A shunt regulator technique may be effectively employed in regulating the charging voltage of a lead-acid battery. Although it has a practical power limitation (< 20W or so), it has the notable advantage of being an LDO (low dropout) regulator (zero voltage dropout in this case) - this is in contrast with the typical series regulator that often has a dropout voltage in the order of 1 to 3V.
Note that 6V solar panels often have minimal voltage headroom that tends to make it difficult to obtain the full charge voltage of 7.2V.
The shunt regulator is connected across the load rather than in series with the load. It regulates the voltage by shunting excess current away from the solar panel. In order for a shunt regulator to function, it must work against high source resistance – in this case, the solar panel is ideal, having a very high source resistance. Shunt regulators tend to be less efficient than series regulators because they dissipate the most power when the output is at minimum – however, in this case the power dissipated is the excess solar panel power that would be otherwise unused.
The circuit is a simple, but practical solar charge regulator – generally simpler than the typical series regulator.
D10 is a schottky battery isolation diode – it is vital in this application so that the actual battery charge never gets shunted by the regulator. The low forward voltage of D10 (approx 0.25V @ low current) means that the solar panel must generate only 0.25V over the battery voltage. The shunt regulator consists of two transistors (Q1 & Q2) in the composite NPN/PNP connection. This has the current gain of a Darlington with the input characteristics of a bipolar transistor. Q2 dissipates approx. 3.5W, so a small heatsink is required. D2 in series with zener D3 (6.8V) and the Vbe of Q1 make up the voltage reference – when the voltage exceeds 7.45V, Q1 & Q2 turn on so that the current is shunted. I had to trim up the voltage of the zener slightly by adding D2 because the zener was running on the low side. The shunt current path is through D4 through D8 – these diodes dissipate approximately half of the power thus taking it away from Q2. D4 through D8 also drop sufficient voltage to turn on LED D9, thus indicating at least 75% charge level.
The circuit may be simplified by eliminating the two LEDs and limiting resistors. D4 though D8 then may also be eliminated. This reduces the component count from 15 to only 6.
In the protoboard photo there is an error – I had the green LED connected to common rather than the cathode of D8.
For the future
Extracting excess power from a shunt regulator for useful purposes