TOPIC:

This subject has come up in several threads over the years so I though I'd post this for anyone interested.

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. As a result, if a battery becomes disconnected on a vehicle while the engine is running, it is possible to damage the electrical system on the vehicle.

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.

Unfortunately, automotive voltage-regulators regulate based on average-voltage. If a battery on a motor vehicle becomes disconnected, the voltage becomes more pulse-like. Since the regulator is measuring average-voltage, the pulses will become larger than when the battery was connected (similar to the pulsed-DC case). This can lead to too much heat being generated in devices like light bulbs and ignition coils.

Post edited by: loudhvx, at: 2007/12/02 04:56

Many thanks for the very excellent explanation.

So am now better understanding why battery is sometimes said to serve as a "shock absorber" in the electrical system.

You definitely explained it better than I could. Thanks!

When I opened this link, I did not expect to read about time as a variable in a DC circuit.

I enjoyed reading it very much.

Anyone who has ever had a voltage regulator fail on their Ford (almost all Ford owners then) can appreciate what your saying.

Writing articles like these is tricky business. Usually what happens is some jerk finds it necessary to tear apart everything you've said to make themselves look good.

Having said that, I'd like to add some thoughts for discussion (after all, thats why we're here, right?)

A couple of thoughts:

E is the measurement of voltage from the power source before it is connected to any circuit.

"(A better rating method for voltage would be RMS voltage, but that is another subject.)"

Why not a Watt meter? The load is constant (assume no impedance or reactance), the voltage applied is the variable. Watt meter isn't really the best choice for the same reasons as the Voltmeter; if a tree falls in the forest, does anybody hear? It would need to be monitored for a pulse.

"One way the two circuits could be easily distinguished from each other would be to measure the AC voltage on them."

I realize your example used basic numbers for clarity (everyone hates math, electricity from an engineering perspective is 80% math).

... a dmm is tuned to measure AC voltage @ 60Hz, your example (F=1/T, F=1/.5 F = 2Hz) is for 2Hz. I'm not certain any steady reading would be achieved.

An analogue meter set to VDC, you would be able to easily watch the needle bounce, most decent Dmm's set on VDC would pick up a pulse on the bar graph as well.

Now, I'd appreciate your thoughts on my thoughts.

Saki Jockey wrote:

I'd like to add some thoughts for discussion (after all, thats why we're here, right?)

Absolutely.

Saki Jockey wrote:

A couple of thoughts:

E is the measurement of voltage from the power source before it is connected to any circuit.

"(A better rating method for voltage would be RMS voltage, but that is another subject.)"

Why not a Watt meter?

That would be great, but I don't know of any cheap watt meters that measure wattage directly. (A home wattage meter is really a current meter. Since the voltage is a constant, wattage can be determined by current.) A true watt meter would have to simultaneously measure voltage and current, multiply the readings, then average the reading over time.
This paper I wrote a little while ago explains why RMS is used for AC voltage and what exactly an AC voltmeter measures (it approximates RMS). This paper contains only my findings and I have yet to confirm it with other sources. It's pretty heavy on the calculus, but it was unavoidable since I wanted to be as precise as possible.:
home.comcast.net/~loudgpz/ACvoltmeter.pdf


Saki Jockey wrote:

The load is constant (assume no impedance or reactance), the voltage applied is the variable. Watt meter isn't really the best choice for the same reasons as the Voltmeter; if a tree falls in the forest, does anybody hear? It would need to be monitored for a pulse.

If the Watt meter measured average-wattage (like the voltmeter measures average-voltage), that would suffice. That is the goal, actually, if we are concerned with blowing up light bulbs and other bike circuits. Average wattage, on a particular load like a bulb, will tell you if it's going to start melting or not.
We measure voltage for most things because the system on a bike is a voltage-regulated system. Generally speaking, for the most part, voltage readings are probably used more often than current readings because most systems are voltage-regulated (regardless if it's AC or DC).

Saki Jockey wrote:

"One way the two circuits could be easily distinguished from each other would be to measure the AC voltage on them."

I realize your example used basic numbers for clarity (everyone hates math, electricity from an engineering perspective is 80% math).

... a dmm is tuned to measure AC voltage @ 60Hz, your example (F=1/T, F=1/.5 F = 2Hz) is for 2Hz. I'm not certain any steady reading would be achieved.

I haven't found the common AC voltmeter to be particularly tuned to a specific frequency, but the frequency range does, obviously, have limits. My study of AC voltmeters is included in the above pdf link. Digital meters, however, do have a sample-hold timing involved which may affect the readings. Naturally, if the frequency of the signal is near that of the sample-hold-readout timing, there will be problems. I'm not sure where you get the 2hz number, though. I did not specify a frequency for the 50% duty cycle. (Frequency has no effect on average voltage, current, or power in a purely resistive load.)

Saki Jockey wrote:

An analogue meter set to VDC, you would be able to easily watch the needle bounce, most decent Dmm's set on VDC would pick up a pulse on the bar graph as well.

I believe the needle movement would depend on the frequency. D'Arsonval movent (anolog meter movement) is damped in order to give an average reading much like a digital meter does. In both meter types, if the frequency is sufficiently low enough, an accurate reading is difficult to obtain. Remember, in both types, the point is to give an average-reading. If the needle were not damped, you would only get an instantaneous reading which is not easy to read for pulsing voltages. (The Flukes I use actually have that bar graph in addition to the number readout. The bar simulates the needle movement seen at low frequencies. At higher frequencies, the bar is not seen moving, much the same as in the D'Arsonval movement analog meters.)

Thanks for the thoughtful post. ...very good discussion.