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| Elliott Sound Products | PSU Capacitor Bleeders |
Main IndexArticles Index As most readers will be aware, none of the power amplifier PSUs (power supply units) on the ESP website use bleeder resistors to discharge the caps when power is removed. This is a deliberate omission, because most amplifiers will discharge the filter capacitors fairly quickly, depending on quiescent current. Adding resistors to make the discharge faster dissipates power, and this is converted to heat. The extra power can increase temperatures inside an un-ventilated case surprisingly quickly.
For example, if you have a power amplifier that draws a quiescent current of 28mA (fairly low by most standards), ±56V supplies will collapse to around 10V within five seconds (assuming 4,700µF capacitors). Mostly, this is quite fast enough to let you work on the amp without having to wait forever for the caps to discharge. However, some people do like the idea of using bleeders, and adding 2k (2 x 1k, 1W in series) will speed this up. However, the bleeder resistors will get quite warm (dissipation is over 1.5W), and it's still rather slow.
The alternative is to use an active bleeder, configured so that it draws close to zero power as long as the mains is present, and it is designed to discharge the caps very quickly when mains power is turned off. Naturally, this requires some circuitry, but it doesn't have to be too complex. It's not difficult to discharge a 56V supply to less than 5V within one second. This can be achieved with any capacitance you like (and any voltage as well).
Note that while you may see references to using a screwdriver to short charged capacitors - Don't! The very high discharge current can damage the capacitor, and it's a risky procedure anyway. If you do need to reduce the stored charge to some low (safe) value, use a high-power resistor with proper insulated probes. Ideally, the resistor will be a value that discharges the cap quickly, but (if you want to be ultra-safe) keeps the current below the capacitor's ripple current rating. Otherwise, a 150Ω 5W resistor will suit most situations and will not damage the cap. Using a screwdriver (or other similar implement) is never recommended by anyone who knows what they are doing.
Probably the simplest way to implement an active discharge system is to use a relay, powered from the 230V (or 120V) mains. When mains power is interrupted, the relay's normally closed contacts connect a discharge resistor. When power is resumed, the relay opens and disconnects the discharge resistor. It's crude, but it can certainly do the job. There are two caveats with this, in that the relay must have a 230V or 120V AC coil, and the contacts have to be rated for the DC voltage in use. This will work well if the DC is less than 30V, but it gets troublesome at higher voltages. DC can cause contact arcing, but provided the current is less than ~250mA (set by the discharge resistor) you should be ok. Have a look at the Relays (Part II) article to see what you can get away with. There's also the issue that you have mains on the relay coil, and supposedly 'safe' DC at the contacts. This makes it a risky proposition unless you are very careful with your wiring. I've been doing mains wiring for most of my life, but this is not the method I'd choose.
The relay coil can also be powered from the transformer secondary, which is a lot safer, as there's no interaction with mains voltages. Finding a relay with a suitable coil voltage may be tricky, as they only come with a limited range of voltages. 12V, 24V and 48V are common, so a series limiting resistor would be needed if the secondary AC is more than 10% higher than the coil's rated voltage. AC coil relays are usually more expensive than DC types, and the relay may cost as much as the parts for an electronic discharge circuit. The relay will have a limited life (especially when switching DC), unlike an electronic circuit.
Note that in all circuits described here, the MOSFET must not be a logic level type. The circuits all rely on the MOSFET needing at least 2V on the gate to turn on, and if it's less, the MOSFET may turn on and off in normal use. The suggested MOSFETs have a minimum threshold voltage of 2V, which ensures that they will remain off when mains power is provided. The MOSFETs shown are only suggestions, you can use anything you wish, provided they have a suitable voltage rating (and aren't logic level). Power dissipation is low, and a heatsink is unlikely to be needed unless you have very high capacitance.
There is little or no consensus as to how quickly filter capacitors should be discharged. It's always a trade-off between speed and dissipation, and with energy costs worldwide increasing all the time, it seems a bit silly to deliberately increase the power consumption of an amplifier or other equipment. It's usually acceptable if the voltage has fallen to about 10% of the maximum within 10 seconds or so, but this isn't always achievable. Some amplifiers will create a large 'thud' through the speakers when the supply collapses, and this has to be considered.
