Re: AC Transformer for HP-33C
Message #6 Posted by Dave Shaffer on 26 Feb 2007, 10:11 p.m.,
in response to message #1 by Jean-Michel
It might be simpler to just get an adapter transformer, as some have suggested.
However, if you are handy with a soldering gun, and/or eager to try something different, then consider this solution.
Since I am perhaps the ultimate cheapskate (a possible challenge to the calculator/geek crowd here - I get a feeling that these characteristics tend to run together!), I once conjured up the following circuit for a trip to Europe so I would not have to buy a transformer and adapter for my electric shaver. The idea is to drop the voltage from 220 to 110 with a series resistor.
The circuit should look something like this
<pre>
R
o---------^v^v^v^v---------o
220 VAC 110 VAC
(European plug) (American socket)
o--------------------------o
</pre>
where the value of R depends on the current to be delivered to the 110 VAC output side.
By Ohm's law, you want the voltage drop to be 110 V across R so that the remaining 110V appears at the output. If you know the current to be delivered to the charger (let's call it I), then you can calculate R from
<pre>
V 110 V
R = ---- = -------- = 5.5 kOhm (as calculated with my trusty 41CV)
I 20 mA
</pre>
I'm not sure what the actual current is. Perhaps Randy can tell us. And, it shouldn't be too critical.
In cases like this, you also need to be a bit concerned about the power that will be dissipated in the resistor. That can be calculated as (I^2 R) or VI, where V is the voltage drop across the resistor. The power here will be 2.2 watts, so you need a modest sized resistor (which will be both heating your house and spinning the electric meter while you are charging your calculator!).
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