Accurate Measurements on pulsed system's harder than you think.

[TinselKoala]

So just for grins I went ahead and set up Woopy's circuit and measured it with my scope's Math.

The CVR I used is the same 1 Ohm 1% tolerance Ohmite non-inductive one I used for the previous testing. The 100 Ohm load I used is a 1 Watt metal-film resistor that measured 100.1 ohms on my meter. I used my regulated bench power supply for the power source, and my F43 FG to pulse the mosfet. I set the power supply to give an amplitude of 12.6 V on the V_drop across the Load resistor, and the FG to give as close to 20 percent ON duty cycle as I could get it. "Ivcr" is the current in the one-ohm CVR, and "LOAD" is the Vdrop across the load resistor. The rest of the readings can be seen on the scopeshot below.

The scope computes the instantaneous power (IP) curve by multiplying the instantaneous values of Icvr and LOAD. It's taking 50 million samples per second, or 500,000 samples per horizontal division at 10ms/div, or 250,000 samples per division per channel, as the "instants" for this operation. _THEN_ the scope is averaging that IP curve, to account for the 20 percent ON duty cycle, and is coming up with the Average Power value displayed in the Measurements section below the traces. Note that the figures are in rough agreement with my calculations using Woopy's numbers done correctly up above. The differences are due to the resistor values, the duty cycle and the mosfet's internal resistance.


@ Woopy, TinMan... don't your scopes have Math capability, so you can do this Ch1 x Ch2 multiplication also?

[TinselKoala]

Thank you TK, two very detailed Posts there! This is the stuff that needs to be saved and referenced at times of taking measurements.



A 0.1 Ohm Current Sense Resistor would have a much lower Voltage drop Across the resistor and have a much lower over all affect on the circuits, if this helps? Also, why not use the CVR or Current Sense Resistor (CSR) on the Negative Rail? It might be easier to have the Scope Grounds closer to the Negative Terminal of the Power Source. Not saying its wrong, but I always think it is a best practice to minimize any possible Ground Loops.

There is always going to be some Inductance in the Current Sensing Circuit, but trying to minimize it as much as possible is also a best practice: 0.1 ohm, 5 W, ± 1%, Open Element, Through Hole Current Sense Resistor - Not a bad investment for $3

Again, you can use ohms law to calculate the Ampere Value flowing through the resistor: I = V/R - 0.05 / 0.1 = 0.5 Ampere's as an example.

TK Thanks again for such a detailed post pointing out the traps. Very nice to see such good info here for all!

   Chris Sykes
       hyiq.org

I have some 0.1 Ohm 1 percent CVRs of the non-inductive type, Ohmite WNBR10FET that I have been using for another project where I can't load the circuit under test very much. They cost a bit less than 3 dollars each (DigiKey lists them at $1.56 each, for example.) These will probably have even less inductance than the straight-wire shunt you've shown, as they are wound by the Aryton-Perry method and come in at under 1 nH at 1MHz.  I even have one set up as a "Kelvin" type probe for insertion into a circuit with greatest accuracy. And for measurement of the total power in a circuit like this, one would normally insert it in the negative rail as you said. However that is not how TinMan first used his resistor, nor is Woopy using it that way. For the measurement of an in-circuit load element as we are doing here, there is nothing wrong with the setup we are using.
 
The 0.1 ohm resistors will of course produce 1/10 the voltage drop for a given current than the 1 ohm units, and for low currents you may be putting yourself in the position of having accurately to measure microvolts of drop. The 0.1 ohm resistor will not disturb the circuit under test as much as the 1.0 ohm resistor will. All of these considerations should be "considered" when choosing a CVR for any particular purpose. It definitely is easiest in most cases to use a simple 1.0 ohm resistor.

You are right to be concerned about groundloops. Most good quality Bench power supplies will not have either of their outputs connected to chassis or Mains ground unless deliberately "strapped" to do so. And many good quality Function Generators, like my F43, can be operated in either mode, fully isolated or with the "Black" BNC shield output conductor connected to chassis (and therefore Mains) ground. Of course I'm using the F43 in isolated mode. So the only other thing to worry about is to insure that both scope Probe references are connected to the same point in the circuit, as mine are.

[EMJunkie]

Another, all be it very small Loss will be Across the Mosfet itself. If the Mosfet has an R_DS(ON) Resistance of 0.550 Ohms (IRF740 for example) the loss in this component is also something to be considered.

   Chris Sykes
       hyiq.org

[tinman]

Let's use the correct terms and values and see what is actually happening here.

In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the V_drop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the V_drop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.

Let us "assume" that the 1 Ohm resistor is accurate, though.

So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the V_drop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?

R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.

The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I²R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.

The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I²R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.

A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)


The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.

as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)

Why are you multiplying the V_drop across the 100R by the "average current"?  What does this value mean? The power dissipated in the 100 R resistor is calculated by I²R, or equivalently V²/R.

This calculation you've made here is totally invalid.

Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.

Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V²/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.

Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)²/94.3 =  1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.

You aren't doing it right !!!!!!!!



The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):

The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.

Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.

TK
I was only posting results shown from Woopy's video the way he was showing them.

In the third place you are going badly wrong in some of your calculations.

Im not going wrong in any of !my! calculation's,as these are woopy's calculations-->not mine.

In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!

This i know. As i said,these were Woopy's calculations from the video--not mine.

(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA"

Aint that a hoot.
So much for using a scope to take accurate measurements.
We'll just adjust everything,so as it all adds up.

In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope.  This is a significant error you have introduced by using the wrong value for the duty cycle.

Once again,i was going on the duty cycle value Woopy gave us.

Please stop saying that these are my error's-->as they are not.


Brad

[tinman]

Let's use the correct terms and values and see what is actually happening here.

In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the V_drop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the V_drop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.

Let us "assume" that the 1 Ohm resistor is accurate, though.

So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the V_drop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?

R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.

The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I²R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.

The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I²R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.

A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)


The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.

(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)

Why are you multiplying the V_drop across the 100R by the "average current"?  What does this value mean? The power dissipated in the 100 R resistor is calculated by I²R, or equivalently V²/R.

This calculation you've made here is totally invalid.

Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.

Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V²/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.

Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)²/94.3 =  1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.

You aren't doing it right !!!!!!!!

In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!

In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope.  This is a significant error you have introduced by using the wrong value for the duty cycle.

In the third place you are going badly wrong in some of your calculations.

The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):

The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.

Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.

And yet if we use your calculated average current of 28mA,then 28mA through a 1 ohm resistor would mean that the 1 ohm resistor is dissipating an average of .784mW,and the 94.3 ohm resistor would be dissipating an average 73.93mW,as the average current flowing through them !is! 28mA

So if we use the peak current value through the 94.3 ohm resistor X 20.9%,we get 134mA through 94.3 ohm's = 1.693 watts X 20.9% = 353mW. But if we use the average current through the 94.3 ohm resistor,we get 73.93mW.

If we multiply the peak voltage 12.6 by the average current of 28mA,we get very close to the 351mW you calculated. But now the resistance has to be 450 ohm's across that circuit.



Brad.