Some power amps (in particular) may use 100,000µF capacitors (or paralleled caps to achieve the same result). Even with 10,000µF charged to 56V, a 330Ω resistor will cause the cap(s) to fall to below 5V in 10 seconds, but it will dissipate close to 10W (x2 for a dual supply), so there's nearly 20W of wasted power. That power is converted directly to heat, and serves no useful purpose. With more capacitance, you either have to accept even more wasted power, or wait longer for the caps to discharge. If you were to use 100,000µF at 56V with a 2k discharge resistor, the voltage will be over 40V for one minute after power is removed, and is still over 30V two minutes after power is turned off.
It should be fairly obvious why I never add the discharge resistors. If you need to keep wasted power to the minimum, the amplifier will almost certainly pull the voltage down faster than (say) a 2k resistor, which will still dissipate over 1.5W as for long as the amplifier is turned on. Discharge resistors were nearly always used with valve (vacuum tube) equipment, because the voltages were much higher than we use now, and valves quickly lose emission as the heater cools. This could easily leave a dangerous voltage across the filter capacitors for several minutes (or longer in some cases).
It's very important to understand that single-supply amplifiers with a speaker coupling capacitor need special attention. If the supply voltage collapses too quickly, the speaker capacitor can force current back through the amplifier, and this can damage output transistors. The amplifier's output must have a diode between the output (before the output capacitor) and the supply rail. This provides a discharge path for the capacitor that doesn't involve reverse biased transistors. Fortunately, such amplifiers are now uncommon, and it should not be an issue.
With modern equipment there's really no need to use discharge resistors, but there will always be constructors who, for one reason or another, prefer to reduce the supply voltage as quickly as possible. It's obvious that using a resistor is not the answer, so we need to add some electronics. The idea is to keep the circuitry as simple as possible, but of course it has to work reliably. Fortunately, this isn't difficult to achieve.
Figure 1 - Relay Bleeder Circuit
The above is an example of a relay based discharge circuit. Bear in mind that some AC coil relays have a slight buzz, which will likely be audible, and if so will be very annoying. This is not the recommended way to make a discharge circuit, but some constructors may find it suits their needs. If you have a dual supply, the relay needs DPDT (double-pole, double-throw) contacts, with the discharge resistors using the NC (normally closed) contacts. When AC is applied, these contacts will open, disconnecting the discharge resistor.
A relay version looks simple, but contact erosion from DC will eventually cause it to become intermittent, or fail permanently. You probably won't know that this has happened until you monitor voltages. If one side of a relay based dual supply discharge fails, you will most likely be rewarded by a loud 'thump' from speakers as one rail falls to zero while the other is still at a higher voltage for a short period. This circuit will work, but it's not recommended.
The essential 'ingredient' is an AC sensing circuit, which detects AC and keeps the bleeder disconnected until the mains is turned off. A simple arrangement using this idea is incorporated into the Project 05 preamp power supply, and is used to activate a muting relay when power is removed. Project 33 uses much the same arrangement, and both are known to work very well.
Once the circuit senses that mains power is no longer available, a bleeder resistor can then be switched into the circuit. Because it's turned off as long as power is available, there's no wasted power, and the bleeder can be a low value to ensure a rapid discharge. While the instantaneous power will be high, it's fairly short-lived, so a 5W resistor will usually be more than sufficient to handle the peak power (which may be 25W or more, depending on the design choices made).
Throwing electronics at the 'problem' isn't quite as bizarre as you may imagine. Some equipment uses mains filters, and the capacitors within can (under some conditions) remain charged. Several manufacturers make ICs designed specifically to discharge the capacitors. The TEA1078 (made by NXP) is one example, but it's by no means alone. In case you were wondering, no, you can't use this IC to discharge big filter capacitors - it's designed to reduce the voltage across a 330nF X2 capacitor to less than 60V in under 300ms. It has minimal current capability.
