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08-07-2026, 01:22 PM
Hi Guys
Another method for creating the boosted supply rails is to use a charge pump for each. This sounds more complicated than it is.
We start with the basic split-rail supply for the output stage. For example, we have +/-50V derived from a 70Vct transformer winding. Consider just the positive rail for the moment: we add a filter cap on top of the +50V, with a bleeder resistor and a zener diode in parallel with this new cap. A half-wave rectifier (single diode) is connected to the positive end of the new cap, cathode to cap+ and anode to to a second resistor tied to +50V. A second cap ties to one end of the 75V winding and its other end ties to the diode anode. This cap is the "charge pump"
As the AC end of the charge pump cap dips down to its most negative extreme, the cap is charging to a voltage equal to the full DC-equivalent of the 75Vac, so about 100Vdc. The cap charges through the second resistor we added tying the diode anode to +50V.
As the AC swings towards a full reversal, the top of the charge pump cap rises above +50V and begins losing charge both through the tethering resistor and the diode, which has turned 'on'. The diode will stay 'on' until the stacked cap charges up to the zener clamp limit. The rest of the charge dissipates through the tethering resistor.
The AC voltage again reverses and the diode turns 'off' and the cap begins to be recharged as its negative end is pulled below +50V.
The second and subsequent charge scoops only discharge into the stack cap when the diode turns 'on' at the point where the sagged stack-cap voltage plus diode threshold voltages are less than the voltage at the top end of the charge pump cap. This is the same as regular cap charging behaviour through a rectifier. The available charge time is then only a small percentage of the AC mains frequency.
On the negative rail, the diode is reversed and the stack cap and zener are mounted appropriately.
Note that the voltage polarity across the charge pump capacitor never reverses, allowing us to use standard electrolytic caps. This saves us money and allows the use of high capacitance values to assure support for the load current of the power amp front-end. Being able to move a lot of charge also compensates for the fact that this is half-wave.
This type of circuit is used in some tube amps to generate the bias supply - amps where the bias supply is an after-thought, as we have discussed in TUTs and elsewhere on this forum.
We can make this full-wave in two different ways:
The first FW method requires duplicating the charge pump cap, tethering resistor and diode and linking the cathodes for the positive rail, or linking the anodes for the negative rail. The new charge pump caps tie to the opposite end of the AC winding referenced to the original charge pump caps.
The second FW method uses integrated bridges for each boost supply. Each bridge has a resistor across its AC terminals, then two charge pump caps, one spanning one end of the winding to one bridge-AC input, and the second cap spanning from the opposite winding end to the second bridge-AC input.
An advantage of the full-bridge approach is seen when we wish to make a much higher supply voltage without adding another PT. Suppose we have a standard solid-state power amp with its split rails and wish to have a tube preamp driving it. We have +/-50V for the PA. using the full-bridge cap stack method, we can stack another 100Vdc on top of the +50V and have +150V for the tube plate supply. We can stack a second circuit on top of the first and have +250V. Obviously this requires fewer stacks if the split rails are much higher to begin with.
Another method for creating the boosted supply rails is to use a charge pump for each. This sounds more complicated than it is.
We start with the basic split-rail supply for the output stage. For example, we have +/-50V derived from a 70Vct transformer winding. Consider just the positive rail for the moment: we add a filter cap on top of the +50V, with a bleeder resistor and a zener diode in parallel with this new cap. A half-wave rectifier (single diode) is connected to the positive end of the new cap, cathode to cap+ and anode to to a second resistor tied to +50V. A second cap ties to one end of the 75V winding and its other end ties to the diode anode. This cap is the "charge pump"
As the AC end of the charge pump cap dips down to its most negative extreme, the cap is charging to a voltage equal to the full DC-equivalent of the 75Vac, so about 100Vdc. The cap charges through the second resistor we added tying the diode anode to +50V.
As the AC swings towards a full reversal, the top of the charge pump cap rises above +50V and begins losing charge both through the tethering resistor and the diode, which has turned 'on'. The diode will stay 'on' until the stacked cap charges up to the zener clamp limit. The rest of the charge dissipates through the tethering resistor.
The AC voltage again reverses and the diode turns 'off' and the cap begins to be recharged as its negative end is pulled below +50V.
The second and subsequent charge scoops only discharge into the stack cap when the diode turns 'on' at the point where the sagged stack-cap voltage plus diode threshold voltages are less than the voltage at the top end of the charge pump cap. This is the same as regular cap charging behaviour through a rectifier. The available charge time is then only a small percentage of the AC mains frequency.
On the negative rail, the diode is reversed and the stack cap and zener are mounted appropriately.
Note that the voltage polarity across the charge pump capacitor never reverses, allowing us to use standard electrolytic caps. This saves us money and allows the use of high capacitance values to assure support for the load current of the power amp front-end. Being able to move a lot of charge also compensates for the fact that this is half-wave.
This type of circuit is used in some tube amps to generate the bias supply - amps where the bias supply is an after-thought, as we have discussed in TUTs and elsewhere on this forum.
We can make this full-wave in two different ways:
The first FW method requires duplicating the charge pump cap, tethering resistor and diode and linking the cathodes for the positive rail, or linking the anodes for the negative rail. The new charge pump caps tie to the opposite end of the AC winding referenced to the original charge pump caps.
The second FW method uses integrated bridges for each boost supply. Each bridge has a resistor across its AC terminals, then two charge pump caps, one spanning one end of the winding to one bridge-AC input, and the second cap spanning from the opposite winding end to the second bridge-AC input.
An advantage of the full-bridge approach is seen when we wish to make a much higher supply voltage without adding another PT. Suppose we have a standard solid-state power amp with its split rails and wish to have a tube preamp driving it. We have +/-50V for the PA. using the full-bridge cap stack method, we can stack another 100Vdc on top of the +50V and have +150V for the tube plate supply. We can stack a second circuit on top of the first and have +250V. Obviously this requires fewer stacks if the split rails are much higher to begin with.
