works, why Boss opted for it, and what benefits it may offer.

[iainpunk]

1) it basically is a filter with a buffer, so its semi-accurate...
the capacitance gets multiplied and the resistance get divided (both impedances go down), the factor of the division and multiplication is the BJT's Hfe

2) don't trust a transistor to protect like a series diode, although it has internal diodes, don't use it for reverse voltage protection. (the 47u still gets reverse bias and blows up, the initial turn on current spike and Base-Collector conduction might fry the transistor creating a short circuit throughout the whole substrate)

cheers, Iain

[aion]

1) it basically is a filter with a buffer, so its semi-accurate...
the capacitance gets multiplied and the resistance get divided (both impedances go down), the factor of the division and multiplication is the BJT's Hfe

2) don't trust a transistor to protect like a series diode, although it has internal diodes, don't use it for reverse voltage protection. (the 47u still gets reverse bias and blows up, the initial turn on current spike and Base-Collector conduction might fry the transistor creating a short circuit throughout the whole substrate)

cheers, Iain

Thank you, that makes sense on both fronts. I'll keep the series protection diode in place so any reverse voltage is cut off before getting to the multiplier.

So bottom line, in your opinion, is this method superior to traditional methods of, say, a 100-220R series resistor and a 470uf capacitor? Theoretically the filtering is better?

[iainpunk]

i strongly believe the capacitance multiplier is better than the resistor and HUGE capacitor for a few reasons:
- smaller size
although there is one more component, the size of the components is smaller than the capacitor alone.
- dissipation
you can if you want throw a 2n3055  and a BD140 in a sziklai configuration * for the fraction of the cost of a single 10W resistor, and the 3055 can handle more than 10 times as much power.
- cost
if you handle high currents, that 3055 and BD140 combo is way cheaper than the appropriate resistor and capacitor combination.

ow, there is a negative side too: loss of headroom
depending on the transistor, you will lose some headroom, 0.6v for a single transistor or a Sziklai pair and 1.2v for a darlington

cheers, Iain

* if you prefer darlington over a sziklai, i recommend a BD139, but sziklai is always better tho

[iainpunk]

[ignore post]

[ElectricDruid]

So bottom line, in your opinion, is this method superior to traditional methods of, say, a 100-220R series resistor and a 470uf capacitor? Theoretically the filtering is better?

+1 agree with Iain. If the capacitance gets multiplied and the resistance gets divided, your filter might have the same cutoff, but the voltage loss due to the current draw will be a lot less. So it's not so much that the filter action is better, but more that you get the same filtering, but with less downsides.

Plus a 100uF is much cheaper and smaller than a 250x100uF = 25,000uF cap, so that's good too.

[Rob Strand]

There is a hidden difference between the RC filter and the active filter.

The RC filter always filters regardless of the size of the ripple on the input.
The output is the average input voltage (ie. the centre of the ripple peaks.) less the IR drop due the load.
So if you have 9V with +/-1V ripple the average is 9V.  If you have 10mA load and a 100ohm resistor,
the IR drop is 100*10mA = 1V so the output ends up at 8V.   

For the active filter only works when the input ripple is small.   It actually *relies* on the 0.6V VBE drop to fend off the ripple.
Imagine the same scenario of 9V with +/- 1V ripple.   

To first approximation voltage at the 2k2 + 47uF is 9V but the base current pulls down the voltage on the base *exactly* like the RC filter.  In reality, for 10mA output current and a transistor gain of 300 the base current is 30uA and that gives 0.066V drop across the 2.2k resistor.  The voltage on cap/base is 8.934V compared to the 8V  for the RC filter circuit.  If the 2k2 is larger then the drop increases.   If you choose the 2k2 low enough it can be ignored.

The output voltage is 0.6V below the base voltage so output voltage is 8.934 - 0.6 = 8.334V

So now the problem.   In order for the filter to work the transistor must not saturate.  That means the VCE voltage must greater tan  say 0.2V.   Since the output is 8.334V that means for the filter to work the input voltage must remain above 8.334V + 0.2V = 8.534V.

In the example the input voltage is dropping to 8V.   That means the transistor will cut-off and not current passes.    No output current means the output ripple is going rise.   In effect the active filter has clipped or dropped-out.

