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What are the peak and RMS values of the voltage of a pulse-width-modulated signal?
Wiki User
∙ 8y ago
A pulse-width modulated (PWM) signal is basically a signal that is turned "on" and "off" at a fixed rate, but is left "on" for varying amounts of time. In the scheme, a fixed maximum value for the pulse is selected, usually by circuit design. That will determine the peak value of the voltage (Vp) of the pulse. The pulse will have some kind of period associated with it. That's the pulse repetition rate (PRR), and it can tell us the frequency of the pulse. We have a time base or clock to set the PRR. The period of a wave (p) which is the time it takes for one complete cycle of the wave, is the inverse of the frequency of the wave. Recall that period equals one divided by frequency (f). We now look at the last factor: the pulse width.
In pulse width modulation, what we concentrate on is the width of the pulse we generate. (Peak pulse voltage and the PRR are almost always fixed.) All we have to do is decide how long to leave the pulse "on" before we turn it "off" at the end of one cycle. The pulse width (the amount of "on" time) could be quite narrow compared to the amount of "off" time. As we increase the amount of "on" time, the "off" time decreases. ("On" time plus "off" time equal the period of our PRR.)
RMS is root mean square, and it's a methematical way to "average" a sine wave. In a voltage sine wave, the area under the curve of a cycle (the RMS value of the voltage for a cycle) is tricky to calculate in a "regular" way. Hey, it's an irregular curve. Not so with a simple PWM signal. It's a modified square wave. Forget RMS. Forget integration. Go with simple math.
A PWM signal is modulated by changing the pulse width (naturally), so a fixed width (tw) must be chosen to find the average value of the output voltage. It's really very simple. Start with the peak voltage. It will have to be known. Then look at the amount of time the pulse is "on" compared to the period of the pulse. A short "on" time will result in a low average voltage. A little longer pulse width will result in a little higher average output voltage. If we reach a point where the pulse is "on" for half the period of the cycle, it will be "off" for the same amount of time. It is "on" for a full half a cycle and "off" for a full half a cycle. If the peak voltage is 100 volts, and the pulse is "on" for half a cycle and "off" for the other half, average voltage is 50 volts. It's a snap to pin it down. The actual average voltage can be calculated by using the following formula:
Vav = Vp x (tw / p)
Let's take a 50 volt peak, 10kHz clocked PWM device. The period of the clock is .1 milliSeconds. (p = 1/f = 1/10kHz = 0.1 mS)
If we have a pulse width modulated signal that has a pulse width (tw) of 1/50,000th of a second (0.02 milliSeconds) we have:
Vav = 50 volts x (0.02 mS / 0.1 mS) = 10 volts average
Finding the average value of a simple PWM device is easy if it's for a fixed pulse width. In a complex application of varying pulse widths, integration is needed, and the calculus is called for. But in the simple application, simple math works.
Not True!
"RMS" is not an "average" by any stretch of the imagination. If it were, a 50% square wave with peaks at 1 and -1 would be 0. (The RMS is actually 1). If it were, the AC from your house receptacles would be 0 volts! (The RMS is actually 120V).
RMS is represents a periodically varying waveform with its equivalent effective DC value. It can be a voltage (V) or a current (I).
For PWM:
VRMS = ( ( C x V12 ) + ( (1-C) x V22) ).5
Where: V1 and V2 are the two peak (top and bottom) voltages in respect to 0, C is the duty cycle in respect to V1 (ie C = .2, duty cycle is 20%). If you plug in the numbers for the 50% square wave above, the answer will agree. It's 1.
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∙ 8y ago
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What is peak detector?
a device which output voltage is almost the real peak valueof an applied signal
Where are Vpp and V rms?
Vpp is Peak-to-Peak voltage, in other words, in AC voltage, the peak-to-peak voltage is the potential difference between the lowest trough in the AC signal to the highest. Assuming the reference to the voltage is zero, Vpp would be twice the peak voltage (between zero and either the highest or lowest point in the AC waveform). Vrms is the Root Mean Square voltage, think of it as sort of an average (it's not quite that simple). For a sine wave, the RMS voltage can be calculated by y=a*sin(2ft) where f is the frequency of the signal, t is time, and a is the amplitude or peak value.
What are the limiting values of a PN junction Diode?
A: heat And PIV(Peak Inverse Voltage) or so called Peak Reverse Voltage is limit the diode characteristics.
What is peak to peak voltage?
