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SPHP Positive

SPHP Positive
SPHP Positive
New
SPHP Positive
SPHP Positive
SPHP Positive
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$69.50
Ex Tax: $69.50
10 or more $54.55
50 or more $45.87
100 or more $39.65
  • Stock: In Stock
  • Model: SPHP Positive
  • Weight: 3.00g
  • Dimensions: 1.50in x 0.50in x 0.50in
  • SKU: SPHPP

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Create unlimited custom product blocks and display them in accordions or tabs or open blocks. Each block can be assigned to all products at once or specific products according to advanced criteria.

SPHP Superpower High Power Voltage Regulator. Up to 1000W output power: 10A at 100V! Use in a power amp, a computer music server, mixers, anywhere a clean, quiet high power source is needed.

  • Positive fixed or adjustable output +5V to +100V
  • ≥10A output current
  • <10mΩ output impedance
  • <5µV/Vout noise
  • Requires custom wiring and mechanical placement for output transistor

 

Download full data sheet (PDF) here.

Parameter

Conditions

Value

Units

Input voltage maximum

 

120

V

Output voltage positive

Adjustable or fixed output

5 to 100

V

Output voltage negative

Adjustable or fixed output

-5 to -100

V

Output Noise

RMS 20Hz – 20KHz (3*)

<1

PPM of Vout

Line Rejection

60Hz, 1Vpk

110

dB

Continuous current

Within power dissipation limits
drop-out voltage
 3V

5V


10
14

A

Maximum power dissipation (2*)

no heat sink
sufficient heat sink

30
160

W

Drop–out voltage (typical)

load current 0 to 1.5A

2A to 4A
4A to 10A

2.0
2.5
3.0

V

Output Impedance

20Hz – 20KHz

50

mΩ

Belleson's SPHP Superpower can deliver up to 10A at 100V. This is enough for most power amplifiers and to replace noisy and slow switched mode power supplies in computer based music servers. High current circuit design requires careful thought about power dissipation, where current flows and other important topics. We'll discuss this diagram of a basic power supply using SPHP: SPHP wiring diagram. To start, let's define some terms:

VIN
Input voltage to regulator
VOUT
Voltage regulator output
VDO
Regulator drop-out voltage: minimum (VIN-VOUT) to keep regulation
IL
Load current from regulator output to its load and back to power source (typically a transformer+rectifier+filter)
PL
Load dissipation: VOUT / IL
PR
Regulator dissipation: (VIN - VOUT) / IL
LDO
Low Drop Out
VRIP
Input Ripple, the AC change at VIN

 

Power Dissipation—Regulator vs. Load

SPHP can provide up to 1000W to a load while the regulator can dissipate 200W. Total load dissipation is thus 1200W. You must choose a power transformer with high enough VA to supply the total power.

Any power used by the regulator is not delivered to the load and is considered wasted. Given regulator power dissipation = drop-out voltage times load current, it's easy to see why low drop-out voltage is important, and why LDO is a standard acronym in regulator data sheets. The closer VIN is to VOUT, the less power is wasted.

Drop–Out Voltage and Supply Efficiency

SPHP drop-out, as for most linear regulators, increases with load current to a maximum of 3V at 10A. Thus it is theoretically possible to make a 1000 Watt power supply with 30W of wasted power, giving an efficiency of 97%. Practically speaking, however, an efficiency of 80% is considered great for a high power linear supply. Why is this?

Looking at the above schematic, AC current is rectified by bridge rectifier BR1 and filtered (smoothed) by capacitor C1. After mains power is applied, C1 charges to its peak DC value and will remain charged until current is requested by the load. Realize that C1 only charges when the peak voltage from BR1 goes above C1 voltage so for much of the AC cycle, C1 can supply current but not receive it.

Input Ripple Voltage and Regulation

While C1 is supplying current and bridge voltage is below C1 voltage, C1 discharges. There is a long time (relative to a power line cycle) when C1 discharges and a short time when it recharges. This is what makes the familiar saw tooth voltage ripple at the input to the regulator. Input Ripple

As you know, i=Cdv/dt and higher load current discharges C1 more quickly. If C1 discharges enough during an AC power line cycle to go below (VOUT+VDO), the regulator stops regulating!

