The K Multiplier - An Economical Supercell Circuit

The basic circuit is at once simple and super-effective. It can be used at any rail voltage, regardless of the transistors' individual voltage limits. Film and low-ESR decoupling can be used with it and it will not oscillate. Is there another circuit that is as effective with a comparable parts count?

Last updated February 7, 2014

In professional circles, designing analog circuits for good supply noise tolerance is the accepted norm.  But home experimenters have found that low-noise power supplies present the ultimate freedom to test out their inexhaustible creativity. The extra degree of freedom gained from clean power allows for less distractions from the ultimate goal of a good-sounding circuit. So audio hobbyists have been hard at work for decades refining the art of quiet, inert power supplies. Along these lines several super-high performance solutions are available from the DIY community. Many of them are very complex, needing performance opamps and many discrete devices. But as is often the case, increasing complexity tends to give diminishing returns. For this reason the simplest solutions are popular to hobbyists. My circuit is a bit simpler than most, but I see few competitors where parts count, cost and ease of assembly are concerned.

I'll use the positive circuit to explain. Q1 and Q2 can be recognized by most poweramp designers as a Complementary Follower Pair, CFP for short. This arrangement provides dramatically reduced output impedance and equivalent current gain to a Darlington arrangement - just what we want for a power source. Both transistors need at least 1.2V Vce to work best, which is what the triple diode string D1 is for. Q2 needs more than Q1, so I've chosen 3 diodes for 1.2V. This may seem wrong to you, but consider that the only current through these diodes is the base current of Q1. Diode Vfb is roughly 400mV in the uA range, following the 60mV/decade rule. Q1 is biased at about 15mA, which generally gives the lowest output impedance. D2 and R8 determine the turn-on time, and limit how much current the circuit will dump into proceeding supply capacitors. This was important for one builder who employed a morbidly gratuitous reservoir array. The first time he flipped the power switch the transistors would blow from the inrush current! It took a few tries for me to realize the problem... If your circuit does not have inrush current over 1A, and this can be verified in simulation, you do not need R8 and can bypass it. D3 discharges C1 when power is lost so it doesn't discharge through Q1 - do not omit! The BC337/327 pair are crucial to this design - they set the upper limit on input rejection. I got input rejection using these of about 66db for the positive circuit and 54db for the negative circuit (measured in real life). BC550C/560C are second-best in this regard. ALL high-voltage transistors I have seen have bad quasi-saturation behavior and should be avoided for low-Vce applications (2N5551, 2SC1845, etc). I think other gain groups may work better but I haven't been able to test and see.

Circuit Specs:

• Input rejection: +66db/-54db or +2k/-500
• Output impedance:
• 112mR at 25mA
• 45mR at 50mA
• 27mR at 100mA
• 20mR at 200mA
• 17mR at 400mA
• Excess output inductance (in inches of wire): at or below 1.
• Output current: 1.4A max continuous. 20-400mA max for linear operation.
• Output voltage: Limited by Q1 Vcemax except in special situations.
• Voltage drop: ~1.8V (can be less but this gives good operation)
• Max input ripple before saturation: 2V pk-pk

If we say that the output impedance of the Kmultiplier is equivalent to that of electrolytics, we establish a baseline to consider its advantages. One incontrovertible advantage over electrolytics is that the Kmultiplier has low impedance into the subsonic range. A comparable lytic in this regard would be huge, and would necessitate soft-start circuitry to prevent the massive inrush from destroying the preceeding wiring. Clearly the Kmultiplier provides an expediency that no capacitor can replace.

The average electrolytic 470uF and up has lower than 50mR ESR. I have measured this figure repeatedly on lytics pulled out of broken modern entertainment devices such as computer monitors, TVs, and power amps. The best have less than 20mR. So the output impedance of this very simple circuit fares very well in practical terms. The fact that this circuit has output impedance comparable to modern lytics AND will tolerate film and low-ESR bypass shows that I have struck a good balance - output impedance is not any lower than is actually needed, but is low enough to augment and seamlessly integrate with a good bypassing and decoupling scheme.

The voltage overhead of the Kmultiplier is low, making it a persuasive drop-in addition to circuits which don't need DC regulated rails but would benefit from heavy power filtering. In this case the Kmultiplier replaces an oversized transformer and banks of capacitors, making it a truly economical and expedient solution.