Conversions of RMS voltage, peak voltage and peak-to-peak voltage. That are the used voltages. The expression "average" voltage is used for RMS voltage.Scroll down to related links and seach for "RMS voltage, peak voltage and peak-to-peak voltage".Answer'Average' is not the same as 'root mean square'. As the average value of a sinusoidal voltage is zero, you cannot convert it to a peak-to-peak value.
Which one of the following values exactly matches the rms value of an AC signal A.Effective B.Average C.Peak-to-Peak D. Peak?
the answer is A effective
What is peak detector?
a device which output voltage is almost the real peak valueof an applied signal
Does a standard multimeter measure ac in peak or peak to peak?
Neither. A.C. voltage, and current, are expressed as root-mean-square (rms) values, and this is what a voltmeter and an ammeter reading will indicate. To measure peak voltage, or peak-to-peak voltage, you will need to use an oscilloscope.
Where are Vpp and V rms?
Vpp is Peak-to-Peak voltage, in other words, in AC voltage, the peak-to-peak voltage is the potential difference between the lowest trough in the AC signal to the highest. Assuming the reference to the voltage is zero, Vpp would be twice the peak voltage (between zero and either the highest or lowest point in the AC waveform). Vrms is the Root Mean Square voltage, think of it as sort of an average (it's not quite that simple). For a sine wave, the RMS voltage can be calculated by y=a*sin(2ft) where f is the frequency of the signal, t is time, and a is the amplitude or peak value.
What are the limiting values of a PN junction Diode?
A: heat And PIV(Peak Inverse Voltage) or so called Peak Reverse Voltage is limit the diode characteristics.
How to calculate the peak-to-peak value of a signal?
A: AC or our line voltage is sinusoidal in nature it goes up to a positive peak returns to zero and proceed to the negative peak. 120V AC is actually swinging from peak to peak. It is 120 volts but the peak is the 120 v times 1.41 or 169.2 volts and since it also go negative then the peak to peak 120 volts times 2.82 or 338.40 volts or twice the peak voltage
What is peak to peak voltage?
Conversions of RMS voltage, peak voltage and peak-to-peak voltage. That are the used voltages. The expression "average" voltage is used for RMS voltage.Scroll down to related links and seach for "RMS voltage, peak voltage and peak-to-peak voltage".Answer'Average' is not the same as 'root mean square'. As the average value of a sinusoidal voltage is zero, you cannot convert it to a peak-to-peak value.
How rms voltage related to peak voltage?
The average voltage is the rms voltage.Volts peak = volts RMS times 1.414Volts RMS = volts peak times 0.7071Use the link below to an RMS voltage, peak voltage and peak-to-peak voltage calculator.********************************The average voltage is not the r.m.s. voltage.The average voltage of a sine wave is 0.636 x the peak value. Conversely, peak voltage is 1.57 the mean or average.
Which one of the following values exactly matches the rms value of an AC signal A.Effective B.Average C.Peak-to-Peak D. Peak?
the answer is A effective
What is the rehation ship of the input amplitude of the voltage at input and out put in the half-wave rectifire?
A 120V AC signal (such as at a power socket) is a sine-wave with a peak amplitude of about 170V and -170V or 340V peak-to-peak. A half-wave rectifier is basically a single diode which will clip off one half of the cycle leaving the other with a slight reduction in voltage. A silicon diode has a forward voltage drop of about .7 (seven tenths) of a volt, so if the input signal is 170V peak (340V Peak-to-peak), the output would be about 169.3V peak.
Use of RMS?
I am not certain what is being asked here. RMS is Root Mean Square which is basically the DC voltage which would produce the same amount of heat in a heating element as the AC voltage in question. Since AC is continuously changing in polarity and voltage, it is handy to use the RMS voltage rather than the peak (169.7V for 120V RMS) or peak-to-peak (339.4V for 120V RMS). The peak or peak-to-peak voltage is handy to know when considering the maximum values such as in rectification.
Why is the peak amplitude of voltage load lower than the amplitude of voltage transformer?
Unless otherwise stated, the value of an a.c. current or voltage is expressed in r.m.s. (root mean square) values which, for a sinusoidal waveform, is 0.707 times their peak value. The output of a voltage (or potential) transformer is no different, its measured voltage will be its r.m.s value which is lower than its peak value.
How do you measure waveforms to determine peak to peak values?
With an oscilloscope. Measure the vertical height of the wave on the screen . Multiply that by the volts per division setting. That will give you its' voltage.