 

Regulator Input Capacitance and Drop Out Voltage

 

Look again at i=Cdv/dt. Rearranging terms gives dv=idt/C to show that ripple is reduced by increasing C. Now rearrange again and substitute frequency f=1/dt to get an equation for C1 given load current and maximum desired ripple: C=i/(2fdv). The factor of 2 is because BR1 is a full wave rectifier and C charges twice per cycle. A half wave rectifier does not have this factor.

 

For 10A load and 1V maximum ripple at a 50Hz power line cycle (worst case, also makes the math prettier), and 1V maximum ripple target, the required C1=10/100, or 100000µF. The 47000µF value assigned to C1 in the above schematic will allow about 2.5V maximum ripple voltage.

Given that the minimum ripple voltage must stay above VOUT+VDO, the voltage at BR1 must go above VOUT+VDO+VRIP to keep C1 charged enough to allow the regulator to function correctly. For a 12V regulator, BR1 voltage must stay above 12+3+2.5=17.5V using C1=47000µF. This explains why a supply efficiency of 97% is unrealistic. For this 12V regulator, efficiency is 100 x 12 / 17.5 = 69%. Even with C1=100000µF, efficiency goes to 75% which wastes 1/4 of the input power. Ripple depends only on load current, so given the same load and VDO, a higher VOUT supply will be more efficient than a lower VOUT supply.

Here is a really good tutorial on building a 10A power supply

General Conclusions

 

  • Power dissipation by the regulator is linear as (Vin-Vout)*(load current)
  • (Vin - Vout) is RMS voltage
  • If Vin is fed from a rectifier+filter cap, Vin is not DC but has ripple
  • Ripple has a linear dependence on load current as
    dv=i/(2fC)
    where dv = ripple amplitude, i=RMS load current, f = power line frequency and C=filter capacitance
  • Minimum peak of ripple must not go below (Vout + Vdropout), otherwise regulator stops regulating

From this we conclude:

  • Larger filter capacitance = lower ripple
  • Lower ripple allows lower Vin
  • Lower Vin allows lower regulator power dissipation

More Overhead

For the chosen 12V supply, the transformer must keep VIN above 17.5V to maintain regulation. With high current demand, transformer secondary voltage tends to sag (decrease). The peak unloaded voltage of the transformer must be increased to account for sag, and also to account for the worst case low primary voltage. Ultimately this transformer must have a 20V to 24V peak output voltage due to these multiple system constraints.

High Current and Physical Design

Now that we've decided on a set of components: SPHP regulator, power transformer, bridge rectifier, large filter capacitor, how do we connect them? Two important factors in wiring an accurate high current power supply are wiring resistance and stability. At 10A, 10mΩ equates to a 0.1V drop. As current changes through an impedance, it will modulate the voltage at the load, so even with a perfect voltage source, a changing high current load can modulate the voltage at the load due to wiring resistance.

SPHP is designed with a small PCB controller and a separate high current output transistor (QN1 in the schematic). This allows the control loop to be independently placed and wired with a maximum current of about 250mA, and QN1 can be separately heat sink mounted with large currents flowing separately through it.

The diagram below shows physical wiring of the above schematic with matching wire colors. Notice the red path of high current from the raw supply to the power transistor to the load and back to the raw supply. Make this path short, from heavy gauge copper trace or wire. The black and green DRV current paths can be smaller, they will have approximately 250mA at full 10A regulator current. The other paths are very low current.

SPHP high current wiring.

 

Two external protection diodes allow stored charge from large capacitance to by-pass the regulator at power-down if input voltage falls faster than output voltage or if VOUT gets pulled below ground by some load related condition (e.g. an inductive load). Even though SPHP has internal protection diodes, larger external ones such as 1N4004 are recommended.

SPHP has no built-in stabilizing capacitor, so an external one must be connected from VOUT to ground as shown here. This should be 100µF or more and have a voltage rating higher than VOUT. Also notice the diagram is not to scale—the capacitor will be bigger than the Superpower :-).

The power transistor supplied with SPHP is NJW3281G, a 250V, 15A, 200W device.

Preventing Damage

With 10A+ available, it's easy to damage a regulator with even the briefest of short circuit to ground. The fast shutdown circuit shown below can prevent this:

SPHP shutdown protection.

 

It's a latching protection circuit that shuts off the internal control loop and prevents any output current from the regulator. To reset, VIN must be powered down and up again.

Good Luck!

Hopefully this is sufficient information to build the power supply of your dreams! Contact us with any further questions and we'll help.

 

ATTENTION! SPHP users please read this.

