power supply software ==> Dummy Load
Dummy Load for power supplies
Homebrewed power supplies need some testing. Of course, you can use power
resistors for that purpose. But if you want to know exactly at which load
your supply is not working any more, you'll need something more flexible.
To be prepared for larger power supplies you'll need a 10 or 20 W
(or even larger) design. That power should be dissipated over a longer
period of time, too, e.g. to test the heat properties of the power supply.
For that we need some larger power equipment.
Schematics of a universal version
The minus plug of the dummy load (GND) shall be tight to the heat sink,
so we can install the power transistor without heat transfer isolation
to the heat sink. For that we need a PNP transistor, where the case is
The dummy load should work from 3 V on, so we can test smaller
power supplies as well. But it should work at 35 V as well, so we
do not need a specially designed dummy load for each of our power
supply devices. Therefore we need a constant voltage that already works
at 3V but also on 35V. This is performed by the FET BF256B and the three
diodes: the BF controls the current flow through the diodes and limits
it to 8.5 mA. The diodes deliver voltages by 0.65, 1.30 and 1.95V
below the positive voltage.
With the 1k potentiometer the voltage can be adjusted to between 0.65
and 1.95V below plus. This is fed into the base of the PNP power
transistor. In its emitter line to plus a resistor is producing a
current dependant voltage. The base-emitter voltage of the transistor
is around two silicon diode voltages, because the transistor is a
darlington. Its hFE is around 10,000, so to regulate a current of 1A
we need less than 1 mA on the base. This one mA does not much
influence the current though the diodes, as held constant with the FET.
Theoretically the base-emitter voltage of 2 * 0.65 V = 1.30 V
gets lost. But, in practice that is not the case: with small base
currents VBE is much smaller than 1.30 V. With an
RE of 1 Ω and at VB of 1.30V already
a current of 0.35 A resultet. Therefore the potentiometer is
attached already to the first of the three diodes.
The Darlington has to dissipate the whole power that the power supply
delivers. If the splly has an elevated power, the transistor gets very,
very hot. It therefore needs a heat sink. In practice the transistor's
temperature has a large influence on the VBE. Therefore the
current increases with time until the transistor has reached its
steady-state-temperature. With large heat sinks this needs longer times.
So prepare to hand-regulate the current very often.
The dimension of the heat sink has to consider
Without a heat sink the transistor has a heat resistance of 1.17°C/W
between the silicon core and the metal case. A finger heat sink of
46-by-46-by-12,7 mm has 7 K/W,, a larger aluminium profile
has 3.5 °C/W, very large sinks have 2.5 °/W. With
7 °/W and a 12V/1A power supply we come to a temperature of
T = 25 + 12 * 1 * 7 = 109 °C.
- the power of the power supply (voltage * current in Watt),
- the maximal temperature of the transistor, e.g. of 100°
Celsius, add the ambient temperature of 25°C and you are at
125°C, permissible would be up to 200°C,
- divide the max. added temperature by the maximum power supply
power and the heat resistance of the heat sink in °C/W.
This here is the mounting on a small heat sink, for small power supplies.
The sink has about 25 °C/W and can dissipate roughly 4 W.
The two schematics in Libre-Office-Draw-Format are
here for Download.
Schematic for a more long-term stable version
If your power supply has at least 5V and you do not want to re-adjust
every 5 seconds, you can build this version. Here, an opamp takes over
the task to drive the transistor's base pin.
Because of the OpAmp this works only with 5V and higher. With currents from
100 mA on the device works long-term stable and only small increases
of the current need to be corrected, if the device gets hotter.
The many 47k resistors translate the voltages down to areas that the OpAmp
can work with (near to the positive supply voltage the 741 is dump).
In practice the regulator starts already with 73 mA and not with
zero Volts. Again, this is due to the OpAmp: it's output pin does not work
at high voltages. So either you can live with that or you place a power
diode between the emitter and the resistor. But in that case keep the
47k on the resistor.
When experimenting with that dummy load care for the temperature of the
heat sink. It might not be a good idea to place the sink onto a piece of
plastic, if the heat sink exceeds 80°C: you'll ruine that piece.
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