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Dual switchover: one leg's pulse into the cell

When a switch leg connects the cell to the supply through its load resistor, how far does the cell charge, what current flows, and what does each pulse cost?

The formula
τ=R·CU=V(1−e−e/τ)q=C·UJ=q·fW=V·J
U
Cell voltage after the pulse, V
τ
Charging time constant, µs
e
Effective on-time, µs
q
Charge per pulse, C
J
Average supply current, A
W
Average power, W
V
Supply voltage, V
f
Pulse frequency, kHz
R
Load resistor, Ω
C
Cell capacitance, nF
LaTeX 
τ = R \cdot C \qquad U = V \left( 1 - e^{-e / τ} \right) \qquad q = C \cdot U \qquad J = q \cdot f \qquad W = V \cdot J

Work it out

V

The B+ the legs switch onto the cell.

kHz

How often a leg fires. Each period holds one leg's pulse.

%

How much of the period the firing leg is on.

µs

The gap after one leg turns off before the other may turn on — taken out of the on-time.

Ω

The resistor in series with the leg, through which the cell charges.

nF

The cell.

or see a worked example with typical numbers

Method

  1. The period is one over the frequency; the on-time is the period times the duty; the effective on-time is that less the dead time, and not less than zero.
  2. The cell charges through the load resistor as an RC circuit: time constant RC, and after the effective on-time it has reached V(1 − e^{−t/RC}) from empty.
  3. The charge moved is C times that voltage. Once per period, so the average supply current is charge times frequency, and the average power is the supply voltage times that.
  4. The peak current is the moment the leg closes: the whole supply across the load resistor.

Assumptions

  • The cell is empty when each leg fires. In the dual switchover the other leg's pulse is meant to drain it, so the model is one charge per period; if the cell holds its charge the second pulse moves almost nothing and the supply current is far lower than this.
  • The switches are ideal: no on-resistance, no drop across a diode, instant edges. A real MOSFET and diode take a volt or two off the supply.
  • The cell is a capacitor and nothing else; its leak through the water during the pulse is not counted here. The step-charging calculation takes the leak.
  • Meyer's claim for the arrangement is that the cell sees voltage while the resistor takes the current. The arithmetic here is what that arrangement does as a circuit; it does not speak to whether it does anything else.