Dual Power Source and Overvoltage Protection

Many applications require a circuit to run from one of two voltage sources. Usually these sources are an ac adapter and battery configuration. This is particularly the case with guitar effect units. It is necessary to supply current to the circuit from the ac adapter when it is present. When this is occurring, the current drain from the battery needs to be minimal. When the battery is the only source present, the voltage needs to be delivered to the circuit without significant voltage drop. It is also desirable to guard against an ac adapter being present that is not the rated to the correct voltage for the circuit (overvoltage protection).

Figure 1 below provides a very basic solution.

Figure 1: Schematic of circuit.

D3 is a zener diode. It is reversed biased and increases the current flow through it once the zener voltage is exceeded (9.1V in the above circuit). The zener works as a voltage regulator, keeping the voltage close to 9.1 volts, even though the ac adapter may be providing much more. Be sure to select a zener diode that has a power rating that will cope with the current flow through it. A 5 watt zener should be sufficient for significant overvoltages, but lower rated zeners can be placed in parallel to provide the same level of protection.

D1 and D2 prevents current flowing from one voltage source to the other. R2 reduces the current flow from the battery when the ac adapter is present. A majority of the current will flow from the ac adapter. R1 and R2 need to be small enough not to cause loading in the main circuit when the battery is the only source present, but large enough to provide significant resistance to current flow when the ac adapter is plugged in. The presence of R1 and R2 will set up a voltage divider with the load resistance (not shown in figure 1), reducing the voltage supplied to the circuit from the battery.

The use of a zener diode for overvoltage protection is not full proof. It will not protect against very large voltage spikes or provide short circuit protection.

Figure 2 demonstrates what happens to the current flow in both of the voltage sources over a voltage sweep. The voltage is increased from 0 - 14 volts in the AC adapter. At 8V current starts to flow from the AC adapter (red) through a 1 k ohm load. At the zener voltage (9.1V) both sources contribute current equally to the load. At around 10V, the battery (yellow) has a significantly reduced current flow. To be effective, this circuit needs to have an AC adapter input of at least 10V. If this is not the case, significant power will continue to be consumed by the battery when an AC adapter is present.

Figure 2: Voltage sweep of AC adapter (0 - 14V, Battery constant @ 9V) with current values of voltage sources (red = AC adapter, yellow = 9V battery).

Also notice in figure 2 how the current of the AC adapter increases in a linear fashion once the zener voltage is exceeded. If 50V is supplied by the adapter, 0.4A flows through the zener. This results in a power dissipation of 3.67W across the zener. One 5W, or four 1W diodes in parallel would then be required.

Figure 3 looks at the output voltage of the circuit. Until 8V, the battery is supplying the output at 7.58V. The 1.42V voltage drop is due to the diode and the resistor divider (R2 and the load resistor). After 10V, the output has started to settle on 8.5V, supplied by the AC adapter. By choosing a lower zener voltage, the difference in output voltage can be largely removed.

Figure 3: Voltage sweep of AC adapter (0 - 14V, Battery constant @ 9V) with output voltage.

See the Matlab .m file below to calculate power requirements and voltage output levels of the circuit.

function [Pz] = zener(Vin, Vout, Rz, Rl)
% ZENER - Power and output voltage of multi source circuit using zener overvoltage protection
% Vin: AC voltage input
% Vout: Required output voltage of the system (zener voltage)
% Rz: Value in ohms of the resistor in serieswith the AC adapter
% Rl: Load resistance
%
% Pz(1): Power dissipation of the zener diode
% Pz(2): Output voltage of the circuit
%
% Example:
% Pz = zener(16,9.1,100,1000)

Vratio = Vout / Vin;

Rb = (Vratio * Rz) / (1 - Vratio);

Iz = Vout / Rb;

Pz(1) = Iz * Vout;

Pz(2) = ((Rl / (Rl + Rz)) * Vout) - 0.45;