The AC detector simply uses the AC from the transformer to turn on a transistor (Q1) 50 or 60 times per second, maintaining a low voltage across a capacitor as long as AC is present. A simplified version of the circuit for a single supply is shown below, so that the various parts can be examined. Some of the component values will be changed, depending on how quickly you want the capacitor to discharge, but the circuit can be used with no changes with DC voltages from 22V up to 100V. The only reason for the 15V zener diode is to protect the gate of the MOSFET, which is vulnerable to ESD (electrostatic discharge) and any voltage above 20V may cause the insulation to fail. The result is a dead MOSFET.
Figure 2 - Basic Active Bleeder Circuit
The discharge switching device is a MOSFET, because they require almost no current to turn on, and they provide excellent switching capabilities. A BJT (bipolar junction transistor) can be used, but it's nowhere near as good, will dissipate more power, and may require a heatsink. The MOSFET will have to handle up to 15W, but it's only for a few milliseconds. Any MOSFET with a suitable voltage rating can be used, provided you leave a 10-20% safety margin. The IRF520 (N-Channel) and IRF9520 (P-Channel) are suitable for supply voltages up to ±80V. This will be enough for the vast majority of applications.
Q1 is the AC detector, and it will keep C1 discharged (typically below 1V) so the MOSFET can't conduct. When the mains is interrupted, the voltage across C1 rises and the MOSFET turns on. This discharges the filter capacitor (Cfilt, shown as 10mF - 10,000µF) via the discharge resistor. With 150Ω as shown, the voltage will drop below 5V in about 2.5 seconds. There is no need to make it any faster, and the 150Ω discharge resistor can be used with any DC voltage. At 80V DC, it will dissipate a peak power of 40W, but that will drop below 5W in less than 1.5 seconds. A 5W resistor should be able to handle that without difficulty. The MOSFET will dissipate up to 10W at 80V, but typically only for less than 10ms, and it will not need a heatsink. D2 ensures that the voltage across C1 isn't discharged by R2 as the supply voltage collapses.
Because the MOSFET's gate has voltage for a considerable time, it can continue to conduct. D2 prevents C1 from discharging through R2, and enough gate voltage is present to ensure conduction until the output voltage has fallen to zero. C1 will discharge via R3 (2.2MΩ), but that will take a while, because R3 is deliberately a high value. This does not affect the circuit's ability to be re-started, as the first AC cycle will cause Q1 to discharge the capacitor so normal operation resumes immediately.
R1 should normally pass a peak current of around 500µA to the base of Q1. It's not critical, and it will work fine with anything from 200µA up to 1mA. The value is determined using Ohm's law, using the DC voltage as the reference. For example, a transformer with a 25V RMS secondary will provide 35V DC, so R1 is determined by ...
R1 = 35 / 500µA = 70k (Use 68k)
Apart from the MOSFET and Rdis (the discharge resistor), the only other value that changes is R2. It should normally pass around 1mA. If the DC voltage is (say) 80V, R2 will be 82k (and R1 should be 150k). With a nominal 1mA charge current, C1 will charge at a rated of 0.1V/ ms, so it takes 10ms for C1 to charge to 1V, or 100ms to 10V. The circuit can also be used with high-voltage supplies (see 'High Voltage' below). Just make sure that the MOSFET(s) are rated for at least 20% more voltage than you'll be using. Compared to a resistive bleeder, this circuit will provide a much faster discharge, and will dissipate almost no power when the equipment is in use - about 33mW with the values in Figure 2.
The discharge time is based on a simple time constant, the filter cap (Cfilt) and Rdis (150Ω). The time constant of 10mF and 150Ω is 1.5 seconds, at which time the voltage will be 37% of the original voltage. After two time constants (3 seconds), the voltage has fallen by another 37%, down to 4.8V (for a 35V supply). This process continues, with the voltage falling another 37% for each additional time constant you add. In theory, this is known as an asymptote, and the voltage will never reach zero. In practice, it's generally considered that 10 time constants is close enough for both a full charge or discharge. After 15 seconds (10 time constants) the voltage is only around 1.6mV (when starting from 35V).
Note that there is a small delay before the MOSFET conducts, because C1 has to charge after power is removed. The delay is about 130ms with the values suggested. This is not an issue, and although it can be reduced, there's no reason to try to do so.
