The Museum of HP Calculators

HP Forum Archive 17

 AC Transformer for HP-33CMessage #1 Posted by Jean-Michel on 26 Feb 2007, 4:28 p.m. Hi, I received recently one HP-33C with it's battery charger ( :-) ), but for 90-120V AC input. I live in France where the voltage is 220-240V AC, and the plugs different from the ones on the original charger. As anyone here solved this problem ? Is it possible to adapt another european charger with 10V AC output ? If so, which kind of charger ? I'm not expert in this and don't want to burn my "new" old HP ! Thanks in advance. Regards.

 Re: AC Transformer for HP-33CMessage #2 Posted by Gerson W. Barbosa on 26 Feb 2007, 5:51 p.m.,in response to message #1 by Jean-Michel Hello Jean-Michel, I guess you have an HP-82087B adapter, input 90-120 V AC, output 10V AC, 1.8 VA (this is the output power in VA). You'll need a 220 to 110 V AC transformer. This 50VA transformer below is a bit exaggerated and rather expensive. Perhaps you'll find a cheaper one (3 to 5 VA is all the power you will need). Specifications from the linked site: ```---------------------------------------------------------------- Entrée: 220-230v 50 Hz équipée d'un cordon secteur avec une fiche Européenne 2 pôles + Terre. Sortie :110-120-130 V 50 Hz équipée d'une prise femelle Type US 2 pôles + Terre Autotransfo. U.S 220/110 50-60 Hz 50W Puissance en VA ou W :50 VA Longueur (mm) :76 mm Largeur (mm) : 90 mm Hauteur (mm) : 57 mm Poids en Kg : 0.73 Kg ---------------------------------------------------------------- ``` Regards, Gerson. Edited: 26 Feb 2007, 5:57 p.m.

 Re: AC Transformer for HP-33CMessage #3 Posted by Walter B on 26 Feb 2007, 6:17 p.m.,in response to message #1 by Jean-Michel Bonjour Jean-Michel, you can split the problem in 2 parts: get a small 230V/110V transformer and look for a simple travel adaptor to let you connect your US charger with this transformer. I bought my transformer on eBay some years ago for a very reasonable price (postage was far more expensive, because this model is oversized by far), so this may be an alternative source for you, too. As Gerson mentioned, some 5VA will be sufficient. More power means more iron and will only increase the weight. Regards, Walter

 Re: AC Transformer for HP-33CMessage #4 Posted by DBAX on 26 Feb 2007, 6:23 p.m.,in response to message #1 by Jean-Michel Well, I wish you had one of the Classic hp calculators, then we could just trade. I just bought a European AC Transformer for my hp 45 and I'm in the U.S. Here's what I've found: There are "step-up/step-down" transformers that will take care of the voltage change needed. Many (most) of these are designed to work in both directions, i.e., convert from 110VAC to 220VAC and also convert from 220VAC to 110VAC. For small wattage devices they are fairly cheap. See for example the VT 100 model at http://www.voltage-converter-transformers.com/step-up-transformer.html for US\$17. Most of these have a switch that determines whether you are "stepping up" the voltage or "stepping down" as in your case. Your AC Adapter should plug right into the 110VAC side of this device.

 Re: AC Transformer for HP-33CMessage #5 Posted by Les Wright on 26 Feb 2007, 6:47 p.m.,in response to message #1 by Jean-Michel North Americans frequently purchase travel adapters so they can use their shavers and hair dryers while travelling in Europe. I actually think I have one around here. Wouldn't they be safe to use as an interface between your North American Spice adapter and your French receptacles? They should be easy to get. Les

 Re: AC Transformer for HP-33CMessage #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
R              o---------^v^v^v^v---------o     220 VAC                                 110 VAC (European plug)                        (American socket)              o--------------------------o
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
V      110 V R = ---- = -------- = 5.5 kOhm (as calculated with my trusty 41CV)      I       20 mA
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!).

 Re: AC Transformer for HP-33C - OOPSMessage #7 Posted by Dave Shaffer on 26 Feb 2007, 10:17 p.m.,in response to message #6 by Dave Shaffer Well, I though I had the
and
OK, but I guess not! I have just been doing some HTML listing, where <> are necessary. But not here, I guess, where [] seem to be needed. So, how about this for the guts of my presentation: The circuit should look something like this ``` R o---------^v^v^v^v---------o 220 VAC 110 VAC (European plug) (American socket) o--------------------------o ``` 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!).

 Re: AC Transformer for HP-33C - OOPS, againMessage #8 Posted by Dave Shaffer on 26 Feb 2007, 10:19 p.m.,in response to message #7 by Dave Shaffer Getting closer! Here's the second formula: ``` V 110 V R = ---- = -------- = 5.5 kOhm (as calculated with my trusty 41CV) I 20 mA ```

 Re: AC Transformer for HP-33C - OOPS, againMessage #9 Posted by Paul Beijer on 27 Feb 2007, 4:13 a.m.,in response to message #8 by Dave Shaffer Quote: Getting closer! Here's the second formula: ``` V 110 V R = ---- = -------- = 5.5 kOhm (as calculated with my trusty 41CV) I 20 mA ``` If you use this method you're risking destroying the calc. The voltage drop accros the resistor is current dependent. If the charge circuit inside the calculator does not draw a constant current the voltage will fluctuate. In the worst case when the charge circuit stops drawing current when the battery is full, the voltage will double. What you need is just a simple 230V to 10VAC transformer of say 3 to 5VA, which should be readily available in any electronics store.

 Re: AC Transformer for HP-33CMessage #10 Posted by Tony Duell on 27 Feb 2007, 5:15 a.m.,in response to message #1 by Jean-Michel The internal charging circuit of the Spice (HP3xE, HP3xC) and Woodstock (HP2x) is much the same, and the charger units have the same characteristics, other than the output connector. My first Spice series also came with the US charger. I cut the output cable and fitted a 2 pin connector to it (the polarity doesn't matter, the output of these chargers is AC). I did the same to a spare Woodstock charger (240V input). I could then plug the Spice cable into the Woodstock transformer unit and charge my Spice that way. Looking at the other replies, it should be OK to run the US charger off a step-down transformer (either a double-wound transformer or an autotransformer), but I'd not use a series resistor. The primary current of the charger is no way constant, you run a serious risk of applying far too much voltage to the transformer primary, damaging it and also damaging the calculator, possibly. Also watch out for 'travel converters' for heating appliances. They are not transformers, they are effectively lamp dimmer circuits (triac, etc) set to give the same power in a resistive load as you'd get if you connected said load to 110V mains. They do _NOT_ work with transformers or electronic equipment, they will do a lot of damage.

 Re: AC Transformer for HP-33CMessage #11 Posted by JLS on 2 Mar 2007, 8:34 a.m.,in response to message #10 by Tony Duell Hi Jean-Michel, just go to an electronic parts supplier (like electronique diffusion in France) and buy a 10V/230V 1.8VA transformer and a small plastic box, the only thing you have to do is solder 2 or 4 wires. I did it for my HP-34C and it costs less than 5€ ! Edited: 2 Mar 2007, 8:38 a.m.

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