## Circuit custom power supply for audio amplifier – Symmetrical

Table of Contents

Power amplifiers that provide reasonable power require a power supply sized according to the circuit. Most amplifier circuits require a symmetrical + -V power supply. The schematic features a traditional non-regulated symmetrical source assembly with transformer, rectifiers, and large filter capacitors.

The diodes for the circuit should be sized according to the amplifier being used. The electrolytic capacitors should be at least 1000Î¼F for each current ampere used in the power supply of the amplifier.
Good filtering of the power supply is what guarantees the excellent sound quality in hi-fi systems.

Transformer for the supply circuit The value of the final voltage, can be calculated using the following formula:
Tf = (1,4142 x Ti) â€“ 0,5
Tf = final voltage (DC)
Ti = voltage at the output of the transformer (AC)
0.5 = Diode voltage drop

##### Get new posts by email:

Transformer voltage from DC power supply
Ti = (TF/1,4142) + 0,5
Transformer power

Example: We will use a 200W Class AB amplifier of full power.We know that AB class will yield about 60%.
So Pt = Pa(Pa x 0,4) = Pt = 200 (200Ã—0,4) = 280 Watts
Pt = Transformer power
Pa = Power of the amplifier

Transformer voltage for our amplifier
Let us assume that the maximum supply voltage is + -40V (80V).
Ti = (TF/1,4142) + 0,5 = Ti = (80/1,4142) + 0,5 = 57,06
As the transformer is double we have 57,06 / 2 = 28,53

We will use the next lower commercial value that this value is found, in this case a 28V + 28V transformer.

Transformer Current By Ohm’s Law -> I = P / V
V = 56 V
P = 280 W
I = 280/56 = 5A

For our 200W Class AB amplifier, we need a 28 + 28V / 5A transformer.

Choice of diodes Parameters for choosing the diode:
Maximum Repetitive Reverse Voltage (VRRM)
IFAV (Average Forward Rectified Current)
In our case
VRRM = Vp
IFAV = ImX1,414
Im = Average load current Vp = peak voltage

In our case study:
VRRM = 40V
IFAV = 1.414 x 3.5 = 4.94 A
So let’s choose to meet these parameters, at least 30% more.

Supply capacitors
Practical formula to find the capacitor:
C = I / (2xF xVr)
C = Capacitor – capacitance at Farad
I = DC current of the circuit
F = Frequency of power in Hertz
2 = Correction factor for frequency in a full-wave rectifier
Vr = Ripple voltage allowed for the circuit.

DC current (I)
Let’s divide the total power consumption to determine the current on each side of the power suply.
280W / 2 = 140W then I = 140W / 40VDC = 3.5A

Voltage of riple (Vr)
Consider:
3 to 5% = Very good
Up to 7% = Good
Up to 10% = Regular
This tension depends on the purpose of your design and your wallet.
Let’s stay with 7%
I = 3.5A
F = 60Hz
Vr = 40V * (7/100) = 2,8V
With the data we have:
C = 3.5A / (2 x 60 x 2.8) = 0.01041F or 10.410Î¼F

We will use capacitor of commercial value of 10.000Î¼F in our project. Note that when it is said that high capacitance is good for bass, it is precisely at the moment that the amplifier is requesting high current from the power supply.

Resistor for led R = (Vsupply – Vled) / Iled
In our case with red led:
R = 40V – 2V / 20 Ã— 10-3 = 1900 Ohms
For power, we use Ohm’s law
P = V2 / R = 40 2/1900 = 0.842 W
We will use 2.2Kohms 1W

 Led Led Color Voltage (V) Current (mA) RED 1,8 – 2V 20mA Yellow 1,8 – 2V 20mA Orange 1,8 – 2V 20mA Green 2 – 2,5V 20mA Blue 3V – 3,5V 20mA White 3 – 3,5V 20mA

Attention: If you use a power supply that provides voltage greater than the amplifier circuit, the power transistors or the integrated circuit will burn. The amplifier circuits usually accept various voltage supply values, so if you use the minimum value you will have less power, you will use a smaller heatsink and capacitors with a lower working voltage.

About power supply circuit Resistors R2 and R5 are safety components as they discharge the large capacitors when the amplifier is switched off. C7 and C8 are bypass capacitors, and the RC networks R4 / C13 and R5 / C14 reduce the parasitic inductance of large electrolytes, this improves the performance of the power supply at high frequencies and the performance of the amplifier in complex loads. The led indicates the operation of the power supply and should be placed on the amplifier panel. For circuit safety, use a fuse in the 220v line at the power supply and at the connection with the amplifier circuit.
The power supply circuit can be placed in the same enclosure where the power amplifier is mounted.

