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137 Threads found on edaboard.com: Peak Rms
If that's 440V down to 5V rms the peak will be (5 * 1.414) - (2 * Vf) or about 5.8V DC assuming you have added a reservoir capacitor. So to drop it to 2.56 you need a resistive divider with a series to parallel ratio of 1:1.264 You could use a series resistor of 13K and a parallel resistor across the ADC input of 10K. It would be wise to add prote
Can someone explain these specs to me, it looks like the latest and greatest oil furnace ignition transformer put's out more power than it uses. Background I have been looking at ways to make my oil fired furnace more efficient. One of the things that caught my eye was the new oil ignition transformer that has a 35 VA input compared to my
I wish to know how to setup 230v source from proteus "Generators " - SINE. I worked with dc well. now when I click SINE there are three fields to fill. (I need 230 v ac, 50 hz and also to work with PIC I need 2.30v , 50hz source) 1.offset ? 2. amplitude/peak/rms ? 3. Frequency (I think this is 50hz); VSINE also works the same way w
No, the diode will fail. The parameter you need to observe is 'PIV' or peak Inverse Voltage. The 220V rms will be swinging between about +250V and -250V at it's peaks. If the rectifier is driving a load that does not store any voltage you need diodes rated at a minimum of 250V. If the rectifier has storage after it, for example a (...)
An "Ideal" sawtooth of amplitude A peak-peak has an rms value Vrms= 2A√3 But your waveform is zero for ~10% of the time, so it will be 10% less.
If this ripple is not periodic, you can use a modern oscilloscope that has a recording capability.The oscilloscope will record this non-periodic waveform in a long term time slot and you will able to measure peak-to-peak ( or avarage,rms etc).
My motor rated current is 3a found in name plate it is rms current are peak current regards kalyan This is thermally rated for rms, whereas peak current will depend on load, acceleration and Coil resistance, Rs. *Added 5x .. Depends on winding resistance Often this is 5x-8x steady rated current
Its a bandwidth thing, with CW you have a DC level, with a pulse, you have not, so if its a proper rms detector then the output is down by the mark/space ratio. If its peak measuring (rms calibrated) then it should be OK, providing the circuits have a wide enough bandwidth to handle the narrow pulse width. Frank
A phase angle meter has a Total button and a In-Phase Button The Total Button measures in rms and is the same AC measurement as a DVM meter in AC mode? The In-Phase measures in rms or peak? The In-Phase measures the input voltage in phase with a reference input voltage of the same frequency The Polarity and Magnitude is (...)
When a digital multimeter is used to measure the AC voltage, does it show the amplitude of voltage or peak-to-peak voltage ?
2A peak into 8 ohms is 1.414A rms which produces an output of 16W into 8 ohms. The voltage swing is 11.3V rms or 16V peak. Phase delay might be a problem at 20kHz if the frequency of the LC lowpass filter at the output has a cutoff close to 20kHz and you use negative feedback from the speaker to the input.
The voltage rating for a capacitor is its maximum allowed peak DC, which is 1.414 times higher than its sinewave rms maximum voltage rating. So if the voltage to a capacitor is 25VAC then the capacitor must have a voltage rating of at least 35V and a 100V capacitor can be used for good reliability. Note that some types of capacitors are polarized a
Unless there you don't mind measuring Signal + Noise, the appropriate method is to convert peak to rms assuming Sine wave and rms=peak/Root(2) or 0.707 x Vp as Audioguru stated. If you want to eliminate noise, then a tracking filter is needed.
If crest factor depends by modulation (peak and removed modulation) how can people spent time defining crest factor for un-modulated signals as: sine, square, triangle? Above is mentioned that: Crest factor is the peak amplitude of the waveform divided by the rms value of the waveform. You mentioned 6d
To deliver 100mA into 100Kohm you need 10000volts and an output power of 1000 watts. That is rms so for a sinusoidal signal you need 14000volts peak, positive and negative. Even 100mA into 1Kohm needs 100volts rms, 140v peak Give a realistic specification for output current and load impedance. Lower maximum current, (...)
If you connect one leg of the transformer secondary to ground, then you can apply the other leg to the ADC input; BUT you will need to drop the voltage, since the peak voltage will be over 14 volts (5 volts rms is about 14 volts peak-to-peak). You can determine what the peak value of the signal is, and (...)