Because the Kmultiplier has no more feedback than needed and is self-limiting in bandwidth, the compensation capacitors are merely those parasitically included with the transistor junctions. This meager amount of capacitive stress means that the circuit is unlikely to glitch even with very fast load transients. Many regulators will emit a highly distorted output signal when stressed with fast load current swings, which can get into circuitry and make things worse than if the supply was not regulated. The Kmultiplier however remains faithful into the hundreds of KHz. Of course, why even risk high-speed signals at the regulator when you can apply liberal decoupling across the load without fear of oscillation?

Design Considerations

• Choosing Components
• Capacitors: R1 and the ESR of C1 form a resistive divider that limits the PSRR from the outset. Luckily, a cap with 100mR ESR as C1 will reduce PSRR by 1/5 in the worst-case scenario. The average cap in the 1000u rage has >50mR ESR, and going below this gives diminishing returns, so most noble brands should be safe.
• Resistors: Resistor type will not affect the performance of the Kmultiplier significantly, unless maybe they are inductive. Film resistors will be fine.
• Dissipation and heatsinking
• Due to the exceedingly low Vce of Q2 it does not need a heatsink; it will hit Icmax before dissipation reaches 2W.
• Controlling inrush current
• An inrush of over 1A can destroy Q2, and often the LED as well. The RC filter gives us a convenient and simple way to manage inrush. At turn-on the full supply voltage minus the LED and diode string voltage is imposed across R8. Thus if you have a 30V input, R8 will see about 25.2V. This gives C1 a max charging current of 25.2V/680R or 37mA. Because the output follows the RC voltage, this means that for every 820uF of capacitance at the output, there will be 37mA of inrush. So say you have 4700uF at the output. (4700uF/820uF)*37mA gives us about 212mA.
• For the circuit above, use this equation to find out what value of R8 you will need to stay below 1A: (Vin-4.8)/(1A/(Cout/C1)) where Cout is your TOTAL output capacitance.
• Output impedance depends on output current
• This regulator exhibits diodic output impedance. This means output impedance decreases with increased loading. The opposite is also true, which is why this circuit should NEVER be used with loads less then 25mA.
• It cannot absorb negative load surges
• The transistors only conduct in one direction - make sure your circuit will not generate negative current surges.
• Keep layout compact
• This is a very fast circuit. It has behaved well for me so far, but long traces at sensitive nodes may encourage it to do things it ordinarily wouldn't.
• It is not a fixed-voltage regulator - the output just lags the input by several seconds
• The output voltage will track the input voltage over time - this can be to your advantage. If your rails can vary by several volts, you won't need the full voltage headroom you would need for a fixed-voltage regulator. The Kmultiplier will always "relax" to 1.8V voltage drop between input and output. However if your circuit needs a precise and regulated supply voltage, this circuit may not be for you.

Troubleshooting

• How do I know when it is working correctly?
• A very fine balance is required to maintain the high input isolation. Therefore if anything is wrong, filtering will be impaired. To verify that you Kmultiplier is working right, follow this procedure:
1. Turn on your Kmultiplier and have an AC voltmeter at hand. Short the probes to make sure it can read below 1mVAC.
2. Measure and record the input AC voltage and the output AC voltage.
3. Divide the measured output by the input. Make sure this matches the specified input isolation.
• The LED should flash when you flip the power switch.
• If the LED is emitting ANY light at all during operation, it means your Kmultiplier is struggling and failing to meet the current demand, or that there is a fault.
• The voltage drop roughly corresponds to 3 diode junctions. Measure the DC voltage between the input and output of your Kmultiplier. It should be within .15V of the specified voltage drop.
• Isolating the fault
• Voltage drop too high
1. Measure the voltage across R2. If it is over 80mV, you're drawing too much current or Q1 is faulty possibly from overcurrent or a backwards discharge through the BE junction. C1 could also be faulty due to aging or exposure to heat over time.
• Voltage drop too low
1. If voltage drop is less than a diode (.6V) Q2 is faulty from overcurrent, possibly inrush. Rarely, if that is not the problem, D4 may be shorted.
• LED does not flash during turn-on
1. Check your input voltage. Most likely a failure in another part of the circuit has blown the LED. This will not change the troubleshooting process.
• Getting help with troubleshooting

Making Changes

If you want to change something, beware that it may change troubleshooting procedures and design considerations. That said, if you are confident you do not need guidance for your application, here are some ideas and helpful information.
• Decreasing voltage drop
• One reason people use C-multipliers is because they can tolerate low input voltages. You can decrease the voltage drop of the Kmultiplier by taking diodes out of the diode string D1. Beware however each time you do this you decrease the ripple tolerance by 800mV pk-pk.
• Input ripple considerations
• If you are increasing voltage drop in order to have more ripple tolerance, pay attention to the LED breakover voltage. If you expect to have more ripple than the specified tolerance for the original circuit, then upgrade to a green LED instead, or a string of red LEDs. Otherwise ripple will intermittently turn on the LED and ruin input isolation.
• The RC time constant will determine how closely it tracks the input voltage
• For better filtering you may try to increase the RC time constant, but beware that the "stiffer" the filter, the more likely that rail sag and power demands will saturate it and possibly cause glitches.

