Direct.Dude Posted February 1, 2016 Share Posted February 1, 2016 So apparently my fly swatter has a capacitor on the output. I measured its lines DC and it read 170VDC. Okay... Now today I was thinking that my shock and the spark was too strong for 170VDC. I measured it at AC and it read 600VAC, and it kept increasing. Can anyone explain to me why this happens and how is it possible to have dc and ac on the same line? Sorry if this is the wrong section. Link to comment Share on other sites More sharing options...
swansont Posted February 1, 2016 Share Posted February 1, 2016 A flyback oscillator circuit is a common charging mechanism for such devices. I think the output has an AC component on top of a DC offset. Link to comment Share on other sites More sharing options...
Sensei Posted February 2, 2016 Share Posted February 2, 2016 (edited) I think you should connect it to oscilloscope to have more useful/reliable data, with time in X axis, and voltage/current in Y axis. Edited February 2, 2016 by Sensei Link to comment Share on other sites More sharing options...
Direct.Dude Posted February 2, 2016 Author Share Posted February 2, 2016 Unfortunately I don't have an oscilloscope. What can I do? Today I measured it again and it read 700VDC and 1KVAC. Link to comment Share on other sites More sharing options...
lance7 Posted July 6, 2016 Share Posted July 6, 2016 (edited) So apparently my fly swatter has a capacitor on the output. I measured its lines DC and it read 170VDC. Okay... Now today I was thinking that my shock and the spark was too strong for 170VDC. I measured it at AC and it read 600VAC, and it kept increasing. Can anyone explain to me why this happens and how is it possible to have dc and ac on the same line? Sorry if this is the wrong section. The fly swatter is wired to produce pulse DC current to store charge in the capacitor. Else an alternating current won't charge the big capacitor. The reading you got is due to the bleeder resistor connected between the output for safety reason. When the switch is off and the capacitor is charged, the current would travel backwards through the resistor into the negative terminal, to 'bleed' out the charge. As it travels backwards, the capacitor would be charged inversely, so the current would travel back and forth like an alternating current until the current is completely gone. Usually it takes just a few second for the current to be dissipated fully. However, when the output is producing too much voltage for the bleeder during closed circuit, the current would alternate between the outputs across the resistor and thus produce an AC component. This is made possible because the output is only half-rectified, so the moment when backward current is opposed by the diode, it travels elsewhere [ through the resistor ]. Cutting out the resistor will get rid of the AC component and would enable you to store charge in the capacitor longer. Edited July 6, 2016 by lance7 Link to comment Share on other sites More sharing options...
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