Why even care about average voltage?

Thread Starter

winmx

Joined May 24, 2007
14
I hope I can get my point across. I know the maths and integrals of average peak and rms voltages and currents. But why for example should one even bother with the average voltage value at the output of a AC/DC rectifier bridge (controlled by scr's or just a plain diode bridge) since the rms voltage is the "real" voltage in the sense that the rms voltage will be the one used to calculate true thermic power on a resistor. (I am aware of the fact that most voltmeters will register the average output voltage of a rectified sinusoidal). Could the answer be that the average power is used to calculate ripple etc? Thank you in advance for any help. I am really stuck with that question... To put it another way - what do i gain by knowing the average voltage or current and what can i use it for?
 

beenthere

Joined Apr 20, 2004
15,819
Knowing average current and voltage can let you figure power over time. That helps in figuring the size of heat sinks.

You might get a better meter. My 20 year old Fluke 23 does a nice job of figuring RMS. My $9.99 Rad Shack doesn't.
 

Ron H

Joined Apr 14, 2005
7,014
Knowing average current and voltage can let you figure power over time. That helps in figuring the size of heat sinks.

You might get a better meter. My 20 year old Fluke 23 does a nice job of figuring RMS. My $9.99 Rad Shack doesn't.
Almost all multimeters (and I'm guessing many bench and rack-mount meters) use a full wave rectifier followed by a lowpass filter for AC measurements, with the scale calibrated for RMS under the assumption that the waveform is a sine wave. The RMS values for other waveforms will be incorrect. Based on the datasheet, I believe this to be true of the Fluke 23 as well:
AC conversions are ac-coupled, average responding, and calibrated
to the RMS value of a sine wave input.
Obviously(?), this won't work on DC with AC riding on it.
True RMS conversions require nonlinear circuitry (analog multipliers) or, as in the HP3400 wideband RMS voltmeter, in which the AC signal heated one of a matched pair of heaters. A DC voltage was servo'ed to drive the other half of the matched pair. These two heaters were thermally coupled to a matched pair of thermocouples. The DC voltage nulled the differential signal from the thermocouples, creating a DC voltage which had the same heating effect is the AC signal, and thus represented the RMS value of the input.
Contemporary RMS voltmeters can use digital signal processing techniques. I'm not sure what the current state of the art is regarding bandwidth.
 

bloguetronica

Joined Apr 27, 2007
1,372
I hope I can get my point across. I know the maths and integrals of average peak and rms voltages and currents. But why for example should one even bother with the average voltage value at the output of a AC/DC rectifier bridge (controlled by scr's or just a plain diode bridge) since the rms voltage is the "real" voltage in the sense that the rms voltage will be the one used to calculate true thermic power on a resistor. (I am aware of the fact that most voltmeters will register the average output voltage of a rectified sinusoidal). Could the answer be that the average power is used to calculate ripple etc? Thank you in advance for any help. I am really stuck with that question... To put it another way - what do i gain by knowing the average voltage or current and what can i use it for?
I personally care with the average voltage and the peak voltage. They are both useful. You should consider peak voltages when determining the diodes reverse voltage or the filter capacitor working voltage. I use for those a safety factor of 4/3. Average voltages are useful when calculating, for example, the nominal dissipation of a voltage regulator (heatsink) (of course you should consider average_input_voltage = peak_voltage - 0.5 x ripple_voltage, since you have a filter capacitor).
 

Thread Starter

winmx

Joined May 24, 2007
14
As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
 

bloguetronica

Joined Apr 27, 2007
1,372
As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
Using RMS values would wield to a wrong result, since RMS (root mean square) applies to sinusoidal waves or rectified waves. Thus, applying RMS: average = sqrt (0.5) x peak_voltage. In a real situation, where a wave is being filtered, you would consider average = peak - 0.5 x ripple, which give an aproximation.
 

Ron H

Joined Apr 14, 2005
7,014
Using RMS values would wield to a wrong result, since RMS (root mean square) applies to sinusoidal waves or rectified waves. Thus, applying RMS: average = sqrt (0.5) x peak_voltage. In a real situation, where a wave is being filtered, you would consider average = peak - 0.5 x ripple, which give an aproximation.
This is wrong. RMS is for any waveform, sine, square, pulse, triangle, noise, ..., when you want to know the heat-generating value of it. It works for AC, DC, DC with ripple, whatever. Average voltage will not give you average power. As an extreme example, consider a pure AC sine wave. The average value is zero, but the RMS value is non-zero.
In another thread, you said
My only concern is to give correct answers if I can, or not answer at all if I am uncertain or don't know.
:rolleyes::rolleyes::rolleyes::rolleyes:
 

Thread Starter

winmx

Joined May 24, 2007
14
This is wrong. RMS is for any waveform, sine, square, pulse, triangle, noise, ..., when you want to know the heat-generating value of it. It works for AC, DC, DC with ripple, whatever. Average voltage will not give you average power.
This is what I have understood after having read a lot on the subject and I understand the importance of knowing the rms and peak values. But in the real world what use is knowing the average value? What can I use it for? :confused:

@cumesoftware Do you mean that once the harmonics have been filtered out the only thing remaining is the average value? (that is the first bit of a fourier series is the average voltage?)
 

beenthere

Joined Apr 20, 2004
15,819
My point about power over time was a bit unclear. I was refering to situations where the duty cycle of a device is less than 100%. In a situation where the duty cycle is under 100% the power dissipated will be reduced, so the size of the heat sink can be made smaller.

