I was testing some very simple circuit's today,and found one that has my up most attention. As the frequency is around 127KHz,i don't trust my DMMs to give me an accurate amp average reading-although all 3 show the same current draw. I have drawn a basic circuit below,which we will use to answer my question.
Using the 3 ohm resistor in the circuit,can i use my scope across that 3 ohm resistor to work out the circuit's power draw,or will the rms voltage across that resistor only tell me what the resistor is dissipating?. (I tried a 1 ohm resistor in it's place,but i get a very noisy signal on my scope-the 3 ohm is much smoother on the scope.)
SG setting's.
Square wave at tuned frequency (around 127KHz)
Duty cycle-38%
VPP-8.2
Off set-4.1v-so as we have 0 volts at the bottom of the wave(62% off time)
Hi Brad,
The rms voltage across the resistor can indeed tell you the dissipation across the 3 Ohm: P=V*V/R. But you can learn about the current that flows in the circuit if you use Ohm's law for the rms voltage and the resistor values, I=V/R
Here I assume the rms voltage value across the 3 Ohm is correct i.e. the scope considers duty cycle.
Finally the circuit power consumption could be calculated by taking the rms value of the 8.2Vpp square wave, 38% duty cycle and divide it by the rms current.
I hope I am correct... :)
Gyula
tinman,
You should do what our lab instructor in school said to do: always draw the signal generator as a circle with the AC signal
for the picture *then* draw the generator's internal impedance as a resistor in series with it. So really your diagram should
have two resistors. Those coaxial cables that connect to
the circuit from the generator are supposed to match the impedance of the generator so as to not reflect any power back.
The manufacturer of the generator does quite a good job making the generator emit a near perfect signal, but he can't
create the impossible - a zero ohm impedance output. The generators output impedance is usually marked near that coaxial spigot.
Probably 50ohms. That step will help you analyzing what is happening. The reason is what you want your current shunt resistor
value to be is a trade off. The other thing is to not forget the AC power information; A two channel scope will let you determine
the phase shift between the voltage signal across the circuit and the current signal through it. You can check whether your circuit
has a PF power factor that is inductive, capacitive, or resistive at the given frequency. Any inductor will have a self capacitance and
some frequency will cause parallel resonance and will allow you to tune to parallel frequency resonance of the inductor + self capacitor.
By the way a sinewave output RF frequency generator is easiest to understand, a pulse generator is somewhat different because
digital circuits require +/- unbalanced signals of fixed voltages creating asymmetric wave signals.
Also gyulason while you power numbers will be close to accurate, you really need to integrate (summate) the powers of the individual
Fast Forrier decomposition (sinewaves) of the voltage waveform with the Fast Forrier decomposition (sinewave) of the current waveform
then use that to calculate the power of each decomposition then add them back together to get total power. One of TB's capacitor
overunity proofs did this wrong and showed overunity in a capacitor where there really was none.
This is why it is better to measure the power after rectification then to try to calculate it.
:S: MarkSCoffman
Looking at the scope shot below(across the SG),which voltage do i use to calculate my P/in-as i already have the I/in.
My DMM is showing a very accurate I/in,as i checked it against my scope using a known value resistor across my SG at the same frequency.
In regards to the scope shot,it dosnt seem right that we should be using the RMS voltage to calculate power. As the duty cycle is 38%,and the vpp is 8.2,how is it that the RMS is 4.72?-which is more than half the 8.2 vpp,and yet only a 38% duty cycle.
The mean voltage is the same as the average voltage,and my DMM also shows 2.8 volts across the SG-same as means and average voltage on the scope.
My I/in is 2.75mA-can you work out the P/in from the scope shot and current(2.75mA)
You have a situation where your input voltage is pulsed. So what you need to do to find the _average_ input power over many cycles, is to compute the VxI value during the pulse, then multiply that by the 0.38 duty cycle. Your voltage varies a little during the pulse, going from 8.2 down to perhaps 7.5 volts. So really you want an "average" here, take 7.8 volts for example. Your scope is computing the "average" and "mean" (here the same thing for this waveform) by doing something like this, taking the "average peak" and multiplying by the duty cycle. So you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power. This won't be as strictly accurate as performing the instantaneous multiplication of the V and I traces and then integrating that resulting trace over time but it will be close.
I think.