As you know, SPHP is a high power voltage regulator. To quote Spiderman's Uncle Ben (who was quoting Sir Winston Churchill), “With great power comes great responsibility.” Please note the following and be careful during design and testing of your power supply:

  • Use extreme caution with high voltage circuits, this circuit can deliver lethal voltage and current. Keep one hand in your pocket!

  • There is no internal limit on output current. With 10+ amps, SPHP is unforgiving of mistakes and is quickly destroyed by an output short circuit or other incidents. If possible, use an inexpensive monolithic substitute such as LT1084 for new designs, then replace with SPHP when design is fully functional. The IN/GND/OUT pins on positive SPHP controller and GND/IN/OUT on negative SPHP match LM78xx and LM79xx pin connections for temporary substitution at lower voltage and current.

  • Output voltage is factory adjusted to 12V during testing. If your system is designed for high voltage Vout and thus has high input voltage, adjust SPHP to the desired output voltage under NO LOAD. In other words, if you power up SPHP under load with 12V output but Vin = 80V (for example) and load current = 5A, that puts the regulator output transistor under 340W load until Vout is adjusted to its correct value. This will destroy the output transistor and probably the regulator controller.

  • Vout adjustment is reverse of convention— clockwise decreases Vout, counter-clockwise increases Vout.

  • For best transient response, place a 100Ω resistor between base and emitter of the output power transistor.

  • For best stability across all load current, place a 0.1µF polypropylene capacitor from Vin to GND pins near the regulator control PCB.

  • SPHP has two internal protection diodes, however they are small SMD devices. If your application has large capacitance at regulator input and output, two external high current protection diodes (e.g. 1N4007) from [Vout to Vin] and [GND to Vout], [given anode to cathode], will improve long term reliability.

  • The Superpower data sheet has a special section discussing SPHP. Also read the tab FAQ:10A Power Supply before requesting support.

 

Belleson's SPHP Superpower can deliver up to 10A at 100V. This is enough for most power amplifiers and to replace noisy and slow switched mode power supplies in computer based music servers. High current circuit design requires careful thought about power dissipation, where current flows and other important topics. We'll discuss this diagram of a basic power supply using SPHP: SPHP wiring diagram. To start, let's define some terms:

VIN
Input voltage to regulator
VOUT
Voltage regulator output
VDO
Regulator drop-out voltage: minimum (VIN-VOUT) to keep regulation
IL
Load current from regulator output to its load and back to power source (typically a transformer+rectifier+filter)
PL
Load dissipation: VOUT / IL
PR
Regulator dissipation: (VIN - VOUT) / IL
LDO
Low Drop Out
VRIP
Input Ripple, the AC change at VIN

 

Power Dissipation—Regulator vs. Load

SPHP can provide up to 1000W to a load while the regulator can dissipate 200W. Total load dissipation is thus 1200W. You must choose a power transformer with high enough VA to supply the total power.

Any power used by the regulator is not delivered to the load and is considered wasted. Given regulator power dissipation = drop-out voltage times load current, it's easy to see why low drop-out voltage is important, and why LDO is a standard acronym in regulator data sheets. The closer VIN is to VOUT, the less power is wasted.

Drop–Out Voltage and Supply Efficiency

SPHP drop-out, as for most linear regulators, increases with load current to a maximum of 3V at 10A. Thus it is theoretically possible to make a 1000 Watt power supply with 30W of wasted power, giving an efficiency of 97%. Practically speaking, however, an efficiency of 80% is considered great for a high power linear supply. Why is this?

Looking at the above schematic, AC current is rectified by bridge rectifier BR1 and filtered (smoothed) by capacitor C1. After mains power is applied, C1 charges to its peak DC value and will remain charged until current is requested by the load. Realize that C1 only charges when the peak voltage from BR1 goes above C1 voltage so for much of the AC cycle, C1 can supply current but not receive it.

Input Ripple Voltage and Regulation

While C1 is supplying current and bridge voltage is below C1 voltage, C1 discharges. There is a long time (relative to a power line cycle) when C1 discharges and a short time when it recharges. This is what makes the familiar saw tooth voltage ripple at the input to the regulator. Input Ripple

As you know, i=Cdv/dt and higher load current discharges C1 more quickly. If C1 discharges enough during an AC power line cycle to go below (VOUT+VDO), the regulator stops regulating!