Hello, I'm building a Fan Control Circuit for controlling the speed of a 380V Fan.(380V rms, ~540V peak-max 150W) using a microcontroller. My main circuit is something similar to this FreeScale Application note : In my circuit, I have a different microcontroller, Op
peak ripple would of course be (Max-min)/2 which is closer to rms noise, but depends on if your interest is thermal noise, clock feedthru or some other noise in the signal. Normally if it is fundamental f or 2f , we call it ripple, otherwise just "noise p-p" covers all types including transient noise.
. It depends on your sampling rate. ADC is linear but your detector is not. Do you want to only track changes in Vac over long term or short term interruptions or do statistics Avearage & deviations ? Choices: quasi rms with peak, P-p or rectified average I would use LPF Noise filter and rectified average with another LPF. 200Hz LPF before br
It doesn't increase - it decreases by the voltage dropped across two of the bridge diodes. The reason you get the apparent increase is because if you fit a reservoir capacitor after the bridge it charges to the peak of the rectified half cycles. If you are using a capacitor, the voltage across it will rise to the rms voltage multiplied by sqrt(2) t
As you know that 220V is basically rms voltage , so in order to get the peak voltage , you have to rectify it (bridge rectifier) followed by a filter stage , which incorporates a inductor and capacitor , or simply a capacitor the output voltage would be 220*root2 = 311v, if you use a large capacitor , the DC waveform would be more smoothen.
Hi folks, I designed a signal amplifier and need to measure the amplitude of the output accurately (less that %5 error) to be independent of oscilloscopes. Since output signal has some sort of distortion, especially at high frequencies, the idea of using rms converters is useless. Furthermore, I tried to simulate other ideas based on fast ECL co
The 200W total is probably the power supply power to the amplifier that causes the output plus heat. 44V peak-to-peak is 15.6V rms. Then the output power is 15.6V squared/8= 30.4 Watts. Amplifier power is measured with a sinewave. If a square-wave is used then the output power is doubled (60.8W for this amplifier) because an additional (...)
-105dB is almost 0.0005% and that is just the peak. The rms value of the distortion will be less. Some very good audio opamps have harmonic distortion at 1kHz of 0.00008% but with a 2k ohm load, a gain of 1 and an output level of 1vp-p. You might be able to hear 0.1% distortion at 1kHz.
A good speaker has a continuous rms power rating and a much higher momentary peak power rating. Cheap speakers have lies instead of true power ratings.
With a 15V supply and a 4 ohm speaker, the original circuit would have an output of about 11V p-p which is 3.9V rms which is only 3.8W. The peak current in the output transistors is 1.38A. The darlington output transistors have a current gain of typically 3000 so the drive current into them is fine. Since there is only two diodes to bias the (...)
I see what u mean, the data sheet is sparse on output power info. it does say that full scale current output is 20.48 ma, and the output impedance is 50 ohms. That leads me to think that the peak output power would be at +/- 10.24 mA into 50 ohms. Poutpk=I^2*R=7.688 milliwatts peak, or 5.4 milliwatts rms. So practically, I would expect (...)
rms is the voltage that has power. A peak may be of a high voltage with a high instantaneous power but a low power in the real sense of the word. Any meter that claims to read power must detect the rms voltage and current. otherwise the meter should be a peak power meter. Frank
Hi, Crest Factor = peak Current / rms Current. In a linear load, the Crest Factor is 1.414. However, in a general purpose SMPS, due to its non linear nature, this factor will be much higher - in the region of up to 4. See if you can measure the current using a CRO/DSO by putting a small resistor but having higher wattage in line. Then measure vol
There is real power and 'market hype' power. What you are trying to measure is continuous rms Watts but the specification may be for "peak music power" or similar. For example, my desktop speakers here each have a single 10cm loudspeakers in them, the back of the speaker is marked 8 Ohms 5W but the outside (...)
Obviously, an active "precision" rectifier will be required. Different AC quantities are available, averaged rectified value, peak value, rms value. Averaged rectified value is giving best results if no true rms circuit is required.
Just some basic estimating.. 50v peak-to peak is 25V base to peak, and staying with 50 Ohms for now.. Then, for sine waves, the rms voltage is 25/SQRT(2) = 17.67 Volts, and Power = V^2/50 = 6.25 Watts For the other extreme, a square-wave of (say) 90% on, 10% off comes to 11.25 Watts. peak power is (...)
For CW-2tones, it should generally be peak value unless it's been mentioned/remarked.
"I want to stepdown 300Vac 50hz to 5v(peak)", 5V peak is 3.5V rms. 220 to 12 V is not a good place to start from. Ignoring any transformer problems, you would get 300/250 X 12 = 14.4 V when you need 3.5, so you will have 4 times as many volts as required!!! If the 3.5V is going to feed a FIXED load (er, LED perhaps?) then a simple series (...)