Quote from: TinselKoala on September 13, 2014, 12:40:11 AM
You have a situation where your input voltage is pulsed. So what you need to do to find the _average_ input power over many cycles, is to compute the VxI value during the pulse, then multiply that by the 0.38 duty cycle. Your voltage varies a little during the pulse, going from 8.2 down to perhaps 7.5 volts. So really you want an "average" here, take 7.8 volts for example. Your scope is computing the "average" and "mean" (here the same thing for this waveform) by doing something like this, taking the "average peak" and multiplying by the duty cycle. So you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power. This won't be as strictly accurate as performing the instantaneous multiplication of the V and I traces and then integrating that resulting trace over time but it will be close.
I think.
Thanks TK.
This is what i asumed,and lines up exactly with my DMM reading's.
Quote from: tinman on September 12, 2014, 02:26:22 PM
I was testing some very simple circuit's today,and found one that has my up most attention. As the frequency is around 127KHz,i don't trust my DMMs to give me an accurate amp average reading-although all 3 show the same current draw. I have drawn a basic circuit below,which we will use to answer my question.
Using the 3 ohm resistor in the circuit,can i use my scope across that 3 ohm resistor to work out the circuit's power draw,or will the rms voltage across that resistor only tell me what the resistor is dissipating?. (I tried a 1 ohm resistor in it's place,but i get a very noisy signal on my scope-the 3 ohm is much smoother on the scope.)
SG setting's.
Square wave at tuned frequency (around 127KHz)
Duty cycle-38%
VPP-8.2
Off set-4.1v-so as we have 0 volts at the bottom of the wave(62% off time)
At 127kHz, you will need to take a little care, not too much to avoid measuring the inductance of your wiring instead of the real current. My first suggestion is that you purchase some low inductance resistors. If you want to be really accurate, buy four terminal Kelvin resistors. If you can afford a small amount of error then Ohmite WNE series Aryton-Perry resistors cost a dollar or two. The thing to remember about current sense resistors is that the frequency at which the inductance adds 41% extra amplitude is: R/(2pi*L). So, the higher value current sense that you can tolerate the less frequency dependent gain the inductance introduces. Here is a link to Digikey's in stock parts.
http://www.digikey.com/product-search/en?pv7=2&FV=fff40001%2Cfff80482&k=wne&mnonly=0&newproducts=0&ColumnSort=0&page=1&stock=1&quantity=0&ptm=0&fid=0&pageSize=500
Quote from: tinman on September 12, 2014, 11:47:51 PM
Looking at the scope shot below(across the SG),which voltage do i use to calculate my P/in-as i already have the I/in.
My DMM is showing a very accurate I/in,as i checked it against my scope using a known value resistor across my SG at the same frequency.
In regards to the scope shot,it dosnt seem right that we should be using the RMS voltage to calculate power. As the duty cycle is 38%,and the vpp is 8.2,how is it that the RMS is 4.72?-which is more than half the 8.2 vpp,and yet only a 38% duty cycle.
The mean voltage is the same as the average voltage,and my DMM also shows 2.8 volts across the SG-same as means and average voltage on the scope.
My I/in is 2.75mA-can you work out the P/in from the scope shot and current(2.75mA)
Anyway: rms voltage or current is the equivalent DC voltage or current that would cause the same amount of heat in a pure resistor. If your circuit is not a pure resistor, then in order to obtain power, you measure both voltage and current and multiply them. Then you can process that to find peak, minimum, and/or average power over some time interval.
The measurement that often goes wrong is the current. If you get some of the resistors I recommended and you follow good hook-up practices with your scope, then you will be OK. Poynt99 published some nice YT videos on measuring current, voltage and power in pulsating circuits.
Quote from: MarkE on September 13, 2014, 04:29:12 AM
Anyway: rms voltage or current is the equivalent DC voltage or current that would cause the same amount of heat in a pure resistor. If your circuit is not a pure resistor, then in order to obtain power, you measure both voltage and current and multiply them. Then you can process that to find peak, minimum, and/or average power over some time interval.
The measurement that often goes wrong is the current. If you get some of the resistors I recommended and you follow good hook-up practices with your scope, then you will be OK. Poynt99 published some nice YT videos on measuring current, voltage and power in pulsating circuits.
Hi Mark
Below is the scope shot across the 3.2 ohm resistor,and second shot is across the SG/circuit. Are you able to have a go at calculating the P/in from these two shot's.No matter what resistor i use,there will always be the wave form you see below,due to the way the inductor and LED's are hooked up.3.2 ohms is the exact value of the 1 watt resistor im using.
@TK-please have a guess aswell,so as i can see how close you and Mark are to what my DMM amp meter read's.
Thanks
It would be a lot easier if you could post CSV files. Otherwise, I will have to extract the values which is possible, but somewhat time consuming.