 

Regulator Input Capacitance and Drop Out Voltage

 

Look again at i=Cdv/dt. Rearranging terms gives dv=idt/C to show that ripple is reduced by increasing C. Now rearrange again and substitute frequency f=1/dt to get an equation for C1 given load current and maximum desired ripple: C=i/(2fdv). The factor of 2 is because BR1 is a full wave rectifier and C charges twice per cycle. A half wave rectifier does not have this factor.

 

For 10A load and 1V maximum ripple at a 50Hz power line cycle (worst case, also makes the math prettier), and 1V maximum ripple target, the required C1=10/100, or 100000µF. The 47000µF value assigned to C1 in the above schematic will allow about 2.5V maximum ripple voltage.

Given that the minimum ripple voltage must stay above VOUT+VDO, the voltage at BR1 must go above VOUT+VDO+VRIP to keep C1 charged enough to allow the regulator to function correctly. For a 12V regulator, BR1 voltage must stay above 12+3+2.5=17.5V using C1=47000µF. This explains why a supply efficiency of 97% is unrealistic. For this 12V regulator, efficiency is 100 x 12 / 17.5 = 69%. Even with C1=100000µF, efficiency goes to 75% which wastes 1/4 of the input power. Ripple depends only on load current, so given the same load and VDO, a higher VOUT supply will be more efficient than a lower VOUT supply.

Here is a really good tutorial on building a 10A power supply

General Conclusions

 

  • Power dissipation by the regulator is linear as (Vin-Vout)*(load current)
  • (Vin - Vout) is RMS voltage
  • If Vin is fed from a rectifier+filter cap, Vin is not DC but has ripple
  • Ripple has a linear dependence on load current as
    dv=i/(2fC)
    where dv = ripple amplitude, i=RMS load current, f = power line frequency and C=filter capacitance
  • Minimum peak of ripple must not go below (Vout + Vdropout), otherwise regulator stops regulating

From this we conclude:

  • Larger filter capacitance = lower ripple
  • Lower ripple allows lower Vin
  • Lower Vin allows lower regulator power dissipation

More Overhead

For the chosen 12V supply, the transformer must keep VIN above 17.5V to maintain regulation. With high current demand, transformer secondary voltage tends to sag (decrease). The peak unloaded voltage of the transformer must be increased to account for sag, and also to account for the worst case low primary voltage. Ultimately this transformer must have a 20V to 24V peak output voltage due to these multiple system constraints.

High Current and Physical Design

Now that we've decided on a set of components: SPHP regulator, power transformer, bridge rectifier, large filter capacitor, how do we connect them? Two important factors in wiring an accurate high current power supply are wiring resistance and stability. At 10A, 10mΩ equates to a 0.1V drop. As current changes through an impedance, it will modulate the voltage at the load, so even with a perfect voltage source, a changing high current load can modulate the voltage at the load due to wiring resistance.

SPHP is designed with a small PCB controller and a separate high current output transistor (QN1 in the schematic). This allows the control loop to be independently placed and wired with a maximum current of about 250mA, and QN1 can be separately heat sink mounted with large currents flowing separately through it.

The diagram below shows physical wiring of the above schematic with matching wire colors. Notice the red path of high current from the raw supply to the power transistor to the load and back to the raw supply. Make this path short, from heavy gauge copper trace or wire. The black and green DRV current paths can be smaller, they will have approximately 250mA at full 10A regulator current. The other paths are very low current.

SPHP high current wiring.

 

Two external protection diodes allow stored charge from large capacitance to by-pass the regulator at power-down if input voltage falls faster than output voltage or if VOUT gets pulled below ground by some load related condition (e.g. an inductive load). Even though SPHP has internal protection diodes, larger external ones such as 1N4004 are recommended.

SPHP has no built-in stabilizing capacitor, so an external one must be connected from VOUT to ground as shown here. This should be 100µF or more and have a voltage rating higher than VOUT. Also notice the diagram is not to scale—the capacitor will be bigger than the Superpower :-).

The power transistor supplied with SPHP is NJW3281G, a 250V, 15A, 200W device.

Preventing Damage

With 10A+ available, it's easy to damage a regulator with even the briefest of short circuit to ground. The fast shutdown circuit shown below can prevent this:

SPHP shutdown protection.

 

It's a latching protection circuit that shuts off the internal control loop and prevents any output current from the regulator. To reset, VIN must be powered down and up again.

Good Luck!

Hopefully this is sufficient information to build the power supply of your dreams! Contact us with any further questions and we'll help.

