The following circuit produces a voltage higher than 12v via a CHARGE PUMP arrangement in which the energy in an electrolytic is fed to a battery to charge it. The circuit produces about 900mA "charge current" and the diodes and transistors must be fitted with heat sinks. The LEDs are designed to prevent the two output transistors turning ON at the same time. The lower output transistor does not start to turn on until the voltage is above 5v and the top transistor does not turn on until the voltage drops 4v from the positive rail. This means both transistors will be turned on when the voltage passes a mid-point-gap of 4v. The transition is so fast that no wasted current (short-circuit current) flows via the two output transistors (as per our test).
The electrolytic charges to about 10v via the lower transistor and top diode. The top BD679 then pulls the negative of the 2200u electrolytic towards the 12v6 rail and the positive is higher than 12v6 by a theoretical 10v, however we talk about the energy in the electrolytic and in our circuit it is capable of delivering a current flow of about 900mA. This energy is passed to the battery via the lower diode. Most batteries should not be charged faster than the "14-hour-rate." This basically means a flat battery will be charged in 14 hours. To do this, divide the AHr capacity by 14 to get the charge-rate. For example, a 17AHr battery should be charged at 1.2A or less. For lower-capacity batteries, the 2200u can be reduced to 1,000u. Charging is about 80% efficient. In other words, delivering 120% of the AHr capacity of a battery will fully charge it.