PSU: How does it convert energy?
Gone are the times when the generic 250W power supply could do our computers justice. Nowadays, its all about the amps, the efficiency and raw power.
This FAQ is to give you a small rundown on how a power-supply actually works. For the purposes of this excercise, we are going to assume, that the PSU we have is a 500W.
The source
The purpose of a power-supply is to provide power to the motherboard and other components (duh). So lets look at how it does this.
A normal house connection point carries a voltage of 240V AC (340V peak). Hence, for our power supply, we can make the following calculations:
Power = 500W
Voltage = 240V
According to the formula P=VI:
Current = 2.08A
This is the current that flows into the power-supply.
Transformer
The process of voltage transformation occurs within a PSU in a similar way to the diagram provided. The transformers function is dependent on a principle involving electromagnetic induction by which the current is transferred from one side of the core to the other (a description is beyond the scope of the FAQ).
The thing to remember is that the coil ratio (coils in primary side:coils in secondary side) is the same as that for the voltage.
The PSU hence has 3 transformers:
240:12 (12v rail), this means a coil ratio of (20:1) is achieved since the input voltage is 240.
240:5 (5v rail), this means a coil ratio of (48:1) is achieved.
240:3.3 (3.3v rail), this means a ratio of (2400:33) is achieved.
For this FAQ, I will look into further depth at the 12v rail.
Ratios
POWER:
Because the power in must equal the power out, it is constant.
ie. Power
ower = 500:500 = 1:1
VOLTAGE:
As we ascertained before, the ratio is 20:1
CURRENT:
As for the amps, a simple calculation must be undertaken.
Remember that P=VI
Therefore, when P = 500W, and V = 12V:
I = P/V = 41.6A
However, remember that power-supplies have two 12V rails;
as a result, the theoretical 12V amperage is 20.8A.
This is not often reflected by power-supplies.
Power loss
Ever wondered why PSUs have some trouble under load?
Well, as power required increases, as does the current.
Again recall that the voltage is a constant 12V.
Since P=VI, the current must increase to increase the power.
A new formula must be used now. Kirchoff's laws stipulate that V= IR.
Inside the power-supplies, there are a large amount of wires, each of which have a small resistance.
Hence we can see that when current increases, the voltage drop across the wires also increases. As the voltage drop increases, the voltage available is less.
This loss of power through resistance happens through the dissipation of power as heat. This is the reason why power supplies require a fan.
ie. 12 – V(drop) = V(available)
Thus, we can see fluctuations in cheaper power supplies.
These fluctuations also determine the efficiency of the power supply.
This we can see through the formula P=I^2R.
Thus, the power-loss is highly dependent on the resistance in the wires. The higher the power-loss, the lower the efficiency of the PSU.
How much does a PSU cost me?
While currencies will carry in different countries, for the sake of this FAQ, we will us $US.
Power is charged in kWh which is the amount of energy used when one kilowatt is dissipated over one hour.
What is kW exactly? 1W = 1 Js.
Therefore, 1kW = 1000 Js.
Hence, one kWh = 3.6MJ
An average of the amount charged per kWh is 6c.
A 500W PSU will use about 0.5kW per hour. Lets assume we are folding 24/7 for a year.
0.5 x 24 x 365 = 4,380 kWh used.
Therefore, we are charged: 4,380 x 0.06 = $262.80 just for power!
I hope you now are more liberal with power usage and hope you have an insight into the inside of a PSU. I don't have a pHd in physics so if there are any mistakes, please tell me.
Gone are the times when the generic 250W power supply could do our computers justice. Nowadays, its all about the amps, the efficiency and raw power.
This FAQ is to give you a small rundown on how a power-supply actually works. For the purposes of this excercise, we are going to assume, that the PSU we have is a 500W.
The source
The purpose of a power-supply is to provide power to the motherboard and other components (duh). So lets look at how it does this.
A normal house connection point carries a voltage of 240V AC (340V peak). Hence, for our power supply, we can make the following calculations:
Power = 500W
Voltage = 240V
According to the formula P=VI:
Current = 2.08A
This is the current that flows into the power-supply.
Transformer
The process of voltage transformation occurs within a PSU in a similar way to the diagram provided. The transformers function is dependent on a principle involving electromagnetic induction by which the current is transferred from one side of the core to the other (a description is beyond the scope of the FAQ).
The thing to remember is that the coil ratio (coils in primary side:coils in secondary side) is the same as that for the voltage.
The PSU hence has 3 transformers:
240:12 (12v rail), this means a coil ratio of (20:1) is achieved since the input voltage is 240.
240:5 (5v rail), this means a coil ratio of (48:1) is achieved.
240:3.3 (3.3v rail), this means a ratio of (2400:33) is achieved.
For this FAQ, I will look into further depth at the 12v rail.
Ratios
POWER:
Because the power in must equal the power out, it is constant.
ie. Power
ower = 500:500 = 1:1VOLTAGE:
As we ascertained before, the ratio is 20:1
CURRENT:
As for the amps, a simple calculation must be undertaken.
Remember that P=VI
Therefore, when P = 500W, and V = 12V:
I = P/V = 41.6A
However, remember that power-supplies have two 12V rails;
as a result, the theoretical 12V amperage is 20.8A.
This is not often reflected by power-supplies.
Power loss
Ever wondered why PSUs have some trouble under load?
Well, as power required increases, as does the current.
Again recall that the voltage is a constant 12V.
Since P=VI, the current must increase to increase the power.
A new formula must be used now. Kirchoff's laws stipulate that V= IR.
Inside the power-supplies, there are a large amount of wires, each of which have a small resistance.
Hence we can see that when current increases, the voltage drop across the wires also increases. As the voltage drop increases, the voltage available is less.
This loss of power through resistance happens through the dissipation of power as heat. This is the reason why power supplies require a fan.
ie. 12 – V(drop) = V(available)
Thus, we can see fluctuations in cheaper power supplies.
These fluctuations also determine the efficiency of the power supply.
This we can see through the formula P=I^2R.
Thus, the power-loss is highly dependent on the resistance in the wires. The higher the power-loss, the lower the efficiency of the PSU.
How much does a PSU cost me?
While currencies will carry in different countries, for the sake of this FAQ, we will us $US.
Power is charged in kWh which is the amount of energy used when one kilowatt is dissipated over one hour.
What is kW exactly? 1W = 1 Js.
Therefore, 1kW = 1000 Js.
Hence, one kWh = 3.6MJ
An average of the amount charged per kWh is 6c.
A 500W PSU will use about 0.5kW per hour. Lets assume we are folding 24/7 for a year.
0.5 x 24 x 365 = 4,380 kWh used.
Therefore, we are charged: 4,380 x 0.06 = $262.80 just for power!
I hope you now are more liberal with power usage and hope you have an insight into the inside of a PSU. I don't have a pHd in physics so if there are any mistakes, please tell me.
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My System
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