12-01-2013, 03:36 PM
DC-DC converter to step up input voltage A simple step-up converter
[attachment=46799]
This step-up converter is intended for use in a '67 Citroen 2CV. This car, and I use the word loosely, has a 6V battery and won't support a modern radio that needs 12V. The circuit described here converts 6V to 12V at 1A sustained load current.
It works something like this:
When the switch is closed an extra current flows through the inductance and stores energy there. The capacitor supplies the load with current during this time.
After the switch closes the capacitor is charged by the energy stored in the inductance and an extra current starts flowing through the load, causing the output voltage to rise (energy is supplied directly from the input source also as long as the diode is forward biased). During this time, the system behaves like a RLC-circuit, so, after a while, the current decreases. The switch is then closed again and the cycle repeats. One could say that charge is pumped from input to output, increasing the output voltage up to the point where there is an equilibrium between the discharging of the capacitor while the switch is closed and the charging by the inductor while the switch is open.
[attachment=46799]
This step-up converter is intended for use in a '67 Citroen 2CV. This car, and I use the word loosely, has a 6V battery and won't support a modern radio that needs 12V. The circuit described here converts 6V to 12V at 1A sustained load current.
It works something like this:
When the switch is closed an extra current flows through the inductance and stores energy there. The capacitor supplies the load with current during this time.
After the switch closes the capacitor is charged by the energy stored in the inductance and an extra current starts flowing through the load, causing the output voltage to rise (energy is supplied directly from the input source also as long as the diode is forward biased). During this time, the system behaves like a RLC-circuit, so, after a while, the current decreases. The switch is then closed again and the cycle repeats. One could say that charge is pumped from input to output, increasing the output voltage up to the point where there is an equilibrium between the discharging of the capacitor while the switch is closed and the charging by the inductor while the switch is open.