classic charger question



#4

i have four questions that i know someone must have already answered while i was not paying attention.

which calculator was the 03502A charger for? the museum just labels it as being for the classic series.

why does the 3 prong classic type plug have two chargers driving it (03502A and 82002A)?

why is the little squarish housing that looks like a ti charger give so much lower voltage than the older unnumbered one (5vdc/3.5vdc vs 14vdc/5vdc) regardless of the number printed on it?

is one of these another one of thoes "has to be checked under load" chargers so i am comparing apples and oranges?

thanks in advance. - d


#5

I don't know. My 67 came with the 82002A which is switched (86/127V to 172/254V). I then obtained a 82002C which is not switched.

tm

#6

You're right about measuring under load but when measuring the unloaded output voltage of a constant current supply, you are basically seeing the unregulated voltage going into the current regulator, which is running "wide open" trying to force 50 mA through infinity (or 10 Megohms or 20 Kilohms per volt, or whatever your voltmeter resistance is).

I don't know about the 03502A but generally, the two sources in the Classic charger are a constant voltage supply to power the calculator while the battery charges (usually seems to be described as 4V, but 3.75 is what the nominal battery voltage is), and a constant current supply to charge the batteries - supposed to put out 50 mA regardless of the load it is connected to (within limits). In practical terms, the highest voltage across 3 NiCads while charging is about 1.5V*3=4.5V. The simplest way to get a fairly constant current is to start out with a voltage considerably higher than the voltage you will see at your load and then put a resistor in series with a value equal to the difference between the high voltage and your load voltage, divided by the constant current you want. Then as the resistance of your load changes, the series resistor is always much higher so it has a greater influence on the current flowing. Similarly, one way to generate a constant voltage is to start out with a voltage considerably higher than you need and put a transistor between that supply and your load - additional circuitry makes the transistor drop whatever voltage difference there is between the input and the load voltage. The higher the voltage across the transistor, the smaller percentage change it undergoes as the input voltage changes (the input voltage is unregulated - it is dependent on the AC line voltage and the load current). And the more constant the voltage across the transistor, the more predictable its gain and other characteristics. It is especially important that the voltage across the transistor doesn't drop to zero - then the regulator stops regulating altogether! The voltage difference that a given regulator circuit needs in order to regulate within specs is called the drop-out voltage - a term borrowed from electromagnetic relays - remember the old relay voltage regulators in cars?

The 82002A recharger starts out with a single unregulated supply of about 14V max. to produce the 4V constant voltage and the 50 mA/4.5V max constant current supply. It doesn't just use a resistor to generate the constant current, it uses two transistors and some resistors including a 13 ohm resistor that the charging current flows through, generating a small voltage drop that controls the behavior of the transistors - in a way multiplying the resistance of the small 13 ohm resistor but doing so in a way that makes the output current less dependent on the variation in the unregulated supply voltage. The extra voltage going into both regulators has the effect of improving their regulation (qualities of "headroom" and "linearity" apply) but since both load currents (calculator and battery) have to pass through devices that are dropping the extra voltage, and power being the product of voltage and current, the extra voltage represents power that is being wasted. In fact, if the input voltage stays near 14V under both loads (it probably doesn't), then approximately 10V is being dropped by the regulators and 4V is going to the loads. Without knowing just what the total of the two currents is, (that current I * 10V) watts is being wasted to generate (I * 4V) watts at the load, so the efficiency is only 4/(10+4) or 29%.

It is possible to get higher efficiency if you waste less voltage by starting with a lower unregulated voltage. It might be that the 03502A recharger was designed to be more efficient than the 82002A - smaller size is consistent with less heat dissipation due to higher efficiency. As long as the unregulated input voltage stays higher than the necessary load voltage by "a certain amount" - maybe as little as 1V for a very careful design - the regulators will work correctly (this applies to both constant voltage and constant current regulators.)

Even at these small power levels it is important to increase the efficiency because the wasted power takes the form of heat inside the recharger which shortens the life of the components. A lot of "wall wart" supplies now-a-days contain switching regulators, which are much more efficient, for this reason, although it is also justified because they have somewhat higher power ouput, like 5V at 1A, for more expensive loads, like pocket computers and digital cameras.


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