XPost: 24hoursupport.helpdesk   
   From: ianREMOVETHISjackson@g3ohx.demon.co.uk   
      
   In message , ian field   
    writes   
   >   
   >"Ian Jackson"  wrote in message   
   >news:4Lvlz7KQMdcKFwkC@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.   
   >   
   >Actually if you have two strings each having the same number of diodes and   
   >put the two strings in  parallel it doesn't matter, when you have a few or   
   >more diodes in series as the variation in Vf for each diode averages out.   
   >   
   Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You   
   don't see many circuits with paralleled diodes (stringed or otherwise).   
   --   
   Ian   
      
   --- SoupGate-Win32 v1.05   
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