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Imagine you have a load which requires 5V at 1A and that you have a 12V supply (similar to this problem).
If you place a 7 ohm resistor in series with the load, you have 7W dissipated in the resistor and 5W in the load.
Now, imagine that you instead have a voltage divider with the bulb connected across the lower leg. You arrange the resistors so that the bulb still gets 5V at 1A. But now there is a resistor acorns the load. some current flows through it (let's call that Ir A. The power it will dissipate is 5 * Ir W. now the upper resistor also needs to carry that current, so it dissipates 7 * (1 + Ir) W. The difference is 7 * Ir W So the total extra power consumed (for no effect) is 12 * Ir W.
Since Ir is always positive if the extra parallel resistor is in place, the extra power is always positive, so the efficiency (5 * 1)/(7*(1+Ir) + (5*Ir) + (5 * 1)) (lamp power / total power) is always going to be lower.
With a linear voltage regulator, you essentially have an electronically controlled resistance in series which is varied to maintain 5V at the output. These have a small operating current and can appear to be a very smart voltage divider with the lower resistor having a high value. Because of this, if the load is constant, a resistor is (very slightly) more efficient. The efficiency approaches Vout/Vin (5/12 = 42% -- over half the energy is wasted)
With both resistors and linear regulators, the input current is essentially equal to the output current, so losses are (at a minimum) the load current multiplied by the difference in voltage between the input and output.
But we also have switchmode regulators. These are a constant power device. The output power is very similar to the input power. Typically they might have an efficiency in excess of 90%.
If we assume an efficiency of 90%, then a switchmode regulator having a 5V 1A output will draw (5/12)/(0.9) A from a 12V supply. This works out as around 0.463 A. Yes, the input current is LESS than the output current! In this case, the switchmode regulator would need to dissipate about 1/5 of the heat that a linear regulator (or resistor) would need to. Efficiencies can easily exceed the 90% I have quoted, so in real life the actual dissipation might be a lot lower.
If you place a 7 ohm resistor in series with the load, you have 7W dissipated in the resistor and 5W in the load.
Now, imagine that you instead have a voltage divider with the bulb connected across the lower leg. You arrange the resistors so that the bulb still gets 5V at 1A. But now there is a resistor acorns the load. some current flows through it (let's call that Ir A. The power it will dissipate is 5 * Ir W. now the upper resistor also needs to carry that current, so it dissipates 7 * (1 + Ir) W. The difference is 7 * Ir W So the total extra power consumed (for no effect) is 12 * Ir W.
Since Ir is always positive if the extra parallel resistor is in place, the extra power is always positive, so the efficiency (5 * 1)/(7*(1+Ir) + (5*Ir) + (5 * 1)) (lamp power / total power) is always going to be lower.
With a linear voltage regulator, you essentially have an electronically controlled resistance in series which is varied to maintain 5V at the output. These have a small operating current and can appear to be a very smart voltage divider with the lower resistor having a high value. Because of this, if the load is constant, a resistor is (very slightly) more efficient. The efficiency approaches Vout/Vin (5/12 = 42% -- over half the energy is wasted)
With both resistors and linear regulators, the input current is essentially equal to the output current, so losses are (at a minimum) the load current multiplied by the difference in voltage between the input and output.
But we also have switchmode regulators. These are a constant power device. The output power is very similar to the input power. Typically they might have an efficiency in excess of 90%.
If we assume an efficiency of 90%, then a switchmode regulator having a 5V 1A output will draw (5/12)/(0.9) A from a 12V supply. This works out as around 0.463 A. Yes, the input current is LESS than the output current! In this case, the switchmode regulator would need to dissipate about 1/5 of the heat that a linear regulator (or resistor) would need to. Efficiencies can easily exceed the 90% I have quoted, so in real life the actual dissipation might be a lot lower.
