Voltage multipilier and 50 V capacitor for 12 V DC

Hi folks, Here is my problem that I believe will have an easy solution, but as a chemist I cannot come up with one myself.

We are dealing with 12 V DC system (car). I have a electric pressure gauge that is connected to "hot in ACC" +12 V. In case, "hot in ACC" means the wire gets powered only with ignition on and the power supply is not interrupted during cranking (some circuits are cut off to limit load to the battery).

How it works. The moment I turn ignition on - the gauge sweeps from 0-100% and returns to current reading (zero for now). When the engine cranks pressure is being built up and gauge will show the reading. It works just fine most of the time. However, from time to time the voltage drops low enough for the gauge to reset. I am not sure how low it gets, but I would presume below 11 V. I could not find specs for it.

Hence I have been thinking of a solution to not allow the gauge to reset.

My thoughts were anywhere from a small cylindrical 12 V battery with a diode or zener diode, but I could not find any small enough. Then I thought to use 25 V capacitor, but the problem I see would be limited charge - it will reach 14.5 V at the max (but will discharge after a while). Hence, I would need a large one... but then it gets big.

So finally I thought - there must be a way to multiply voltage in order to charge, say 50 V capacitor, which would drain through a gate that would be set at 11 V or so. I hope that makes sense.

Here is a circuit, something showing my thoughts... Possibly very wrong, but the best I could come up with.

I do not want the whole setup to be big. The smaller foot print, the better.

Any help appreciated. cheers,

by PLP
December 30, 2018

What current does the gauge take? Is it actually 1A to 1.45A as shown? How long could you expect to be cranking on a cold day?

by mikerogerswsm
December 30, 2018

Correction: 0.1A to 0.145A

by mikerogerswsm
December 31, 2018

IThanks for your quick response.

The cranking takes no more than 2-3 seconds in the worst case scenario. We could assume 4 as the max time.

As for the current draw - I did not measure it yet. I estimated it to be no more than 0.1 A. My reasoning is - how much power you would need for this small motor and some controlling board? I understand it will make a difference for the capacitor discharge time and I will try to get the answer soon.

by PLP
December 31, 2018

Oops. I think by removing answer (I should have posted a comment, sorry, my bad) I deleted a comment from other user.

Is there a way to bring it back?

But in order to address his/her statement - no, I am not excluding any scenario/solution. I did not even think of adding SMPS... as I did not even think of it.

I thought of a capacitor. That is my limited knowledge here. My very first thought was a small SLA battery that I would add there with a few diodes and a relay to not discharge it while cranking and to not power up the gauge when car is off.

As for SMPS - I am not sure how low it would go. I am not sure if power gets interrupted how long it would sustain power.

And finally, I did not know one can get 12 V to 12 V power supply...

I am open for any solution. The easier the better.

by PLP
December 31, 2018

Oh, and why 50 V capacitor? To keep it smaller (less expensive) and get more power out of it. I used some online capacitor discharge calculators and I came up with about 50 V, 10 mF as enough charge to supply current for about 4 seconds.

by PLP
December 31, 2018

It is quite easy to make a buck/boost SMPS, that is to say one which will reduce or increase the input voltage as necessary.

Because it runs at high frequency it uses very small components.

by mikerogerswsm
December 31, 2018

So something like that would work? https://www.amazon.com/eBoot-Converter-Voltage-Adjustable-Step-up-x/dp/B06XWSV89D/ref=sr_1_3?ie=UTF8&qid=1546275906&sr=8-3&keywords=dc+step+up+voltage+converter

Even if voltage drops to say, 8 V (God forbid - that means I got some other issues!!) it would still regulate back to set value, right?

by PLP
December 31, 2018

Yes, that looks right.

Here's some more information:


by mikerogerswsm
December 31, 2018

Wow, 40 pages of specs... :) But I was going one step further. This piece would get my low voltage problem solved. Sounds good.

But once the car is up and running, voltage would jump to about 14.2 V. When I set the module to step up voltage to 12 V to prevent shutdown, how will it behave once the input is higher then the output?

A regulator would be more in place... Correct me if I am wrong.

Or I bypass it by using a zener diode or so.

by PLP
December 31, 2018

Well spotted!

You will need buck/boost:


by mikerogerswsm
December 31, 2018

Another boost/buck:


by mikerogerswsm
January 01, 2019

I ordered one. I will update how it works once I receive it. I might actually expand it to power other items.

Anyway, if not the regulator, would my initial idea be a reasonable one? Or even doable? It would work similarly to the boost/buck with "battery backup" due to the capacitor...

by PLP
January 02, 2019

As you ask again for comment on your original circuit I've taken the liberty of redrawiing it:

Yes, if completed with a pump signal and voltage multiplier chain this unconventional circuit might work. But you might want to add regulation and protection.

by mikerogerswsm
January 02, 2019

Just for additional background you might use a Dickson voltage multiplier:


by mikerogerswsm
January 03, 2019

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