Elenco Basic Electronic Experiments - Manual (Page 38)

38
EXPERIMENT #25: BATTERY IMMUNIZER
Connect the circuit according to the Wiring Diagram and schematic. Note that the collectors of the center and right
transistors are not connected although their wires cross over each other in the schematic. Connect the loose wire to (+)18
or any of the (+) holes in the same row (which are connected to the battery); the LED is bright. Now connect the loose
wire to the emitter of the left transistor (holes f15, g15, h15, i15, or j15) as shown in the schematic; the LED is just as bright.
So we made a change and nothing happened, does this seem like a dull experiment? It may seem dull but the important
idea here is that we made a big change to the circuit but nothing happened to the LED.
Take a look at the schematic. The circuit to the left of the loose wire reduces the voltage to 4.7V. You connect the loose
wire to either the 9V battery voltage or the modified 4.7V. The circuit to the right of the loose wire creates a fixed current
to the LED, which will not change even if the voltage (9V or 4.7V) to the circuit changes. So when you changed which
voltage the loose wire was connected to, you didn’t see any change in LED brightness.
In case you’re not convinced by this, let’s change the circuit to prove it. Place a second LED in series with the 3.3k
Ω resistor
(reconnect the 3.3k Ω so that it is between d20 and f20, add an LED into holes e20 and f21 with the LED’s flat side in f21).
Now connect the loose wire to the two voltages as before and you should see the new LED change between bright and
dark while the old one remains bright as before.
You could use a circuit like this when you don’t want your performance to be affected as your voltage drops, perhaps due
to a battery weakening over a long period of use. So you could say your circuit is immune to (protected against) a weak
battery.
+9V
3.3k Ω
470
Ω
33k
Ω
(LOOSE WIRE)
DIODE