These circuits are classified as LOW IMPEDANCE circuits and this means
the input needs to have a capacitor so the circuit can “push
against the capacitor” via the positive rail and produce the required
output. If a capacitor is not present, the circuit relies on the
low-impedance of the battery. (The output needs to have a capacitor to store
the pulses and deliver a DC output.)
This means you cannot put a milliammeter (this is the correct wording for a
milli-amp meter) in the positive (input) line and expect the circuit to work
correctly. The high-resistance of the meter will cause the output to
decrease considerably - because the circuit simply will not work. It needs a
low-impedance supply because it requires a very high current due to the fact
that the output voltage is being multiplied about 3 times - and the current
is further multiplied by two because the inductor is being "charged" for 50%
of the time and delivering for 50% of the time. To maintain a
low-impedance supply while taking current-readings, a 1 ohm resistor
(brown-black-gold-gold) is placed in the positive line and a multimeter
(200mV or 2v range) is connected across the resistor. This means each mV
reading will represent 1mA current.
For the PHONE CHARGER circuit, the idle current was less than 0.1mA and
could not be read on the meter. The output voltage dropped to 5v (in idle
mode) and kept the 1u ceramic capacitor charged.
To turn-on the PWM controller chip requires a reasonably–high current for a
very short period of time. This is achieved by the 10u tantalum capacitor
and when the chip draws current, a voltage of 5.1v is dropped across the
zener. The 10k resistor is needed to quickly discharge the 10u if the power
is removed and re-applied. If the 10k is removed, it takes more than 10
seconds to discharge the 10u so it can re-apply the spike of current.
We placed a 47R across the output to check the regulation. It got HOT
(200mA) but the output voltage remained at 10v. This is classified as “100%
regulation” and is an amazing achievement for microscopic components.
The photo shows the surface-mount components added to the top of the PC
board. The SS14 Schottky diode is turned 90 degrees so the additional
components can be fitted. This is shown in detail in the second photo above.
If you are connecting a 9v battery snap to the output, remember it is being
used “in reverse” and the negative lead must be connected to the positive
output on the board so the
snap can be connected to another snap on a radio or other device needing 9v.
MODULE 2:
The second module is a 3v to 5v PWM boost circuit with an output current of
500mA (or more under ideal conditions).
It can be purchased on eBay for less than $3.00 posted:
DC-DC Power Supply Converter Step Up Boost Module 1A 3V to 5V
It can be "jacked up" to 12v by changing the value of R1 or R2.
There are a number of modules that look exactly like the
photos above but the voltage-divider will have different values.
Our module had 20k (203) and (180) where the "180" surface mount resistor
was a special value and measured about 150k.
By increasing the value to 270k the output voltage was 8.6v and 330k
created an output voltage of 10.5v. You can adjust this value by using a
higher value and placing a 1M or 2M2 on top or altering R2. Don't omit R1or
the output voltage will rise above 30v.
FB is the "Feedback" pin or Vreference pin and it needs to see about 1.2v.
Rather than going into the formula to determine the output voltage, it is
much easier to change the values of R1 and R2 and measure the output.
Increasing R1 will increase the output voltage. And placing a 220k across
the 20k resistor will increase the output from 8.6v to 9.4v. It
is much easier to put a resistor across R2 than replace R1.
The 1206 surface mount resistor was the only size available and a small length of fine tinned copper wire was needed to
join one end to the pad.
A 0805 surface mount component would fit perfectly.
The module fails to regulate if the input voltage falls below 2.5v and this
will limit the output current when 2 cells are used as the output voltage
will very easily dip to 2.5v if a high current is drawn. The input voltage must be 2v below the output
voltage, so a 6v supply could be used for 9v output.
A point to note:
This module drives an AM radio without any background hash but the FM
reception is impossible to receive.