Electronics > Mechanical & Automation Engineering

Describe to me how this relay voltage doubler works

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Qmavam:
OK, it's a simple circuit, but I can't understand how the negative side of the capacitor goes to negative 6v. (I assume that's what happens)
Also, explain the function of the diode to ground. I understand if you tied it to ground it wouldn't work, but what are the currents in the diode.
  Thanks, Qmavam


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TimFox:
The 220 uF cap and the lower 1N4148 diode form a negative "DC restorer" (q.v.).  Regardless of the DC level on the positive terminal of the capacitor, the negative swings at the positive terminal will pass through it and the diode establishes the baseline as near ground.  (The diode keeps the negative end of the capacitor from going substantially positive with respect to ground, and the charge in the capacitor will hold the voltage across the capacitor somewhat constant.) 

H.O:
I'll take a stab at this, hopefully I won't make a fool out of myself :-)

When the BC547 is "off" so is the BC557. The capacitor gets charged thru the 2k2 resistor and the bottom diode. If you measure the voltage between "ground" and the positive terminal of the capacitor you'll see around 6V. The voltage across the capacitor itself will be slightly less (~5.5V) due to the bottom diode.

When the BC547 turns on it A) turns on the BC557 which "connects" the high side of the relay coil to the 6V  rail and B) "connects" the high side of the capacitor to ground. Remember how the voltage across the capacitor was roughly 5.5V. Now, with the capacitors positive terminal pulled to ground there's still 5.5V across the capacitor and therefor its negative terminal will be roughly 5.5V below ground and there will be roughly 11.5V across the relay coil - for a short time.

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