Mark,
This is what I did. (http://www.overunityresearch.com/index.php?topic=2603.msg41493#msg41493)
Quote from: poynt99 on September 15, 2014, 10:22:45 PM
Mark,
This is what I did. (http://www.overunityresearch.com/index.php?topic=2603.msg41493#msg41493)
I am not a member of OUR.
I don't think OUR is blocked to read, is it? Let me know if it is.
QuoteHere's what we can do manually.
Eyeballing the current wave form, an equivalent pulse wave would be about 4mV (1.25mA) peak. Now we can simply multiply it out:
1.25mAp x 6Vp x 0.38 = 2.85mW. Add our 78uW for the 50 Ohm, and we have 2.93mW for Pin.
WARNING!
Keep in mind that scopes exhibit terrible accuracy (and resolution or precision) down in these low millivolt levels. The offsets in the inputs is often 5mV (or more), so there could be 100% or more error here.
Quote from: poynt99 on September 15, 2014, 10:39:44 PM
I don't think OUR is blocked to read, is it? Let me know if it is.
QuoteWarning!
The topic or board you are looking for appears to be either missing or off limits to you.
Please login below or register an account with OverUnity Research.
Quote from: poynt99 on September 15, 2014, 10:22:45 PM
Mark,
This is what I did. (http://www.overunityresearch.com/index.php?topic=2603.msg41493#msg41493)
Hi poynt99,
I also have problem to read your post at your above link. After login I receive this message :
" An error has occured! The topic or board you are looking for appears to be either missing or off limits to you." Why is that I wonder?
Gyula
Quote from: MarkE on September 15, 2014, 01:47:11 AM
It would be a lot easier if you could post CSV files. Otherwise, I will have to extract the values which is possible, but somewhat time consuming.
What are CSV file's Mark?
Ah, sorry guys.
This particular thread is not viewable to non-registered, non-enabled readers.
Quote from: tinman on September 16, 2014, 09:22:08 AM
What are CSV file's Mark?
Comma separated variables. It is a common format that most digital scopes can export, and any spreadsheet can import.
To keep it simple and almost precise - I would simply add a driver stage - and measure the DC power consumption of the driver-stage.
With a modern scope you can get the power directly using math function by measuring voltage on DUT and voltage on current sensing resistor.....
__BUT__ a scope is not a precise instrument - and you multiply the errors in that case - and if you derive the power from a single period - this error is multiplied by 172.000 within a second for 172kHz.
... and you can export the data using csv or whatever
- but believe me - thats lots of effort - and will never reach the precision you got with 2 cheap meters measuring DC amps and DC voltage.
rgds.
Some people are of the mind that a decoupling filter spoils whatever magic they think might be going on inside their circuits. When they see that DMMs don't show OU, and neither do their scopes on the DC side they figure the problem is the filter and not their circuit. Former prof. Steven Jones claimed 8X overunity for awhile with a Joule Thief derivative. His results were all measurement error that the kind of filter you propose would instantly expose.
Hi MarkE,
I didn´t propose nor mention a filter (because a proper driver stage must have one anyway;-)))
Cynical people connect the rise of OU with the availability of cheap DMMs.....
Anyway, if a setup meant to be used to measure efficiency of something hooked up to a generator - the efficiency of the generator should be part of the loop.
rgds.
Quote from: fritz on September 17, 2014, 04:50:03 AM
Hi MarkE,
I didn´t propose nor mention a filter (because a proper driver stage must have one anyway;-)))
Cynical people connect the rise of OU with the availability of cheap DMMs.....
Anyway, if a setup meant to be used to measure efficiency of something hooked up to a generator - the efficiency of the generator should be part of the loop.
rgds.
The dream of something for nothing is much older than DMMs. I suppose that having a numeric readout may may measurements seem more legitimate to some than dancing needles.
Quote from: MarkE on September 17, 2014, 05:15:59 AM
I suppose that having a numeric readout may may measurements seem more legitimate to some than dancing needles.
Some day (1995) I attended a conference hosted by power companies with presentations of master thesis projects of the local university.
Some guy presented a mock up to measure photovoltaic networks. Obviously his approach to measure current ended up in an almost unusable noisy signal.
2/3ds of his thesis covered the clever ways he invented to smooth and integrate this noisy signal in the digital domain.
I didn´t even dare to ask why he didn´t use low noise opamps or appropriate circuitry - because he explained to me that on measuring a signal - the last digit of the DMM jumps anyway - so there is too much noise on any signal - and thats the legitimation why his master thesis is so important. OMG.
One step at a time. The end goal is to have an onboard SG, so as to get an accurate P/in measurement-probably just a fet fed by a cap and triggered by the SG. Then we just measure the DC power flowing into the cap.