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The application notes of several boost-converters, e.g. from TI (TPS61070, p.18) or LT (LT3464, p.14), show similar unregulated auxiliary output circuits for doubling or inverting the (regular) regulated output.

The following schematic shows how the invertingg auxiliary amplifier topology is combined with the boost converter. Diodes should be schottkies.

schematic

simulate this circuit – Schematic created using CircuitLab

Do these circuits have a specific name? I could not find them in literature about switch-mode-power-supplies, but would like to learn more about it.

As far as I can tell, the auxiliary output is mainly practical for supplying small loads e.g. for operation amplifiers or logic-level-conversion ics.
Are there any noteworthy trapdoors or design considerations when using this auxiliary output for such applications? E.g. can this introduce a negative impact on the boost stage or overall efficiency?

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Do these circuits have a specific name?

Cockcoft Walton voltage multiplier is the name. Here's a multi stage version: -

enter image description here

Sometimes they are loosely called voltage multipliers but that does give rise to some confusion.

Are there any noteworthy trapdoors or design considerations when using this auxiliary output for such applications? E.g. can this introduce a negative impact on the boost stage or overall efficiency?

The forward diode voltage drop is the main limiter to efficiency but other than that they can be used for quite high powers (I've designed one for 200 watts) and they can certainly be used for generating HT voltages (50 kV for example in some X-ray machines).

Here's Dave: -

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  • \$\begingroup\$ Nice, I was not expecting a video. Extending this with multiple stages truly extends possible applications. Gonna try that with a custom transformer albeit not in the kV domain. \$\endgroup\$ – Grebu Sep 1 '16 at 9:14
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The auxiliary circuit you provided is an example of negative voltage generator (similar configuration exists using a voltage doubler to achieve a positive auxiliary voltage, including some with isolation by transformers). During the time Toff (M1 turned-off), capacitor C1 is charged with a voltage Vboost - 2 * Vd (Vd is the voltage drop across the diode). During the Ton time, this voltage is transferred to capacitor C2, reaching a value of -((Vboost - 2*Vd)- Vd) = -Vboost + 3*Vd.Capacitor C2 must provide current to its load (auxiliary) during the time Toff. So, for a switched converter (boost, in this case):

$$D =\frac{T_{ON}}{T_{ON}+T_{OFF}}$$ and $$t_{OFF} = \frac{1-D}{f_{SW}}$$

In order to estimate the value of capacitor C2 (equal to C1), based on tolerated Vripple and maximum output current, Iaux:

$$i\cong C\frac{\Delta v}{\Delta t}$$ Doing $$\Delta v=V_{RIPPLE}$$
$$\Delta t=T_{OFF}$$ and $$i=I_{AUX}$$ results in $$V_{RIPPLE}=\frac{I_{AUX}(1-D)}{Cf_{SW}}$$

Also, a term can be added to Vripple expression regarding to voltage drop contribution for the capacitor Equivalent Series Resistance (ESR): $$I_{AUX}ESR_{C}$$

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    \$\begingroup\$ In power electronics (switched), the most widely used name for this configuration is CHARGE PUMP; evidencing the use of capacitor in contrast to others inductor based switched converters. \$\endgroup\$ – Dirceu Rodrigues Jr Aug 31 '16 at 19:28

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