As discussed in the design group thread, a low dropout regulator looks good.
> Path: eternal-september.org
> From: amdx <am...@knology.net
> Newsgroups: sci.electronics.basics
> Subject: Re: Use resistor instead of regulator?
> Date: Wed, 10 Feb 2021 19:08:48 -0600
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> On 2/1/2021 5:26 AM, John Doe wrote:
>> Rich <ri...@example.invalid> wrote:
>>> John Doe wrote:
>>>> Rich wrote:
>>>>> John Doe wrote:
>>>>>> I can put a resistor after a voltage that's too high for the circuit?
>>>>>> Of course the resistor must handle the current.
>>>>>> A decoupling capacitor after the resistor.
>>>>>> Any problems with that?
>>>>> If your load has a variable current draw, then you will also have a
>>>>> variable voltage on the load side of the resistor.
>>>> If it's a big capacitor, the voltage variation is minimal?
>>> No, the size of the capacitor simply determines the rate at which the
>>> voltage changes.
>> Then your answer should be "Yes", given the same amount of time.
>> Why are you fighting this?
>> Do you think it makes you look smart?
> ¶ÿNow you have made yourself look ignorant, which you are, otherwise you
> would not have ask the question.
> ¶ÿSay your device needs 9V and you have a 12V source. At some point in
> time you measure the current your device draws and it is
> 20ma, so you decide you need a series 150 ohm resistor.¶ÿ You have it all
> figured out. Then at some point it fails to work.
> At some point the motor in this device (the one you didn't explain )
> turns on and the voltage drops to 4 volts causing the circuit to fail to
> ¶ÿDo you want to calculate the capacitance needed to keep you supply
> voltage within 5% of your required 9v?
> Or do you want to just admit it is a big ass cap and a regulator may be
> On the other hand you could give us details of the circuit, so it would
> not be a guessing game.
> ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ¶ÿ Mikek