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 1000μF 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) – 0,5**

Tf = *final voltage (DC)*

Ti = *voltage at the output of the transformer (AC)*

0.5 = *Diode voltage drop*

**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 (200×0,4) = 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 1.6) = 0.01041F or 10.410μF

We will use capacitor of commercial value of 10.000μF 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 × 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.

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 ?

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 10,410μF How do I get that 1.6 in this formula?

Thanks

thanks