The dual version uses a mirror-image for the negative supply. Q3 and Q4 are PNP and P-Channel devices respectively, and there's no longer a requirement for D1 shown in Figure 2, because the PNP transistor clamps the negative voltage for Q1 and vice versa. One could try to be clever and make the negative discharge circuit a slave to the positive version, but that would end up needing more parts. Everything involved is cheap, and the two circuits will be complementary. Small differences are inevitable, but they should not cause any problems with a sensibly designed circuit.
Figure 3 - Dual Supply Active Bleeder Circuit
Components are calculated in the same way as for the Figure 2 circuit, and nothing is particularly critical. Naturally, all parts need to be rated for the voltage being used, and if you don't need a fast discharge, the value of Rdis can be increased. The only down-side of the dual version is that P-Channel MOSFETs are usually a bit more expensive than their N-Channel counterparts, but the difference should be very small in practice (a few cents at the most).
A useful change would allow the circuit to discharge the supply of a valve amp. This may be 450V or more, and using a bleeder is highly recommended. They are often incorporated into the power supply anyway, because they also act as 'balancing' resistors to ensure the same voltage across each cap. While not strictly necessary (for reasons I won't go into here), 100µF electros will typically use a 220k resistor, with two such pairs in series as shown below. This will discharge to 37% of the original voltage in 22 seconds, not including any current drawn by the valves (which is usually the case, unless they have been removed during testing!). Without valves, the voltage can remain hazardous for much longer than we'd like.
Figure 4 - Active Bleeder Circuit For High Voltage
Using a high voltage MOSFET and with the guidelines shown in section 2, the discharge time can be reduced to under one second, with almost zero wasted power. The discharge resistor should be increased to around 4.7k, and even though the instantaneous power is over 40W, a 5W resistor should be able to handle this with ease (peak current with a 450V supply is just over 96mA). R1 should be around 820k, and R2 should be 470k. Ideally, both will be 1W, not because of power dissipation, but to ensure they can handle the voltage. The voltage across C1 cannot exceed 15V, but 10µF, 63V electros are so common that you wouldn't use anything else.
While further improvements are possible, there appears to be no good reason to add any more parts, because it works just fine as it is. If it were for a military system I'm sure that the extra parts count would be of no consequence, but for 'normal' usage by hobbyists and others who need a discharge system, it's already more than acceptable.
There is one other circuit I found, which was patented in 1996 [ 1 ]. It appears that it will work (at least in the simulator). I have reservations about the original though, for a number of reasons. Only one diode was used as originally patented, and it also required C1 to be a reasonably large electrolytic capacitor (which is undesirable for many reasons). Using two diodes as shown reduces the ripple voltage across C1 to about 620mV peak (vs. 1.3V with one diode), which is a better option. The major change is from a BJT to a MOSFET, and this allows C1 to be much smaller, which means you can use a film cap.
Figure 5 - Active Bleeder Circuit (Based on Patent by Fluke Corporation)
The diodes keep C1 discharged (to within a few hundred millivolts of the supply voltage), biasing off Q1. When the mains is interrupted, C1 rapidly charges via R1, Q1 turns on, and the supply is discharged. The patent drawing showed C1 as an electrolytic cap that was subjected to a small reverse polarity when AC is present, which is not optimal. As shown above, C1 has +100mV (relative to the main supply) during normal operation. The peak charge current is beyond what I'd like to see if only one diode is used (it's a cheap addition, and a single diode isn't recommended).
The original design used a BJT as the discharge 'switch', and that required C1 to be much larger than the value shown. Using a P-Channel MOSFET means far lower dissipation in the operating state, because R1 is a much higher value than can be used with a BJT. If you want to see the original, look up the patent document. While the circuit is clever and uses the absolute minimum number of parts, it's not the one I'd recommend. I like the simplicity, but not the compromises. The original only used four parts, but has many more likely problems than the modified version shown, which saves only two parts over my suggested versions. Still, it's the only viable alternative circuit I could find, indication that active capacitor discharge circuits probably fall into the 'esoteric' category.
When servicing equipment, and especially valve guitar amps and SMPS, high and possibly lethal voltages may be stored in filter caps. Making contact with 400V or so isn't fun, and it's something that is ... shall we say 'best avoided'. This final circuit is manual - it's leads are attached to the filter cap with clips or soldered on, and it needs a well insulated case. The wiring must be rated for the likely voltages you'll encounter.