I think the equation you are trying to remember is VAC * 1.414 = VDC. The peak value of an AC waveform is the square root of 2 higher than the rms value. An AC voltage is normally given as rms for power distribution. Root 2 is 1.414 not 1.141. You multiply, not subtract. If you want to take account of the diode drop it is typically 0.7V (...)
I used TINA-TI to model a circuit and it predicted a Total Noise that is negligible below 300 KHz and flattens out at 300 mV above 60 MHz. I built the circuit and want to correlate what I see on my scope with what the model predicted. My analog scope has a 250 MHz bandwidth so I believe it should have negligible effect on the readings. Wit
Use an A/D converter or a peak detector.
I am designing an inverter for a diy project. This inverter should be transformer-less and microprocessor based. I am planning to use a DC-DC converter, but should i used a the peak voltage or the rms for the DC-DC output, and how do i configure the circuit to handle the power i have in mind? How could I limit the device to....say 250W , with a 24v
36V is the rms voltage of the transformer. Not the DC voltage you get after rectification. For a sinusoidal waveform, rms can be converted to its peak value by multiplying by 1.41, so 50V is about right. It will likely be higher even than 50V because the transformer will be rated to give 36Vrms at its maximum current. (...)
Hi to everyone, I'm working on a three phase sinewave converter project for a 3 kW squirrel cage motor (230/400 V AC). The converter's topology I have to realize is shown in the simplified scheme in the attachment. I've some doubts about the right selection of IGBT's current ratings. Assuming for the motor a power factor cosφ = 0.8, I determ
If it is a peak value, P = V^2/2R = 1^2/100 = 0.01W = 10mW 10mW = 10*mW As we have powers, convert "10" to dB via 10*log10 = 10*1 = "10dB" 10 mW = 10dB*mW. As we are lazy sometimes, we write this mostly as 10 dBm. When it is an rms value, the power V^2/R = 0.02W = 20mW, converting "20" to dBs gives "13dB", so together with the mW:
But what's with the 36V? The wall plug gives 24V rms, so the peak voltage should be 24 * sqrt(2) = 34V. There's a bit of voltage lost across the diodes, so you should get about +-33.4V DC unregulated. So it's only 6% higher than expected. Two possible reasons: With light loads, a transformer's output
Hi Sir, Can u please give me the circuit diagram of SIDAC testing..... I am giving the following details of SIDAC Its name is SIDAC K-300 Silicon Bilateral Voltage Triggered Switch Breakover Voltage- minimum 270V and maximum 330V Blocking Voltage- 215V peak off-state current at VDRM- 10uA Continuous On-state DC or rms Current- 1A Holding cu
I know Current Transformers have been discussed here before but i have a slightly different angle that i cant find the answer too, I am currently desiging a circuit to measure current from say 0-30A, I have the current transformer, load resistor etc and that is all ok, I intend to feed the variable DC output into a Micro and use AD converters in
Hi all! I know voltage gain is given as Vout/Vin...I want to know that whether the values of voltages used for this calculation are rms or peak to peak? Thanks in advance!
Hello guys I'm doing some signal processing/data analysis. Hope to get some advice on something. I collected some vibration data just the other day while metal cutting and I'm wondering what is the correct (or better) way to do this. These are what the signals look like. Sampling rate of the data is 10000 hz.
First convert voltage from peak-peak to rms, then calculate power. Impedance is usually 50 ohm for RF. Vrms = Vpp/(2*sqrt(2)) Power = Vrms^2/impedance
Measure the voltage with a true-rms meter and a non-true-rms meter. If they are the same then the voltage is sinusoidal. If you don't have a true rms meter, but the voltage reads right (what it is supposed to be) then you have a sinusoidal output. Non true-rms meters read the average voltage which is (...)
Welll, first of all, you are going to need a 53 Watt resistor there; that'll get warm!. Your output voltage is going to be 0.75 V (not sure how you got 660mV). And that .75V is rms, which means you've got a little more than 1 volt peak, which exceeds the range of the MCP3909(+/- 1) by a little bit ; you probably want to have a little headroom.
have used a development board to create a small circuit containing a 12V supply plugged into an analogue development board which contains a 100mA fuse. After this i have attached an oscilloscope across the two inputs to complete the circuit and this created a sine wave image on the scope. When looking at the wave it has a peak voltage of 20V when