glue on one side; you print on the glue side and soak in
warm water to remove the paper.
The problem is that 400 degrees F is not hot enough
to melt the toner, causing it to be porous. If you do melt
the toner, it spreads. Pulsar has a solution called Green
TRF Foil. After transferring the pattern, put a sheet of
Green Foil over the pattern and heat again. The green
plastic sticks only to the toner and seals it.
You can print using an inkjet printer, and then run it
through a laser copier to make the transferrable pattern.
Don’t forget to make it mirror image. Measure the copy to
make sure it is 1:1; most printers don’t do X and Y to the
NIMH BATTERY CHARGER
This is a design that I did recently that I want to share.
The objective is to make a fast charger that will allow the
fully charged battery to float indefinitely. Since the charge
voltage is not a good indicator of state of charge, battery
temperature is used to indicate full charge. The circuit is in
Figure 5. The battery is 10 AA cells in series for a nominal
12 volts. The charge current is one amp but the charge
voltage is limited to 15 volts.
In Figure 5, the low dropout (LDO) regulator, IC1, is
configured as a constant current source of one amp: I =
Vref/ (R5+R6) = 1.25/ (1.24) = 1.008 amps.
If the terminal voltage of the battery should exceed 15
volts ( 15. 6 volts at R8), the current through R8 is
15.6/(110K+ 39.2K) = .105 mA, and the voltage across R8
is .105 mA* 39.2K = 4.099 volts. That is enough to turn on
IC2B and reduce the current.
The thermistor, R4, monitors the ambient temperature;
the other thermistor, connected to J1, is embedded in the
battery. When the battery temperature rises 10 degrees C
above ambient, the output of IC2A switches, turning off
the current source.
Positive feedback through R3 produces hysteresis to
prevent the charger from turning on until the battery cools
down. The thermistors are 10K at room temperature, 1%
tolerance, Mouser part #71-01C1002FP.
The design of the IC2A circuit with hysteresis
proceeds this way: I want the current through R2 and R1
to be low so as not to heat the thermistor. If R1 is 100K,
the current is 18V/(110K) = .164 mA. That puts the
input (pin 2) at 1.64 volts. The other input wants to be
lower at room temperature ( 25 degrees C) and higher
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