What is the best method for dropping the voltage of a device that has a range of 8 to 12V powered by a car battery(so max, say 14V) to bring it within spec? The device uses from 100uA to 3mA depending on state.I was thinking I probably could just use a diode or two and that should be fine but it would take about 4-6 to do it though and that assumes the battery is fully charged.Seems like a regulator is overkill though(as I do not need regulation). I pretty much want the cheapest solution that will work.I'm thinking of a simple zener regulator, say at 9V with a ~700 ohm current limiting resistor but this doesn't seem to work all that well over the whole supply range.Basically I'd like something that will drop 2V when the supply is at about 14V and 0V when its at 8V. I imagine this is to much trouble and probably I won't find a better solution than just two diodes in series? Although this won't let me run the device if the voltage drops down past 10V ;/).The biggest problem is that with low current draw(100uA) and high voltage(14V) it will drop, say, 1V, but at high current draw(3mA) and low voltage(8V) it will drop 2V ;/ So seems like I can't have my cake an eat it too ;/The zener regulator also has a similar problem unless I want to waste a lot of power ;/Any other methods I should look into that won't be to expensive? I thought about DC-DC converters but I'd like to avoid the additional cost and all the extra components.(essentially the zener diode will work if I could somehow remove it from the circuit as the supply voltage decreased. Maybe a transistor could accomplish this?)Thanks,Jon

Devices should generally be protected up to 16 volts in that invironment. Somecharging specs go up to 15. 5 volts, but just over 14 would be more common. I douse diodes many times. How about diodes and a zener?greg

That must have been a 6 volt system, or possibly the voltage at the starter terminals. Of course, in 1908 many (most/all?) cars had a manual crank, and probably a magneto ignition, with no battery.The lead-acid battery was invented in 1859 The first practical electric starter was invented in 1911 The first battery powered ignition was introduced in 1910 A simple way to avoid the effect of low voltage during cranking, in this case, would be to add a diode and a fairly large storage capacitor. 10,000 uF should give you at least 10 seconds before the capacitor voltage dropped below 8 volts (starting from 12 V).Paul

Best method for dropping supply voltage a volt or two for low power device?

Apr 22, 2008 26 Replies

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Paul E. Schoen 18 years ago

That must have been a 6 volt system, or possibly the voltage at the starter terminals. Of course, in 1908 many (most/all?) cars had a manual crank, and probably a magneto ignition, with no battery.

The lead-acid battery was invented in 1859

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The first practical electric starter was invented in 1911

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The first battery powered ignition was introduced in 1910

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A simple way to avoid the effect of low voltage during cranking, in this case, would be to add a diode and a fairly large storage capacitor. 10,000 uF should give you at least 10 seconds before the capacitor voltage dropped below 8 volts (starting from 12 V).

Paul

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Jim Thompson 18 years ago

I tend toward things of this sort...

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scaled to larger voltages, of course.

...Jim Thompson

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legg 18 years ago

good one, but high bias. Swapping in a zener and lowering 5V6 to 2V7 gives submilliamp bias and 'no' headroom.

. Vin . | . .----+----------. . | | | . | | | . | | | . [10K] | [47K] . | | | . | |< | . +--| | . | |\\ | . | | | . | '---+------|-----+-> out . | | | | . | _ | | . \\| ^ 6V2 | [2K7] (3mA load) . |------|------+ | .

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Fred Bloggs 18 years ago

Yep- that's the only way you're going to get LDO reverse protection and deal with heavy loads during load dumps, but you can't have the thing burping every time an overvoltage short duration transient hits, and you can't use 1000V MOSFETs, so some kind of TVS is necessary...

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Fred Bloggs 18 years ago

If you're satisfied with the regulated output being somewhere in the

8-12V , probably more on the low side, then that will do it...my SPICE shows your output at 7.2V with 500mW zeners...

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John Larkin 18 years ago

Hey, I have a brilliant idea: use a voltage regulator!

John

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John Fields 18 years ago

I think you\'re right. I was thinking about it in the shower earlier (for me, random noise is a place where some great ideas come from) and it seems to make a lot more sense to open up the circuit when the voltage rises past some threshold, except that if you open it up too fast you\'ve got that pesky E = Ldi/dt waiting to punch through the switch. So, it seems to me, in order to keep that from happening there needs to be a Q spoiler in there somewhere which will keep the voltage from rising too high. Plus, in this application, there\'s the drop across the switch. Plus, we don\'t even know if it\'s an automotive application since all the OP said was that he wanted to run it from a 12 volt battery.

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