The circuit above will work with the vast majority of applications within the specified current range. It's very one-size-fits-all. However for some applications the intent designer will want to adapt it to better suit the application. For instance, in vacuum tube designs with several hundred mA of current draw and very high voltages, the Kmultiplier will excel. However an 820uF 350V cap is very uneconomical for the average hobbyist. Therefore you would want to raise R2 to up to 100k, and lower C1 to 8.2uF. This would contribute to about 5V drop across the resistor itself, but for a 350V supply this is reasonable and common practice. Don't forget to account for inrush current with C8 and as Vce will be much larger, make sure Q2 can handle the extra dissipation.

But what if we want to use the Kmultiplier for lower currents? Say for filtering the frontend of a power amp? This is a good idea. Even under high power demands the K-multiplier will not dropout or clip, because it will track the rail voltage rather than resist it. For this purpose I have used variations on the circuit below. It has not quite been tweaked as much as it could, because it really depends on the application. Linear to 20mA.

Is there a Kmultiplier I can use for a power amp?

Adapting the Kmultiplier for higher currents is not so simple. An increased number of active devices is required, and new feedback loops must be added. This creates a pandora's box of new difficulties which are not in the original design simply because of its simplicity. A more powerful version needs compensations and very careful dimensioning to achieve the benefits of low voltage drop, low output impedance and compatibility with well-decoupled designs. The resulting circuit is not so elegant in its presentation or its performance, but I think it can be done, and I may have done it. I will link to it tentatively.

Theory of Operation

The design of this circuit begun when I was thinking about the capacitance multiplier circuit block, shown below. The C-multiplier is such a useful, underused circuit. It can replace banks of caps, given enough voltage headroom. Furthermore its input isolation can be a big bonus for RF filtering provided you select transistors carefully. However it seemed to me no one had thought about this circuit for decades. Relegated to ancient history, no one seemed to grasp it's potential when revamped with modern transistors and design expertise. Here is the whole story.

It is called the capacitance multiplier because, (according to theory) it appears to the load as if C1 has been multiplied by the Hfe of Q1. This particular transistor has an Hfe of 150 or so, which means our load sees a supply capacitance of about 1500uF.

Those who have gotten this far probably understand that the reality is less pretty. After all, the BE junction of Q1 acts like a silicon diode in series with the load. This means the output impedance of this filter circuit is very high, and very nonlinear. The impedance of general purpose silicon follows the rule R=.033/Id, where R is the small-signal resistance of the junction and Id is it's forward bias current in the given application. So if our load draws about 40mA, then our filter has an output impedance of tentatively .825 ohms. This is abysmal.

As seen by the load, R1 is also divided by the Hfe of Q1, and this is in series with the "diodic" output impedance - 825mR+(100R/150Hfe)=1.492R. Even worse. But this is followed by a sigh of relief when we realize that at AC the transistor's base sees a short through C1, returning the output impedance to 825mR.

