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  #1  
Old 11-25-2004, 04:23 PM
ledjr ledjr is offline
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Is there a formula or ratio for finding out what happens to voltage when frequency is changed? thanks

ledjr
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Old 11-26-2004, 01:12 AM
ellea ellea is offline
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Hi ledjr,
Now here is the formula for v (notice lower case which means time varying)

v=Vp Sin (2*pi*f*t)

Now as you can see the bigger the f the smaller the (sin (2*pi*f*t))
And you also know that that value does not exceed one so the smaller it is the smaller your v
so if you want your v big a high frequence your better off starting off with a high V peak

I hope that answers your question.

As for the ratio. Man that's a long euation to solve :unsure: :unsure:
Now if you got time start with v/Vp= Sin (2 pi f t) and derive the f

Good luck
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Old 11-26-2004, 11:56 AM
Perion Perion is offline
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Quote:
Originally posted by ellea@Nov 25 2004, 09:12 PM
Hi ledjr,
Now here is the formula for v (notice lower case which means time varying)

v=Vp Sin (2*pi*f*t)

Now as you can see the bigger the f the smaller the (sin (2*pi*f*t))
And you also know that that value does not exceed one so the smaller it is the smaller your v
so if you want your v big a high frequence your better off starting off with a high V peak

I hope that answers your question.

As for the ratio. Man that's a long euation to solve* :unsure:* :unsure:
Now if you got time start with v/Vp= Sin (2 pi f t) and derive the f

Good luck
First of all, by voltage, it wasn't clear whether ledjr was referring to instantaneous or RMS? I'm guessing RMS. Secondly, was he asking about ANY periodic AC voltage in general or the common sinusoidal one? I'm guessing sinusoidal only.

So, for some pure, symetric sine wave voltage variation (with no DC offset), let its amplitude (peak value above/below the zero reference) be equal to some value Vp. The voltage for some given frequency varies as a function of time described by the following equation:


V(t) = Vp*sin(w*t) = Va*sin(2*pi*f*t)

where:

V(t) = the instantaneous voltage at any time t
Vp = peak voltage value (i.e. "amplitude")
f = frequency of the AC signal (hertz or cycles/sec)
w = angular frequency (radians/sec) = 2*pi*f
t = 1/f = period of one oscillation

The form of this equation assumes that at time T0, the instantaneous voltage is 0 volts (i.e. it's not phase shifted). You can see that changing the frequency only affects when in time any specified voltage occurs. At any time t the absolute value of the voltage never exceeds Vp value since the sin(wt) is always less than or equal to 1 and greater than or equal to -1.

Now, I'm guessing that ledjr probably was curious about what he'd measure with an AC voltmeter if the frequency (only) is changed. That is, he'd like to get some average or effective measurement comparisons for different frequencies. So he needs to know how AC voltage measurements are obtained.

Therefore, before proceeding, I think ledjr really should read the article http://www.allaboutcircuits.com/vol_2/chpt_1/3.html - especially the discussion concerning "average" voltage and "the problem of trying to describe the changing quantities of AC voltage in a single, aggregate measurement". He should notice that what we normally mean by "voltage" as used to represent some meaningful AC aggregate measurement value is either a "true" or cleverly simulated RMS (Root Mean Squared) value . This "Vrms" is nothing more than the equivalent DC voltage that would produce the same quantity of heat dissipation (do the same amount of work in a given amount of time) in a purely resistive load. So, the AC voltmeter is designed to give us a "DC equivalent" or "effective" voltage reading. The RMS voltage for a sinusoidal voltage variation is given by:

Vrms = Vp/sqrt(2)

[sqrt(2) = "the square root of two" or about 1.414... ]

The main thing that ledjr should notice is that the Vrms effective voltage is NOT affected by the sine wave's frequency. This result is fairly counter-intuitive however, it should be qualified since it can give rise to some revealing hypothetical problems when pressed to extremes. As usual, things aren't as simple as they seem (as my sig implies ). For example, suppose he lowers the frequency from say 60Hz to 1 alternation per year. It's still technically an AC voltage but it would only be changing about one electrical degree per day! In that case the common AC voltmeter and normal measurement procedures would be totally inadequate for doing any meaningful comparisons. The voltage measurement apparatus and procedures would have to be radically modified in order to preserve the meaning and intention of the RMS definition. I'll leave it for the readers to ponder some of the problems that would need to be addressed under such extremes. But, this does raise the practical issue that for any particular AC RMS "voltmeter" instrument design, there exists some frequency threshold(s) for obtaining measurements within some minimal range of acceptable error. For a brief study of problems in AC voltage measurements see AC Voltage Measurement - Errors in Digital Meters.

Lastly, I should note that effective AC voltage measurements for non-sinusoidal waveforms, non-symetrical waveforms, waveforms containing a DC offset, etc, present many other difficulties which I haven't time to get into. The above linked PDF touches on some of these issues.

Yeehah!

Perion
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  #4  
Old 11-26-2004, 02:08 PM
ellea ellea is offline
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Now ledjr I can see that Perion gave you a more detailed and more efficient answer. You've got all you need there so good luck and enjoy

Thanks Perion
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