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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

Electronic Elementary Questions :: 04-12-2015 13:57 :: betwixt :: Replies: **3** :: Views: **103**

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

Power Electronics :: 03-28-2015 21:41 :: FlapJack :: Replies: **6** :: Views: **443**

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

Professional Hardware and Electronics Design :: 03-23-2015 04:38 :: sunil21 :: Replies: **2** :: Views: **227**

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 **peak**s. 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 (...)

Power Electronics :: 03-12-2015 16:52 :: betwixt :: Replies: **2** :: Views: **222**

An "Ideal" sawtooth of amplitude A **peak**-**peak** has an **rms** value
V**rms**= 2A√3
But your waveform is zero for ~10% of the time, so it will be 10% less.

Power Electronics :: 02-20-2015 16:45 :: SunnySkyguy :: Replies: **9** :: Views: **286**

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).

Analog Circuit Design :: 12-14-2014 10:11 :: BigBoss :: Replies: **3** :: Views: **230**

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

Microcontrollers :: 12-13-2014 12:08 :: SunnySkyguy :: Replies: **1** :: Views: **236**

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

RF, Microwave, Antennas and Optics :: 11-02-2014 06:52 :: chuckey :: Replies: **1** :: Views: **254**

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 (...)

Electronic Elementary Questions :: 10-14-2014 19:15 :: ParkerMike :: Replies: **9** :: Views: **589**

When a digital multimeter is used to measure the AC voltage, does it show the amplitude of voltage or **peak**-to-**peak** voltage ?

Robotics and Automation Forum :: 09-20-2014 10:48 :: pavan garate :: Replies: **3** :: Views: **862**

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 s**peak**er to the input.

Analog Circuit Design :: 09-03-2014 16:57 :: Audioguru :: Replies: **10** :: Views: **449**

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

Analog Circuit Design :: 08-05-2014 11:40 :: Audioguru :: Replies: **2** :: Views: **262**

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.

Analog Circuit Design :: 07-01-2014 17:04 :: SunnySkyguy :: Replies: **11** :: Views: **695**

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

RF, Microwave, Antennas and Optics :: 06-21-2014 10:10 :: vfone :: Replies: **9** :: Views: **774**

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, (...)

Analog Circuit Design :: 06-16-2014 06:45 :: throwaway18 :: Replies: **23** :: Views: **1330**

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 (...)

PC Programming and Interfacing :: 06-10-2014 15:28 :: barry :: Replies: **2** :: Views: **694**

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

Power Electronics :: 05-31-2014 03:06 :: memarian :: Replies: **1** :: Views: **412**

Analog Circuit Design :: 05-13-2014 20:32 :: SunnySkyguy :: Replies: **5** :: Views: **477**

. 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

Electronic Elementary Questions :: 05-11-2014 23:37 :: SunnySkyguy :: Replies: **3** :: Views: **340**

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

Electronic Elementary Questions :: 04-21-2014 04:45 :: betwixt :: Replies: **3** :: Views: **249**

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.

Power Electronics :: 04-09-2014 15:05 :: --BawA-- :: Replies: **3** :: Views: **316**

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

Analog Circuit Design :: 01-22-2014 01:53 :: vahid_ff :: Replies: **2** :: Views: **196**

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 (...)

Electronic Elementary Questions :: 12-07-2013 23:14 :: Audioguru :: Replies: **20** :: Views: **752**

-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.

Analog Circuit Design :: 11-30-2013 18:57 :: Audioguru :: Replies: **3** :: Views: **313**

A good s**peak**er has a continuous **rms** power rating and a much higher momentary **peak** power rating.
Cheap s**peak**ers have lies instead of true power ratings.

Analog Circuit Design :: 11-21-2013 12:17 :: Audioguru :: Replies: **12** :: Views: **364**

With a 15V supply and a 4 ohm s**peak**er, 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 (...)

Analog Circuit Design :: 11-05-2013 23:43 :: Audioguru :: Replies: **16** :: Views: **642**

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 (...)

RF, Microwave, Antennas and Optics :: 09-30-2013 12:14 :: biff44 :: Replies: **1** :: Views: **269**

Electronic Elementary Questions :: 08-12-2013 14:41 :: chuckey :: Replies: **5** :: Views: **446**

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

Analog Circuit Design :: 07-17-2013 04:52 :: ArNvC :: Replies: **3** :: Views: **389**

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 s**peak**ers here each have a single 10cm louds**peak**ers in them, the back of the s**peak**er is marked 8 Ohms 5W but the outside (...)

Power Electronics :: 07-11-2013 12:52 :: betwixt :: Replies: **8** :: Views: **412**

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.

Analog Circuit Design :: 05-30-2013 11:07 :: FvM :: Replies: **93** :: Views: **4108**

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 (...)

RF, Microwave, Antennas and Optics :: 05-14-2013 19:16 :: Darktrax :: Replies: **6** :: Views: **1728**

For CW-2tones, it should generally be **peak** value unless it's been mentioned/remarked.

RF, Microwave, Antennas and Optics :: 05-07-2013 13:18 :: BigBoss :: Replies: **4** :: Views: **489**

"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 (...)

Electronic Elementary Questions :: 04-10-2013 14:49 :: chuckey :: Replies: **6** :: Views: **395**

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 (...)

Electronic Elementary Questions :: 03-28-2013 02:52 :: keith1200rs :: Replies: **3** :: Views: **351**

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

RF, Microwave, Antennas and Optics :: 01-20-2013 13:03 :: PeteC :: Replies: **1** :: Views: **259**

Use an A/D converter or a **peak** detector.

Analog Circuit Design :: 12-16-2012 13:14 :: crutschow :: Replies: **3** :: Views: **472**

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

Power Electronics :: 10-10-2012 00:26 :: fadapow :: Replies: **17** :: Views: **2188**

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 36V**rms** at its maximum current. (...)

Analog Circuit Design :: 10-11-2012 06:08 :: FoxyRick :: Replies: **8** :: Views: **996**

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

Power Electronics :: 09-09-2012 09:55 :: andrew87 :: Replies: **2** :: Views: **730**

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:

Electronic Elementary Questions :: 09-04-2012 06:28 :: WimRFP :: Replies: **1** :: Views: **405**

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

Analog Circuit Design :: 08-26-2012 10:45 :: godfreyl :: Replies: **3** :: Views: **569**

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

Electronic Elementary Questions :: 08-09-2012 09:48 :: kalamkarvivek :: Replies: **0** :: Views: **374**

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

Analog Circuit Design :: 05-27-2012 15:15 :: ste2006 :: Replies: **17** :: Views: **1631**

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!

Electronic Elementary Questions :: 05-19-2012 23:34 :: dlock :: Replies: **1** :: Views: **350**

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.

Digital Signal Processing :: 05-19-2012 04:37 :: Luppy :: Replies: **4** :: Views: **678**

First convert voltage from **peak**-**peak** to **rms**, then calculate power. Impedance is usually 50 ohm for RF.
V**rms** = Vpp/(2*sqrt(2))
Power = V**rms**^2/impedance

RF, Microwave, Antennas and Optics :: 04-23-2012 10:23 :: pstuckey :: Replies: **3** :: Views: **337**

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 (...)

Power Electronics :: 04-10-2012 06:49 :: steveelliott :: Replies: **5** :: Views: **1568**

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.

Analog Circuit Design :: 03-21-2012 15:55 :: barry :: Replies: **19** :: Views: **1067**

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

Electronic Elementary Questions :: 02-01-2012 22:38 :: yellowmania :: Replies: **4** :: Views: **478**

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