### Power supply 01 – Using diodes TO-220

Power Supply Schematic

Printed Circuit Board Suggestion

#### Part list

 Parts Valor Resistors R1 10K 1/4W R2, R3 2.2K – Resistor 3 at 5W R4, R5 1 Ohm 2Watts Capacitors C1, C2, C3, C4 10n – Capacitor ceramic C5, C6, C9, C10 3300ÂµF a 4700ÂµF- Capacitor Electrolytic (Optional) C7, C8, C13, C14 100n – Capacitor ceramic C11, C12 10.000ÂµF – Capacitor Electrolytic Semiconductors D1, D2, D3, D4 MUR860 or equivalent LED1 Red Led 3mm Connectors DC 3 Pin Screw Terminal Block Connector – DC Output AC 3 Pin Screw Terminal Block Connector – AC Input Miscellaneous Printed circuit board, solder, wires, transformer according to design, Box, etc.

Download files in PDF
Mirror

### Power Supply 02 – using rectifier bridge

Power Supply Schematic

Printed Circuit Board Suggestion

Part List

 Components Value Resistors R1 10K – Resistor 1/4W R2, R3 2.2K – Resistor 3 at 5W R4, R5 1 Ohm 2 Watts Capacitors C1, C2, C3, C4 10n – Capacitor ceramic C5, C6, C9, C10 3300ÂµF a 4700ÂµF- Capacitor Electrolytic (Optional) C7, C8, C13, C14 100n – Capacitor ceramic C11, C12 10.000ÂµF – Capacitor Electrolytic Semiconductors B1 Bridge Rectifier 10A /300V LED1 Red Led Connectors DC 3 Pin Screw Terminal Block Connector – DC Output AC 3 Pin Screw Terminal Block Connector – AC Input Miscellaneous Printed circuit board, solder, wires, transformer according to design, Box, etc..

* The bridge rectifier must be equipped with a heatsink

### Power supply 03 – With diodes 10A10

Power Supply Schematic

Printed Circuit Board Suggestion

Part List

 Components Value Resistors R1 10K – Resistor 1W R2, R3 2.2K – Resistor 3 at 5W R4, R5 1 Ohms 2 Watts Capacitors C1, C2, C3, C4 10n – Capacitor ceramic C5, C6, C9, C10 3300ÂµF a 4700ÂµF- Capacitor Electrolytic (Optional) C7, C8, C13, C14 100n – Capacitor ceramic C11, C12 10.000ÂµF – Capacitor Electrolytic Semiconductors D1~D4 Diodes 10a10 or equivalent LED1 Red Led Connectors DC 3 Pin Screw Terminal Block Connector – DC Output AC 3 Pin Screw Terminal Block Connector – AC Input Miscellaneous Printed circuit board, solder, wires, transformer according to design, Box, etc..

How useful was this post?

Click on a star to rate it!

Average rating / 5. Vote count:

No votes so far! Be the first to rate this post.

As you found this post useful...

Share on the social networks!

We are sorry that this post was not useful for you!

Let us improve this post!

Tell us how we can improve this post?

About the author
Xtronic.org blog author. Electronics technician for the technical school of Brasilia - Brazil. Interested in electronics, circuits and technology in general.
Share:

### 8 thoughts on “Circuit custom power supply for audio amplifier – Symmetrical”

1. Circuit custom power supply for audio amplifier â€“ Symmetrical 2x40v

Hi, Toni!
Thanks for your very interesting article.
I am going to make two power supplies for two separate single power amps (mono).
I have two toridial 2x24v (nominal) 170va transformers.
Should I change something in the power supply wiring diagram? The example wiring diadram is for 2x28v transformer.
I like to reduce the ripple. Should I change something in the wiring diagram if adding more filter caps as shown, perhaps up to 18000 microFarads?

Regards,
Mr. Seppo Rapo
[email protected]

Reply
• Hi Mr. Seppo Rapo,

Thank you for your kind words and I’m glad you found the article interesting. Regarding your question about the power supply wiring diagram, you can definitely use your 2x24v transformers instead of the 2x28v transformers mentioned in the example wiring diagram. The nominal voltage difference should not cause any major issues, but it’s always a good idea to double-check the specifications of your specific transformers to ensure compatibility with the circuit.

In order to reduce ripple, adding more filter capacitors can be a good solution. Increasing the capacitance, as you mentioned, up to 18000 microFarads can help in smoothing out the output voltage and reducing any residual ripple. However, it’s important to note that adding excessive capacitance may also introduce additional stress on the rectifier diodes during startup, so it’s recommended to consider the voltage and current ratings of the diodes in order to avoid any potential issues.

Overall, modifying the wiring diagram to accommodate your 2x24v transformers and increasing the capacitance of the filter capacitors should be feasible. Just ensure that the components you use are appropriately rated for the expected voltage and current levels.

If you have any further questions or concerns, please feel free to ask. Good luck with your power supply project!

Reply
2. Hi Tony, what PCB software did you use in this project? Thank you in advance for your attention!

Reply
3. The calculation and design are very helpful. In some audio amplifiers the supply is single. In that case, can we use only the half of this circuit or anything else to be added ?

Reply
• Hi Tony
I can follow your calculation but I do not get the 1.6 from this: C = 3,5A / (2 x 60 X 1,6) = 0,01041F of 10,410Î¼F How do I get that 1.6 in this formula?
Thanks

Reply
• Hello FHJ Beiler
Errata: correct value is 2.8. (VR)