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"Ian Jackson" <> wrote in message news  ...> In message <AC3cm.163781$2>, ian field > <> writes >> >>"Ian Jackson" <> wrote in message >>news:RK8$... >>> In message <Bqfam.7029$>, 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. |
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In message <HAhcm.73167$2>, ian field
<> writes > >"Ian Jackson" <> wrote in message >news ...>> In message <AC3cm.163781$2>, ian field >> <> writes >>> >>>"Ian Jackson" <> wrote in message >>>news:RK8$... >>>> In message <Bqfam.7029$>, 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 |
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In message <t->, Zootal
<> writes > >"Ian Jackson" <> wrote in message >news:... >> In message <HAhcm.73167$2>, ian field >> <> writes >>> >>>"Ian Jackson" <> wrote in message >>>news ...>>>> In message <AC3cm.163781$2>, ian field >>>> <> writes >>>>> >>>>>"Ian Jackson" <> wrote in message >>>>>news:RK8$... >>>>>> In message <Bqfam.7029$>, 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 |
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"Ian Jackson" <> wrote in message news:... > In message <t->, Zootal > <> writes >> >>"Ian Jackson" <> wrote in message >>news:... >>> In message <HAhcm.73167$2>, ian field >>> <> writes >>>> >>>>"Ian Jackson" <> wrote in message >>>>news ...>>>>> In message <AC3cm.163781$2>, ian field >>>>> <> writes >>>>>> >>>>>>"Ian Jackson" <> wrote in >>>>>>message >>>>>>news:RK8$... >>>>>>> In message <Bqfam.7029$>, 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. |
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In message <k9ocm.127797$2>, ian field
<> writes > >"Ian Jackson" <> wrote in message >news:... >> In message <t->, Zootal >> <> writes >>> >>>"Ian Jackson" <> wrote in message >>>news:... >>>> In message <HAhcm.73167$2>, ian field >>>> <> writes >>>>> >>>>>"Ian Jackson" <> wrote in message >>>>>news ...>>>>>> In message <AC3cm.163781$2>, ian field >>>>>> <> writes >>>>>>> >>>>>>>"Ian Jackson" <> wrote in >>>>>>>message >>>>>>>news:RK8$.. . >>>>>>>> In message <Bqfam.7029$>, 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 |
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In message <>, Zootal
<> writes > >"Ian Jackson" <> wrote in message >news  ...>> In message <k9ocm.127797$2>, ian field >> <> writes >>> >>>"Ian Jackson" <> wrote in message >>>news:... >>>> In message <t->, Zootal >>>> <> writes >>>>> >>>>>"Ian Jackson" <> wrote in message >>>>>news:... >>>>>> In message <HAhcm.73167$2>, ian field >>>>>> <> writes >>>>>>> >>>>>>>"Ian Jackson" <> wrote in >>>>>>>message >>>>>>>news .. .>>>>>>>> In message <AC3cm.163781$2>, ian field >>>>>>>> <> writes >>>>>>>>> >>>>>>>>>"Ian Jackson" <> wrote in >>>>>>>>>message >>>>>>>>>news:RK8$.. . >>>>>>>>>> In message <Bqfam.7029$>, 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 > >Larger power supplies will have diodes in parallel, and they will even put >the output trannies in parallel. You can get away with it with diodes but >you have to be carefull to use the same type, preferably from the same >batch. Otherwise you end up with some saturated and others barely turned on, >and you get a high failure rate. > During 50 years 'in electronics', I can't immediately recall seeing any power supplies with diodes in parallel. I'm not saying it's never done but, if it is, there should also be some current-balancing resistance in each path. This could be low-value resistors, or even the resistance of the secondary windings of paralleled transformers which feed each set of diodes. I don't think that having to use diodes from the same batch is a very good design criterion! -- Ian |
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"Ian Jackson" <> wrote in message news:... > In message <>, Zootal > <> writes >> >>"Ian Jackson" <> wrote in message >>news ...>>> In message <k9ocm.127797$2>, ian field >>> <> writes >>>> >>>>"Ian Jackson" <> wrote in message >>>>news:... >>>>> In message <t->, Zootal >>>>> <> writes >>>>>> >>>>>>"Ian Jackson" <> wrote in >>>>>>message >>>>>>news:... >>>>>>> In message <HAhcm.73167$2>, ian field >>>>>>> <> writes >>>>>>>> >>>>>>>>"Ian Jackson" <> wrote in >>>>>>>>message >>>>>>>>news . ..>>>>>>>>> In message <AC3cm.163781$2>, ian field >>>>>>>>> <> writes >>>>>>>>>> >>>>>>>>>>"Ian Jackson" <> wrote in >>>>>>>>>>message >>>>>>>>>>news:RK8$. .. >>>>>>>>>>> In message <Bqfam.7029$>, 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 >> >>Larger power supplies will have diodes in parallel, and they will even put >>the output trannies in parallel. You can get away with it with diodes but >>you have to be carefull to use the same type, preferably from the same >>batch. Otherwise you end up with some saturated and others barely turned >>on, >>and you get a high failure rate. >> > During 50 years 'in electronics', I can't immediately recall seeing any > power supplies with diodes in parallel. I'm not saying it's never done > but, if it is, there should also be some current-balancing resistance in > each path. This could be low-value resistors, or even the resistance of > the secondary windings of paralleled transformers which feed each set of > diodes. I don't think that having to use diodes from the same batch is a > very good design criterion! > -- > Ian A company I used to work for used paralleled rectifiers in ultrasonic cleaner generators up to 1500W, I suspect reliability might not have been all it could have been but they weren't exactly dropping like flies. |
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