Even so the nearly doubled output resistance at DC looms over us.
Lets set up a comparison to give a sense of reality and scale to these figures. This circuit tries to emulate a 1500uF capacitor. A standard electrolytic capacitor in this range has around 50mR or .05 ohms. Furthermore, the transistor only conducts one way, unlike the capacitor which can absorb negative current surges. So as you can see, the circuit does not really compare to a real capacitor. They are so different they cannot be treated like equivalents. The drawbacks are listed as follows:
• High output impedance
• potentially much more than an equivalent electrolytic and even the rectifier and transformer winding.
• Leakage
• Early effect causes Vbe to modulate with Vce, injecting wideband input noise into the output. Transistor selection can reduce this, especially at lower output currents.
• No negative surge tolerance
• The transistor only conducts current one way.
That aside, the circuit still has one considerable benefit over a bare 1500uF electrolytic. This is ripple rejection and input isolation. As Q1 is in the emitter follower configuration, the emitter follows the base voltage, and the base is fed by an RC lowpass filter of 100R*100uF.  We don't get the benefit of this RC arrangement with just a 1500uF capacitor - where the only R is that of the rectifier diodes and the transformer winding. The input isolation is limited only by the RC corner frequency and transistor leakage. Depending on the transistor, you may expect AC input rejection from 40-70db. This is where transistor selection makes all the difference. I won't go into detail on all the ways to improve this dinosaur, but here are a number of ways for you to consider if you don't really need the performance that can be gained from an extra transistor:
• Use a second-order RC filter
• Much improves isolation near the corner frequency, and at higher frequencies up to the leakage limit.
• Use a diode string or LED in place of the resistor
• This will make turn-on instant and dynamically adjust resistance to keep Vcb equal to the diode breakover. But watch the input ripple and your voltage drop.
• Use an FET or MOSFET
• This lets you make the resistor exceedingly large. What you gain in corner frequency you may lose in leakage, voltage drop, and output impedance. Perhaps this concept is more applicable to active audio filters.
At this point there are many, many things we could try to tailor the performance in many directions. Of these, the options adding another active device tend to become a bit less flexible. If we replace Q1 with a Darlington pair for instance (diagram below), you will need to draw enough current at all times to keep the driver transistor on. An extra diode drop is added, but possibly offset by a lower R1 voltage drop. Even so, the performance gains can be dramatic. Due to the greatly decreased base current, we can increase R1 by several times. This allows output filtering to the subsonics, or alternatively less resistive output impedance at DC. However AC output impedance can be made worse. Say for instance our driver transistor is biased at 2mA and our output transistor has an Hfe of 100. The load draws 102mA. Following the diodic output impedance rule, Q1 defines most of the output impedance at 330mR. It's base current is 100mA/100, 1mA. The diodic resistance of Q2's emitter is divided by the output's Hfe like in the original C-multiplier, and comes to 165mR. So the total output impedance, neglecting R1, is 495mR. Often times designers neglect to give the driver any bias current at all except the base current of the  output. Because the output's base current rises proportionally with load current, and the diodic emitter resistance decreases proportionally with emitter current, the net result is that proportions cancel, and the Darlington output resistance gains the nonlinearity of 2 diodes in series - 660mR. Neat, right? Of course, you had better decouple the supplies well, because any fast load signals will saturate and pump the driver transistor and result in nasty glitching.

Ultimately, the Darlington C-multiplier still leaves us wanting more. Many designers don't feel like pushing the limit -  the C-multiplier was never the Rolls Royce of supply solutions anyway. Why not just use an LM7812? But because of reasons mentioned at the beginning of this page, even that is not a satisfying option. Is there a middle ground between simplicity, expediency and performance?

Most designers already know about the benefits of a CFP over a Darlington, even though few seem to have thought to apply it to a C-multiplier. The CFP has higher transconductance in common emitter form, which translates to lower output impedance in common collector form. However the general consensus on the CFP from amplifier designers is that it is unstable and risky. Many early amplifier designs featuring CFP output stages were found to up and blow up one day for no apparent reason. Eventually it came out that the cause was that it has a tendency to oscillate, causing the transistors to dissipate a lot more power than they should have. For well-trained amplifier designers, the problem is just a matter of engineering, but the circuit in question must be measured in the prototype; RF parasitics depend so heavily on wiring that it is easier just to probe with a signal generator. Truth be told many designers don't have the background necessary to understand the problem.

I can go into detail on what the problem is in another article, but here I will discuss the principle of operation.

Here again is the Kmultiplier diagram for you to refer to throughout my explanation:

Another name for the CFP could be the "G-multiplier pair". In essence, the output conductance is the conductance of Q1 multiplied by the  current gain of the Q2/R1 arrangement. Let's analyze the situation and get an idea of how this works.

Q1 is biased into it's most linear range by R1. So Ic(Q1) is roughly .68/47=~15mA. This is a bit low to accommodate for a nominal max of 5mA Ib(Q2). In the nominal range of loading Ic(Q1) ranges from 15mA to 20mA. So, once more following the diode rules, the output resistance of Q1 varies between 2.2R and 1.65R. Now lets include Q2. Lets say our load current is 116mA. This sets the Ic of Q1 and Q2 to 100mA and 16mA respectively, accounting for Q2's Hfe of about 100. At 100mA, Q2's Re is about 330mR. For every 1mA of loading, Vbe(Q2) increases 330uV. This increase in voltage across R1 results in a 7uA increase in it's current. This is added to the increase in Ib(Q2) of 10uA and we get an increase in Ic(Q1) of 17uA. 17uA across Q2's Re of ~2.2R gives us a final 37uV output drop per 1mA. 37uV/1mA gives us 37mR as the entire arrangement's output impedance. This is close enough for horseshoes to the measured values.

Back to Home

Page Views: 588

Background images generated by the E-volved Cultures Screensaver. Other content copyright me, unless otherwise indicated.