Complete newbie here with a circuit question

Discussion in 'General Electronics Chat' started by daryl_gustafson, Mar 11, 2015.

  1. daryl_gustafson

    Thread Starter New Member

    Mar 11, 2015
    I have a electric fence charger that I am trying to troubleshoot.

    The unit uses an AC transformer that takes 120 volts in and delivers 21 volts AC to the triggering circuit and 230 volts AC to the capacitor and step-up transformer that connects to the electric fence.

    I have removed the small PCB board from the unit to focus on the trigger circuit and to avoid the large charging capacitor and step-up transformer. I have connected a 9VAC power supply to the triggering circuit for troubleshooting and analysis. And the very first thing that the 9VAC sees in the circuit is an 1n4006 diode. The triggering circuit consists of 1 diode, 3 capacitors, 10 resistors, an 2n6028 thyristor, an LM324 op amp, and an LED.

    Here's my question....

    Measuring the AC voltage just BEFORE the diode (relative to the other other line from the 9 volt power supply) shows 9.1VAC on my multimeter. That makes perfect sense. But, measure the DC voltage on the other side of the diode shows 12.3VDC. I was expecting something less than 4.5VDC as the AC voltage is now a half-wave rectified voltage and only the +9VAC half of the sine wave is getting through the diode and the -9VAC portion is ZERO. Connected to this (DC) side of the diode is a 100 microFarad 35V electrolytic capacitor (whose negative side is connected to ground). Does this make any sense? Is it possible to have 9VAC come in to a diode and have 12VDC appear on the other side?

    Please remember I am a complete newbie in all of this. But I have always been fascinated by what electronics can do. Black magic that I hope (someday) I might begin to understand.

  2. crutschow


    Mar 14, 2008
    Not only is it possible, but it is expected. ;) You misunderstand the relationship between the AC voltage and the peak voltage.
    The 9.1VAC measured is an RMS value so the positive and negative peak values (not peak to peak) of the sinewave are 1.4 * 9.1V = ±12.74V.
    Subtracting a few tenths of a volt for the diode forward drop allows the capacitor to charge up to the 12.3Vdc you measured.
  3. #12


    Nov 30, 2010
    crutschow has the right answer.
  4. daryl_gustafson

    Thread Starter New Member

    Mar 11, 2015
    THANK YOU! That makes perfect sense. I forgot about the RMS thing. Now I can move on......