In a real circuit there's yet another mechanism that comes into play during drop out.    Any caps on the *output* of the filter will help hold up the supply rail.    If those are very small then current is supplied from the 47uF filter cap via the base-emitter diode.   That will only hold-up if the 47uF filter cap is large enough.  The load on the 47uF will causes cap voltage and the target output voltage to drop so the filter kick in but not at the minimum points of the ripple.    The main point to realize is in the drop-out condition the active filtering is no longer working.   We are relying on big caps to hold up the rail ---->  In some ways no better than the RC filter case, in fact worse because there is *no* current supplied from the input supply during drop-out and it all comes from supply caps.

So the conclusion is the active filter can only handle about +/-0.5V input ripple maybe a bit more if the transistor gain is low.  The RC filter does not have a poop-out point it filters regardless of the size of the input ripple.

The only way to fend off higher input ripple is to have a bigger VCE voltage.  That means more voltage drop so there is a direct trade-off between maximum ripple and drop-out.

One way to increase VCE is to make the 2k2 larger.  The drop depends on the transistor gain so it's not a reliable method.  The other method is to put a resistor across the 47uF effectively forming a divider with the 2.2k resistor.  That way we can consciously trade- off drop-out and maximum ripple.

[iainpunk]

yes, Rob is right, didn't think of that, however, i have never had the situation that a system had a large ripple like that, especially in today's world where switch mode is taking over the big old hunk of iron transformers of yesteryear, utilizing way smaller capacitances and smaller transformers for the same power output.
but in the 50Hz tranny situation, i guess a Darlington pair beats the Sziklai after all.

also, in context, its a pedal power supply for a fuzz that uses less than 1mA, i don't think we should be worried about big ripple.

also, Aion, are you going to do a FZ-2 kit? i have been working on a cool mod that makes the treble control more usefull, its currently a mod daughter board that's less than 15mmx15mm, a single transistor 2 caps and 3 resistors
it gives it a more HM-2 style treble control instead of the Achilles heel of the pedal that's the original treble control. its a gyrator that replaces the cap and resistor combo that's there originally. it boosts or cuts a notch at 1.2kHz, giving that classic Swedish Chansaw, but fuzz instead of distortion, or in the boost mode, add it to any dirt pedal and/or amp you have.

cheers, Iain

[Rob Strand]

yes, Rob is right, didn't think of that, however, i have never had the situation that a system had a large ripple like that, especially in today's world where switch mode is taking over the big old hunk of iron transformers of yesteryear, utilizing way smaller capacitances and smaller transformers for the same power output.

For effects pedals and plug-packs I'd have to agree.     Most transformer plug-packs have 1000uF caps, a small few have 470uF caps.   If we have a peak to peak ripple of Vr_pp =  I_load * (T/2) / C   a 1Vp-p ripple implies 100mA load  which is less than most effects pedals.

So we can see why we don't see the problem with pedals.   In general the problem is lurking in the background.

[bms19]

could this design be adjusted to deliver 9V output?
Benoit

[antonis]

If you could find a 9V7 Zener diode placed in parallel with C36, why not..??

[cordobes]

This is the only Boss pedal I'm aware of that uses one of these.

Not true, it seems to me that you have not checked carefully. Checking my files, at least these designs carry capacitance multiplier: CE3, CE5, CH1, VB2, DD20, WP20, PS3, DD7, DM3, RV3, RV5, PH3, OD2, BD2, FW3.

[R.G.]

A capacitance multiplier is not so much a filter or multiplier as it is an emitter follower with a base fed from an R-C filter. Its action is like Rob said, limited to the transistor being unsaturated and not in cutoff.

I've always thought of these as power supply "polishers" that help when the big energy storage and raw filtering has been done elsewhere. They can remove some amount of any remaining wobble on the input voltage, and also can be thought of as removing cross-contamination of the incoming power from other pedals sharing the power wires. 

The typical emitter-follower capacitance multiplier has limits on how much ripple reduction can be done, and on the saturation voltages, etc. Changing the pass device to a PNP helps lower the voltage overhead at the cost of another transistor or two to supply loop gain and feedback to make a higher performance regulator.

I'm not overly fond of either the reverse diode or series schottky for reverse polarity. I like a P-channel MOSFET. 
http://www.geofex.com/article_folders/mosswitch/mosswitch.htm