# Scoposcopy: Pulsed Circuit Anomaly from TinMan ?

UPDATE: Read the end of the Description for the proper way to calculate power in the bulb.

Here’s my replication of the effect that TinMan noticed with his Bedini SSG. I set up a pulsed circuit that didn’t need a rotor or selftriggering, just using a mosfet switched by my Function Generator, to achieve the same effect.

The light bulb “should” be brighter with more average current flowing through it when the capacitor is connected, but if the circuit resistance is right, it does the opposite.

The scope shows the current through and the voltage drop across the bulb, and computes the average values for those measurements during both tests, with and without the capacitor connected.

You can see that the current measurement is quite a bit higher when the cap is connected but with the 21.3 ohm resistor, the bulb is actually dimmer than it was when the cap is disconnected. With no resistor, the bulb is much brighter when cap is connected, as you would expect from the average current reading. And when the resistor is 18 ohms, there is virtually no change in brightness, even though the current is nearly doubled when the cap is connected.

The schematic and the two scope traces appear at the end of the video.

Thanks to TinMan for reporting this interesting bit of Scoposcopy anomaly… it might make a difference when measuring pulsed circuits. But don’t be misled by the “Red Herrings” !
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Now posted to the thread on ou dot com:

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, Vdrop across the bulb, and duty cycle of the pulse train.