# What do you mean by a circuit

## Forum

### Circuit for halving the voltage (electronics)

written by hws, 59425 Unna, October 22nd, 2007, 6:09 pm
(edited by hws on October 22, 2007 at 6:12 pm)

“It should come out half the tension

This is what you want it to be.

»Because I have two capacitors in series
»Load, that's why every Kondi has half the tension,

That is correct, T2 blocks (i.e. it does not exist at all)
And T1 directs. The capacitors are charged in series.

»And me the
“Wanted to discharge capacitors again in parallel.

Yes, but not with the circuit.
T1 is blocked (i.e. not available) and T2 conducts, so it can be discharged through a wire bridge.
What is it now? The capacitors are still in series and also discharge in series.
The fact that diodes are in the reverse direction over the C's has no effect as long as the capacitors are of the same size and discharge immediately.

If one capacitor were much smaller than the other, it would discharge more quickly and be recharged again in the opposite direction by the larger one. This prevents the diodes (down to -0.7V)

The circuit is totally wrong.

"I have no basic control (sounds stupid, but that's how it is)

no, it's not like that

“The bases of T1 and T2 are connected directly to output 3.

Well then they are controlled via output 3 - or what do you mean by that.

“That's the complete plan and the transistors are like the picture
»Connected.

but in the picture the bases are not connected

»Explain a little?

Take two of the same capacitors. One remains empty, the other is charged on Ux.
Now both capacitors are connected in parallel. It occurs at the capacitors Ux / 2.

Now let's calculate the energy = 1/2 C u²
Once before with Ux and C and once with Ux / 2 and 2 * C.
What comes out

hws