I've run strain gauge bridges with 30 volts excitation. that is 3 times the rating. I could do it because the duty cycle was only 1%, so no significant heating took place in the foils.
 

Thread Starter

winmx

Joined May 24, 2007
14
So if I understand correctly you used the average voltage as a "tool" to roughly calculate the power.

PS I would like to thank everybody for their responses so far :)
 

bloguetronica

Joined Apr 27, 2007
1,372
This is wrong. RMS is for any waveform, sine, square, pulse, triangle, noise, ..., when you want to know the heat-generating value of it. It works for AC, DC, DC with ripple, whatever. Average voltage will not give you average power. As an extreme example, consider a pure AC sine wave. The average value is zero, but the RMS value is non-zero.
In another thread, you said :rolleyes::rolleyes::rolleyes::rolleyes:
Ok, the average voltage of a digital square wave, varying form 0 to 5V (peak) is:
average = 0.5 x peak = 0.5 x 5V = 2.5V
Where did I applied the RMS factor, that is, sqrt (0.5)?

Of course, we have sawtooth waves, from -Vpeak to Vpeak, where the factor is 0.5, square waves from -Vpeak to Vpeak where the factor is 1. If you have doubts, just calculate the integers of these waveforms considering an interval of half cycle (or considering the integer of abs(f(x)) for the entire cycle, being f(x) the waveform). I personally calculated the average voltage (in module) of a sinusoidal and came to the value of sqrt (0.5) .
 

Ron H

Joined Apr 14, 2005
7,014
Ok, the average voltage of a digital square wave, varying form 0 to 5V (peak) is:
average = 0.5 x peak = 0.5 x 5V = 2.5V
Where did I applied the RMS factor, that is, sqrt (0.5)?
To calculate the RMS value,
1. Square the peak value: 5^2=25
2. Take the long-term average: 25/2=12.5
3. Take the square root: Sqrt(12.5)=3.53V. This the RMS vaue of a 0 to 5 volt square wave.

Of course, we have sawtooth waves, from -Vpeak to Vpeak, where the factor is 0.5, square waves from -Vpeak to Vpeak where the factor is 1. If you have doubts, just calculate the integers of these waveforms considering an interval of half cycle (or considering the integer of abs(f(x)) for the entire cycle, being f(x) the waveform). I personally calculated the average voltage (in module) of a sinusoidal and came to the value of sqrt (0.5) .
I didn't understand this part.
 

Thread Starter

winmx

Joined May 24, 2007
14
Good! It seems we have sorted out the peak voltage and the rms voltage! :)
@Ron H What do you have to say on my original question on average voltage?
 

Ron H

Joined Apr 14, 2005
7,014
Good! It seems we have sorted out the peak voltage and the rms voltage! :)
@Ron H What do you have to say on my original question on average voltage?
I believe that the average dissipation of constant current sources (and sinks) is proportional to the average voltage across them. Linear voltage regulators with non-varying loads look like current sources or sinks. Accordingly, the power dissipation will be the product of the current through the regulator multiplied by the average (not RMS) voltage across it. RMS voltage is useful in calculating the dissipation of linear resistive loads.
There may be other applications for average voltage, but I can't think of any.
 

bloguetronica

Joined Apr 27, 2007
1,372
To calculate the RMS value,
1. Square the peak value: 5^2=25
2. Take the long-term average: 25/2=12.5
3. Take the square root: Sqrt(12.5)=3.53V. This the RMS vaue of a 0 to 5 volt square wave.
So, if you have a current with that waveform going across through a resistor, the power dissipated in the resistor is the same when it is crossed by a current with a constant potential of 3.53V? Sounds a little strange, so with a 5Ohm resistor:
Power when crossed by 5V: 5W
Average power (square wave regime): 2.5W
Power when crossed by 3.53V: 3.53^2 / 5 = 2.492W

Makes sense, but that is not the average voltage. So RMS is related to power, that is, RMS voltage is the constant voltage that causes the same amount of dissipation as the waveform. Right?

This confuses me, since some transformers say 230V RMS instead of 230V AC, and they are refering to the average voltage, because I know the peaks are about 325.22 (230 x sqrt (2)). Unless RMS and average voltage are the same for a 230V AC waveform. Also, this definition of average voltage for AC current exists on Malvino (Peak = Average x sqrt(0.5)). I calculated myself the average voltage of the module (I'm refering here to abs, because in Portugal we call it módulo, and I´m not sure about around here) of a sin function. For example the integer from x=0 to x=Pi of sin (x) is 0.707/2. So the integer from x=0 to x=2Pi of abs(sin(x)) would be 0.707, the value of sqrt(0.5).
 

recca02

Joined Apr 2, 2007
1,214
average emf is used for flux calculations (remeber using them for derivation in transformers)
but then again u can always use the form factor cant u?
 
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