XPost: 24hoursupport.helpdesk   
   From: ianREMOVETHISjackson@g3ohx.demon.co.uk   
      
   In message , Zootal   
    writes   
   >   
   >"Ian Jackson"  wrote in message   
   >news:GvAqKUAuZbcKFwgH@g3ohx.demon.co.uk...   
   >> In message , ian field   
   >>  writes   
   >>>   
   >>>"Ian Jackson"  wrote in message   
   >>>news:oQqRlSBFmUcKFw3i@g3ohx.demon.co.uk...   
   >>>> In message , ian field   
   >>>>  writes   
   >>>>>   
   >>>>>"Ian Jackson"  wrote in message   
   >>>>>news:RK8$DzIdHMcKFwRq@g3ohx.demon.co.uk...   
   >>>>>> In message , rf   
   >>>>>>  writes   
   >>>>>>>Roger Dewhurst wrote:   
   >>>>>>>>> You can get simple to use regulator chips that drop the voltage   
   >>>>>>>>> down, you only need a couple of components to make a working   
   >>>>>>>>> voltage   
   >>>>>>>>> regulator. Ask on News:sci.electronics.basic - the folk on there   
   >>>>>>>>> are   
   >>>>>>>>> usually very helpful and should explain all you need.   
   >>>>>>>>>   
   >>>>>>>>>   
   >>>>>>>> Why not just drop the voltage through a few diodes?  Very simple.   
   >>>>>>>> Very   
   >>>>>>>> cheap.   
   >>>>>>>   
   >>>>>>>A *few* diodes at a couple of ten cents per each. A single 7809 for   
   >>>>>>>less   
   >>>>>>>that a $.   
   >>>>>>>   
   >>>>>> More like "a *few* diodes at a couple of cents per each".   
   >>>>>>   
   >>>>>> 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required.   
   >>>>>> More   
   >>>>>> than good enough for the job.   
   >>>>>> --   
   >>>>>> Ian   
   >>>>>   
   >>>>>The forward conduction knee curve on diodes isn't *that* sharp,   
   >>>>>depending   
   >>>>>on   
   >>>>>current draw and rating of the diode the drop can be as low as 0.55V and   
   >>>>>as   
   >>>>>high as 1.1V.   
   >>>>>   
   >>>> For most 'normal' Si diodes, that isn't really the case. The actual   
   >>>> voltage drop does, of course, increase with current, but at 'sensible'   
   >>>> currents, you can reckon on around 0.65V per diode. How much current is   
   >>>> the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should   
   >>>> work fine in this application. I've used this non-elegant 'KISS'   
   >>>> technique   
   >>>> on several occasions, and haven't found any problems.   
   >>>> --   
   >>>> Ian   
   >>>   
   >>>A potential danger with a cassette recorder is the difference in current   
   >>>draw between motor on and motor off. In the condition of low current draw   
   >>>(and low diode drop) supply decoupling electrolytic capacitors can charge   
   >>>to   
   >>>a higher voltage which is then dumped into the circuit when switched to   
   >>>play.   
   >>>   
   >> True, true. But I reckon that a momentary short burst of a   
   >> rapidly-decaying additional 3V won't hurt too much.   
   >> --   
   >> Ian   
   >   
   >It won't even be noticeable. The capacitors won't charge up that high to   
   >start with, and they don't "dump" into the circuit, they just quickly   
   >discharge down to the lower voltage that is present at the output of the   
   >last diode - how fast this happens depends on the size of the caps and the   
   >load. I wouldn't even call it a surge. A resistor from the last diode to   
   >ground will prevent them from charging more then a few tenths of a volt and   
   >is a good idea.   
      
   Indeed. A bleed of a few mA will prevent the off-load voltage rising to   
   12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W   
   will do).   
      
   >And the capacitor doesn't need to be that big.   
   >   
   Which capacitor do you mean?   
      
   >I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible   
   >to failure. Use 2 or 3 amp diodes, they are cheap and readily available.   
   >They are bigger and will run cooler and won't fail as easily.   
      
   1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will   
   run a bit warm, so maybe a physically larger (higher current) diode   
   might be better. But it depends on how much current the Tardis takes!   
      
   >Or use two   
   >strings of 1n400x in series, that is good enough for an app like this.   
   >   
   >So - 12v -> diode -> diode -> diode -> diode -> ~9.2v Output   
   >   
   That's only one string of diodes in series. Did you mean parallel? If   
   so, no, you shouldn't parallel diodes. As you suggest, use higher   
   current diodes.   
      
   >From the output run a 470 ohm half watt resistor through a standard 1/4" LED   
   >to ground (I like the LED so you can see when the circuit is on). In   
   >parallel to this connect a capacitor, say 100uF. Presto, ~9.2V, adequately   
   >regulated, and minimal voltage increase when the load is off.   
   >   
   The cap isn't a bad idea, but 470 ohms will give you around 14mA through   
   the LED (allowing 2V for the LED). Anything between 470 ohms and 1k   
   should be fine.   
   --   
   Ian   
      
   --- SoupGate-Win32 v1.05   
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