Practical_Electronics-May_2019

Your project is fi nished and ready to go,

but the job isn’t done until you’ve found

an appropriate source of power. This

could be as simple as choosing a suitably

rated mains adapter or as complex as

designing a switched-mode power supply

with multiple outputs and battery backup.

Our latest series – Teach-In 2019 – is

here to help, and will provide you with

insight into all aspects of powering your

electronic projects and designs.

Last month, we examined switched-

mode power supplies (SMPS) and

explained the basic principles of

buck (step-down) and boost (step-up)

switching regulators. This month, we

will be revisiting some of the rectifi er

circuits that we fi rst met in Part 2,

and show how we can easily produce

multiples of the basic DC output voltage

with voltage doublers, triplers, and

quadruplers. These are useful circuits

that can be built using just a handful of

low-cost components.

We will also provide some important

advice and guidance on how to select

and use electrolytic capacitors.

Our Practical Project takes the form

of a simple voltage doubler designed for

producing a low-current 15V DC output

from a small 9V battery. This project is

ideal for supplying power to 15V CMOS

logic circuits.

+

V mA W

+

+

by Mike Tooley

The simple half-wave and full-wave

rectifi ers that we met in Part 2 can be

easily modifi ed and extended to produce

multiples of the basic DC output voltage.

For example, Fig.6.1(a) shows a simple

half-wave voltage doubler arrangement.

On the fi rst negative-going half-cycle of

Powering Electronics

Par t 6: Vol tage mul tipliers

Teach-In 2019

Vol tage mul tipliers__

Fig.6.1. Simple half-wave and full-wave

voltage doublers.

Fig.6.2. Voltage tripler and voltage quadrupler

arrangements.

the input AC supply voltage, D1 will

conduct and charge C1 to the negative

peak secondary voltage (less the forward

0.7V voltage drop associated with D1).

This positive voltage appears

at the anode of D2, which then

conducts on the positive-going

half-cycle, charging C2 to double

the applied peak voltage (less the

forward voltage drop associated

with D1 and D2). With

a sinusoidal AC input,

two 1N4001 diodes,

and an RMS secondary

voltage of VS, the peak

DC voltage appearing

across the load (RL) will

be approximately:

Vout = (2 × 1.4 × VS)

• (2 × 0.7V)
  = 2.8VS – 1.4V

Note that the 1.4 factor

converts the RMS

value to a peak value

(strictly speaking the

factor is √2 = 1.4142...).

So, with a transformer

delivering 12V RMS

from its secondary, the

no-load output voltage

can be expected to be

approximately 32V (and

also at low load currents

of up to about 50mA).

An alternative voltage

doubler arrangement is

shown in Fig.6.1(b). This

circuit offers better performance to that

shown in Fig.6.1(a) since both diodes are

conducting for part of every half-cycle of

the secondary input (its action is that of a