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NOTE! This calculator only applies to a linear transformer+rectifier+filter cap power source. It is not accurate if the raw source is a SMPS.

Use this calculator to select a suitable heat sink for your Superpower regulator. The Max Heat Sink value is the highest thermal resistance allowed for the given conditions. The bigger the thermal resistance, the smaller the heat sink.

Calculator fields have only minimum validation so if the Vrms result has something bizarre (like "NaN"), recheck your input values. If Vin is negative, Vout must also be negative, otherwise the calculations are incorrect.

Superpower type
(select to set drop-out)
Drop out voltage:

Regulator DC input voltage (volts)

Regulator output voltage (volts)

Max Load Current (amps)


Regulator Dissipation (Watts)

Max heat sink °C/W

Heat Sink

The value in the Max heat sink °C/W box shows the maximum thermal resistance for a heat sink on a Superpower with the given Vin. The heat sink calculation assumes a 75°C temperature increase of the regulator.

Assumptions

  • Input voltage is DC or DC equivalent in Vrms
  • Select SP for current < 500mA, SPJ or SPL for higher currents
  • Calculator only works to 3A
  • Heat sink allows 75°C temperature rise due to power dissipation

Can this calculator be used for any voltage regulator?

It can be used for any series voltage regulator if you know the drop-out voltage. For Superpower Type choose Custom regulator and enter the drop-out value for your regulator at the given load current.

Superpower Transformer Calculator

Use this calculator to decide the best transformer to use for your Superpower supply. Given the values you enter, it computes the ripple, decides the drop out voltage based on selected Superpower type and load current, sums everything and calculates the minimum Vrms of the transformer.

Calculator fields have only minimum validation so if the Vrms result has something bizarre (like "NaN"), recheck your input values. If Vrms is negative, Load current exceeds the capability of the selected Superpower type.


Superpower type
(select to set drop-out)
Drop out voltage
Regulator output voltage (volts)
Max Load Current (amps)
Line frequency
Rectifier Diode Drop (volts)
Rectifier Filter Capacitance (µF)
Line voltage variation (in %)
Safety margin (in %)  

Transformer 2ary Vrms
Transformer min. VA
  Regulator input Vpeak
Regulator Dissipation (Watts)
  Max ripple (Vpk-pk)
Max heat sink °C/W

Heat Sink

The value in the Max heat sink °C/W box shows the maximum thermal resistance for a heat sink on a Superpower with the Vin shown in Regulator input Vpeak with the other values as given. To see the heat sink needed for a different Vin, change the value in Line Voltage Variation until the Regulator input Vpeak equals the Vin you will use in your application. The heat sink calculation assumes a 75°C temperature increase of the regulator.

Assumptions

 

  • Linear power supply with transformer/rectifier/filter caps/Superpower
  • Transformer has sufficient power that it does not sag under load (use safety margin to account for sag)
  • Vdc of rectifier output is minimum value + safety margin
  • Full wave center tapped rectifier follows the transformer.
    For a bridge with no center tap, double the diode drop
  • Transformer output voltage is specified as Vrms
  • Filter capacitance is entered in µF
  • Select the power line frequency for your locale
  • Regulator dissipation assumes nominal line voltage but allows for a drop of line variation % without losing regulation
  • Rectifier diode drop allows entry of Si, SiC or other diode drop
  • Heat sink allows 75°C temperature rise due to power dissipation

 

Special thanks to a customer whose suggestions helped us improve this calculator...you know who you are!

FAQ

How can Vrms be less than Vout?

Transformers are specified as Vrms. Full wave rectified and filtered transformer voltage is, with no load, approximately Vpeak, which is Vrms X sqrt(2). So Vrms is lower than the Vpeak required at the regulator's input, and with low output current requirements, may be lower than regulator Vout.

Why is Vrms negative and almost 1000?

The calculator does this when Load Current exceeds the capability of the selected Superpower type.

What kind of capacitors should I use for a rectifier filter?

Use the electrolytic capacitor of your choice. The most important issue for regulation is to have sufficient capacitance to prevent ripple that goes below Vout+Vdropout.

Should I bypass the filter capacitors with ceramic?

Yes, a 0.1µF ceramic cap at the Superpower Vin terminal helps reduce high frequency noise and RF. This amount or more capacitance should be placed at the Vin terminal to ground to prevent possible low level oscillation at some load current values. This does not affect the calculation very much.

How much filter capacitance should I use, can I use too much?