The choices for the SCR are many, and the BT151-800 or BT152-800 are common and reasonably priced. There are many others (too many to list) so a search of your local supplier's website will turn up something suitable. Mostly you won't need the 800V rating, but it's better to have it and not need it, than to need it and not have it 
. Naturally, you can use a lower voltage rating if you don't expect to use more than (say) 600V or less.
Figure 6 - Manual Discharge Circuit
Make sure that the 'Discharge' button is either recessed or needs some force to activate. An accidental press could damage the power supply if it's still working, and will also cause the discharge resistor to get very hot, very quickly. With a 1kΩ resistor as shown and a 400V supply, the resistor will try to dissipate a little over 160W. You may choose to use a higher value, and 2.2k will dissipate only 80W.
The SCR (S1) is normally off, and the neon lamp (NE1) indicates that the voltage is above ~90V. R1 and R2 should be 1W resistors, or use two 220k resistors in series. This isn't for power handling, but ensures that high voltages don't cause the resistors to fail. C1 will charge to 15V, and is discharged into the gate of the SCR to turn it on. The current is limited by R3 to prevent gate damage. When the voltage has fallen to the point where it's lower than the SCR's holding current, S1 turns off again.
While the drawing shows test clips, the leads can be soldered in place if preferred. Make absolutely sure that they are connected with the right polarity. The circuit will not work if they are the wrong way around, so care is needed. Make sure that the button is never pressed while power is applied, as the connected circuitry may be damaged. If you are lucky, all that will happen is the fuse will blow, but if it's used with a valve rectifier it may be damaged.
With 400V DC and a 220µF filter capacitor (for example), the voltage will be reduced to around 20V within less than 300ms. It's unlikely that it needs to be any faster than that for normal use.
Personally, I'd rather use the Figure 4 circuit, as it only needs three leads, but is automatic - the capacitor will be discharged as soon as mains power is interrupted. However, I'm not entirely sure I'd be happy using it on a SMPS, because everything is at mains potential. If carefully made, the manual discharge circuit will be safe to use, but as noted above, the push-button must be protected against accidental activation.
Because I don't consider a discharge/ bleeder circuit essential (or even necessary), it's hard for me to recommend using the circuits shown here within an amplifier chassis. However, there may be occasions where you find that, for whatever reason, a rapid voltage reduction is needed. Should that be the case, the circuits shown will do the job, and you can select the discharge speed based on your needs.
The high voltage version is recommended for valve amps and other circuits that use a high voltage but can't discharge the filter caps quickly. Leaving high voltages lurking within a chassis is always somewhat dangerous (particularly for service technicians), and by ensuring a rapid discharge means that you are far less likely to get a nasty surprise when working on it. MOSFETs are readily available for most voltages encountered, and the circuit is so simple that it will add little to the build cost, nor will it occupy much space. It can even be made in a small box, with three leads - chassis, transformer and DC, allowing it to be attached while working on an amplifier.
The circuits shown are by no means the only way that an active discharge circuit can be made. There are other possibilities, but most will be more complex. The principles don't change, as it's still essential to detect that the AC has been turned off, and use the detector to turn on the discharge transistor (BJT or MOSFET). The circuits shown here are about as simple as they can be, consistent with good, reliable performance.
As it turns out, I have just the place for the dual supply version of this project. For most high-power amplifier tests, I use a supply that I call the 'monster'. It uses a 1kVA power transformer, and has around 20mF (20,000µF) filter caps for each rail. It's always powered via a Variac so I can set the voltage to whatever is needed. The maximum voltage is around ±90V, and that can do some serious damage. Provided it's powered off with an amplifier connected, it will discharge fairly quickly (typically in about 20-30 seconds or so), but without an amplifier or other load, it holds the voltage for a considerable time. It can be very embarrassing to connect an amplifier to a 'live' power supply, and the dual supply discharge circuit is an ideal addition.
There are no other references, as the circuit I developed appears to be unique. There are a few attempts shown on-line, but none (other than the reference above) that I saw will work very well (some won't work at all, or are poorly executed).
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