Placing blocking capacitor in series with load ,blocks DC, pass AC,
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
This would be true if the capacitor was in parallel with the load. But the OP is talking about a capacitor in series with the load. And he asked why the voltage drop goes down, not up.A larger capacitance value loads down the signal source causing its voltage to drop.
Title: Understanding Basic Electronics, 1st Ed.Placing blocking capacitor in series with load ,blocks DC, pass AC,
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
That is a bit of a blind statement.Placing blocking capacitor in series with load ,blocks DC, pass AC,
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
It is not clear if you are asking about the voltage drop across the capacitor or across the load, whatever it is. Or ar you asking about the voltage at the source of the AC signal. And that matters a whole lot. So ket us know which voltage and if it is AC or AC + DC. Then you can get an answer that is not a guess.Placing blocking capacitor in series with load ,blocks DC, pass AC,
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
WHICH VOLTAGE!!!! There are three voltages here, the voltage of the source, whuich evidently also has some DC component, the voltage across the capacitor connected between the load and the source, and the voltage across the load, whatever that might be. So without more explanation of which the TS is asking about there is no reason for folks to be guessing.Placing blocking capacitor in series with load ,blocks DC, pass AC,
If you add a larger uF Capacitor, why does or how does AC voltage drop go down?
thanksBecause impedance goes down.
Voltage drop across capacitorWHICH VOLTAGE!!!! There are three voltages here, the voltage of the source, whuich evidently also has some DC component, the voltage across the capacitor connected between the load and the source, and the voltage across the load, whatever that might be. So without more explanation of which the TS is asking about there is no reason for folks to be guessing.
There is a very serious hazard associated with using a capacitor to drop the mains voltage, which is that if an alternative power source is used instead of the regular utility supplied power, there will probably be a lot more harmonics present, and hence more harmonic energy will pass through the capacitor. So one person tried to use one of those "Kill-O-Watt" analyzers on a portable generator output and immediately the "Kill-O-Watt" was destroyed. The current delivered by the capacitor was too much for the zener diode shunt regulator and so that device failed open.The TS is most probably referring to AC mains 'voltage dropping' capacitors as in the compact mains charging circuits of rechargeable battery-operated flashlights and shavers or as in LED 'mains on' indicators.
Regards,
Nandu.
Hi MisterBill2,There is a very serious hazard associated with using a capacitor to drop the mains voltage, which is that if an alternative power source is used instead of the regular utility supplied power, there will probably be a lot more harmonics present, and hence more harmonic energy will pass through the capacitor. So one person tried to use one of those "Kill-O-Watt" analyzers on a portable generator output and immediately the "Kill-O-Watt" was destroyed. The current delivered by the capacitor was too much for the zener diode shunt regulator and so that device failed open.