More filter capacitance is better, it reduces ripple. When the room lights start to dim as you switch on the power supply, you may be reaching the point of "too much." Or maybe you should run a separate mains wire for your audio system :-).

Can this calculator be used for any voltage regulator?

It can be used for any series voltage regulator if you know the drop-out voltage. For Superpower Type choose Custom regulator and enter the drop-out value for your regulator at the given load current.

Superpower regulator

Dissipation & heat sink calculator

Works out how much heat the regulator has to shed at your operating point, and the largest thermal resistance a heat sink may have to hold the rise near 75 °C. Figures assume a linear supply — transformer, rectifier and filter cap. A switching supply upstream makes them meaningless.

Operating point

V
V
V
A

Readout

Regulator dissipation
---.--W
Max sink thermal resistance
---.--°C/W

Heat load

0 W10 W20 W+
Waiting for inputEnter an input voltage, output voltage and load current. The readout updates as you type.

Max sink °C/W

The highest thermal resistance a heat sink may have here. Bigger number, smaller sink. Anything at or below the figure shown holds the rise to roughly 75 °C in free air.

What the maths does

Dissipation is (Vin − Vout) × (Iload + 10 mA), the extra 10 mA covering the regulator's own draw. Sink resistance is 75 °C divided by that wattage.

Drop-out

Vin − Vout must stay above the drop-out figure or the regulator stops regulating and passes ripple straight through. Choose Other series regulator to enter the drop-out from your own datasheet.

Superpower regulator

Transformer calculator

Sizes the transformer for a Superpower supply. It works out the ripple, picks the drop-out voltage from the regulator and load current, adds the headroom you allow for line sag and safety margin, and returns the minimum secondary Vrms along with the heat the regulator will have to shed.

Regulator

V
A
V

Supply

V
µF
%
%

Readout

Secondary Vrms, minimum
--.--V
Transformer rating, minimum
--.--VA
Regulator Vpeak--.--V
Ripple, pk-pk--.--V
Dissipation--.--W
Max sink--.--°C/W

Heat load

0 W10 W20 W+
Waiting for inputEnter an output voltage, load current and filter capacitance. The readout updates as you type.

Reading the heat sink figure

Max sink is the highest thermal resistance a sink may have at the Vpeak shown, holding the regulator's rise to roughly 75 °C. Bigger number, smaller sink. To see the sink for a different input voltage, adjust line variation until Vpeak matches the figure you plan to run.

Where the numbers come from

Ripple is Iload ÷ (C × 2f). Vpeak stacks output, ripple and drop-out, adds the safety margin, then divides by the line variation you allow. Vrms is Vpeak ÷ √2 plus the diode drop. Dissipation is taken at the ripple trough, and sink resistance is 75 °C divided by it.

Assumptions

  • Linear supply: transformer, rectifier, filter caps, Superpower.
  • The transformer is stiff enough that it does not sag under load. Use the safety margin to cover sag.
  • Rectifier output Vdc is the minimum value plus the safety margin.
  • Choosing a bridge doubles the diode drop you enter, since two diodes conduct in series.
  • Transformer output is specified as Vrms; capacitance is entered in µF.
  • Dissipation assumes nominal line voltage but tolerates a sag of the line variation percentage without losing regulation.
  • The heat sink figure allows a 75 °C rise.

Questions

How can Vrms be less than Vout?

Transformers are specified in Vrms. Once rectified and filtered, the voltage sits near Vpeak, which is Vrms × √2. So the secondary Vrms is lower than the peak arriving at the regulator, and at low output currents it can even fall below the regulator's output voltage.

What kind of capacitors should I use for the filter?

Any electrolytic you like. What matters for regulation is enough capacitance that the ripple trough never dips below Vout plus the drop-out voltage.

Should I bypass the filter capacitors with ceramic?

Yes. A 0.1 µF ceramic at the Superpower Vin terminal cuts high frequency noise and RF. Put at least that much from Vin to ground to head off low level oscillation at some load currents. It barely affects the calculation.

How much filter capacitance, and can I use too much?

More is better, it reduces ripple. When the room lights dim as you switch the supply on, you may be approaching too much. Or perhaps the audio system wants its own mains run.

Can this be used with any voltage regulator?

Any series regulator, as long as you know its drop-out voltage. Choose Other series regulator and enter the drop-out from the datasheet at your load current.

With thanks to the customer whose suggestions improved this calculator. You know who you are.