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

[TinselKoala]

OK, are we all done chasing the Red Herrings now?

Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)

With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms.  And the power being dissipated by the bulb is I²R = (0.082)² x 34.1 = 0.224 Watt.

With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V.  So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I²R = (0.163)² x 38.8 =  1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.

Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.

Please check my math and assumptions, and insert your own values for input current, V_drop across the bulb, and duty cycle of the pulse train.

[tinman]

TK, Ok I will take your word for it but certainly looks deceiving. Thinking about this, it is correct and Brads Mean Value is right on Ch1.

Still, this still means we only have a Potential Voltage Difference of 3.374 Volts as opposed to the 8 Volts with the Cap disconnected. Being that Power is V x I when in phase, we can see a clear difference in total power through the Components in question.

8 x 0.096 = 0.768
3.374 x 0.144 = 0.485856

   Chris Sykes
       hyiq.org

Why do you not see the problem in your own calculations Chris?

[TinselKoala]

@TinMan:
Putting a DMM as ammeter inline with the CVR input is problematic as the burden of inserting the meter in series can vary with range, meter brand, etc. A better way is to use the DMM as a voltmeter and measure across the CVR just as one does with the scope and convert to current by Ohm's Law. And of course in the case of the pulse train, it will give the same "average" value as the scope indicates, pretty much. But as I've shown, the "average" value of the current and the "average" Vdrop across the bulb are not the correct values to use in the computation of power dissipated by the bulb. They are Red Herrings of the reddest, fishiest kind.

[tinman]

OK, are we all done chasing the Red Herrings now?

Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)

With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms.  And the power being dissipated by the bulb is I²R = (0.082)² x 34.1 = 0.224 Watt.

With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V.  So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I²R = (0.163)² x 38.8 =  1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.

Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.

Please check my math and assumptions, and insert your own values for input current, V_drop across the bulb, and duty cycle of the pulse train.

TK

Dose your scope have math function?,if so,what value dose that give you as in power value in each scenario ?.

[tinman]

@TinMan:
Putting a DMM as ammeter inline with the CVR input is problematic as the burden of inserting the meter in series can vary with range, meter brand, etc. A better way is to use the DMM as a voltmeter and measure across the CVR just as one does with the scope and convert to current by Ohm's Law. And of course in the case of the pulse train, it will give the same "average" value as the scope indicates, pretty much. But as I've shown, the "average" value of the current and the "average" Vdrop across the bulb are not the correct values to use in the computation of power dissipated by the bulb. They are Red Herrings of the reddest, fishiest kind.

How dose a scope calculate power using the math trace?.
A x B ?