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Pulsed DC vs. Straight Line DC

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**TonyWoodside**:

Just something for you guys to think about ;)

**CaptainKirk**:

Tony,

Excellent Picture...

But the last line on the left, "Current Per Pulse" begs a question... You are comparing apples/oranges (you dropped the dimensional analysis Seconds)...

I think Joules (Watts * Seconds = V*A*S in this example) is more clear.

This way, left hand side = 12V * 1A * 0.250 Seconds = 3 Joules vs. the right side = 12V * 1A * 1S = 12 Joules

Which makes perfect sense. 1/4 of the power is used by having the power ON, 1/4 of the time. (1/2 of a pulse and 1/2 of the pulse side that is on)

Otherwise, it feels wrong, because you took 1A / 0.250 Sec and DID NOT GET: 250mA/Sec, CPP = 41.7mA/Sec (keeping the units)

Kirk Out!

PS: For the beginners, this is WHY pulsing is used in HHO production. Running pure current will heat the water. And that just wastes electricity.

**sebosfato**:

12v1a1s is not = to 12v 1a 1/4 of seconds

it would be 12v 4amps and 1/4 of second...

**TonyWoodside**:

The purpose is to illustrate the difference between pulsed DC voltage and steady DC voltage. Pulsed DC voltage can dissipate much more energy in a load than a steady DC voltage of the same average value.

For this illustration, a few equations will have to be used. First is Ohm's Law. It says the voltage (V) on a resistor is equal to the current (I) through the resistor times the resistance (R) of the resistor. That is written as V = I x R. Through algebra you also get I = V / R.

Next, the power equation will be needed. That equation says the power (P) converted to heat by a resistor is equal to the voltage on the resistor times the current through the resistor. That is written as P = I x V. (Often, E is used here to represent V, but for clarity, V will be used.)

Let's consider a very simple circuit. Let's say there is a power source connected to a 2-ohm resistor. The voltage on the power source equals the voltage on the resistor, and the current through the power source equals the current through the resistor. Let's say the power source can be steady or pulsed DC voltage. Now let's consider two cases:

Case 1

The power source is a steady 12v DC. A voltmeter on the power source would show 12v.

Using I = V / R, the current through the resistor is 12volts / 2ohms = 6amps.

Using P = I x V, the power converted to heat by the resistor is 6amps x 12volts = 72watts. The resistor must dissipate 72 watts.

Let's say the resistor is a 100-watt resistor. It will get warm, but won't get damaged.

Case 2

Now the power source is a pulsed DC voltage with a 50% duty cycle. That means it's on half the time and off half the time. In order to maintain the same average voltage, it will have to be at 24v when it's on, and 0v when it's off. A voltmeter would register 12v, the same as in the steady DC case.

Using I = V / R, when the source is on, the current is 24volts / 2ohms = 12amps.

When the source is off, there are 0amps. The average current is therefore 6amps, same as in the steady DC case.

Now let's calculate the power.

Using P = I x V, when the source is on, the power is 12amps x 24volts = 288watts.

When the source is off, there is 0watts converted to heat. Since it's a 50% duty cycle, the average power is half of 288watts. That means the average wattage the resistor must dissipate is now 144watts.

This is double that of the first case. Therefore the 100-watt resistor will go up in smoke!

These are two cases where, if the circuit were analyzed by a DC voltmeter/ammeter, they would look identical. The meter would show the same voltage and the same current for both circuits. However, in one circuit the resistor would be warm, and in the other, the resistor would be on fire.

The problem, here, is the way the voltmeter measures voltage. It is measuring average voltage, but average voltage is not necessarily a good predictor of power-transfer to a resistor. (A better rating method for voltage would be RMS voltage, but that is another subject.) One way the two circuits could be easily distinguished from each other would be to measure the AC voltage on them. The steady DC circuit will show no AC voltage, while the pulsed DC circuit will show very high AC voltage on it (even though the voltage is actually DC).

It should be noted, if it's not already apparent, the voltage measured by a DC voltmeter cannot automatically be used as the voltage in the equations given above. The voltmeter measures average voltage, but the equations use instantaneous voltage. (The same goes for current.) If the voltage is steady DC (for the most part), then the average reading is the same as the instantaneous reading, and then the meter reading can be used in the above equations.

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