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5VDC 2A Power Supply for Vehicle Electronics

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bradosaurus

New Member
I am building an in-dash USB power supply for my personal electronics in my car. The power supply on the vehicle is dirty (meaning AC leakage/RF leakage/ground loop interference), and I would like to clean up the power supply.

Drawn in the picture is what I've come up with so far.

9Ew0v-1.jpg


The diodes d1 and d2 will be on the primary 14.1 V supply (to clean up stray AC). The output (to USB connector) should be 5 VDC from the Zener diode, d3.

I am not very experienced with DC electronic circuits, which is why I'm asking for help.

Does this system seem reasonable? The capacitors (C1 and C2) and inductor (L) are (to my understanding) there to smooth the waveform and reduce the ripples, and the Zener diode and resistor are there to reduce the voltage from 12 to 5.

I would really like to truly understand the *why* of selecting the correct diodes, capacitors, etc., but any help I can get with this would be appreciated.

What specific diodes should I buy for D1 and D2?
I can't find >250mOhm capacitors in 12V. Do I need larger capacitors for a 2A system?
What inductor should I get? (and why, if you feel like sharing)
What resistor should I get? (I somehow calculated 4 Ohm, 14000mW)
What Zener should I get?

What effect will the power drop from each of the components have? I would like to target 5VDC output as best as I can.

Thank you very much for your help!
 

Diver300

Well-Known Member
Most Helpful Member
A zener regulated supply is very simple, and very robust but will generate a lot of heat. It will take far too much power to be left turned on when the engine is stopped.

The resistor should be smaller than 4 ohms if you want 2 A with the engine off. You should also use a larger power rating of the resistor if isn't going to overheat when the engine is running and the supply rises to 14 V or so. I would suggest at least a 25 W resistor, and a 50 W one would not run so hot.

https://uk.farnell.com/2009343

I can't find a suitable zener diode, but you would be looking at a stud mounted diode on a heat sink.

For capacitors, the value is measured in Farads, so I don't understand why you are looking for resistance larger than a 0.25 Ohm. Normally the ESR, equivalent series resistance, of a capacitor is better if it is smaller.

The capacitor and inductor values depend on how well you need to filter noise, which isn't well defined. You may have to try it and see.
 

bradosaurus

New Member
Thanks, Diver300. I couldn't find an appropriate zener diode either. Would it be necessary to have multiple zener's in parallel to achieve desired load? Or perhaps zeners in series to step down the voltage?

As for the capacitor, I mis-wrote my units. I can't seem to find a capacitor greater than 250 mF for a 12V system. Can I use a capacitor rated higher than 12V? Say, for example, 100V? (I've read up on capacitor theory, I just don't understand all of the technicals).

I haven't measured the noise/ripple in the incoming power, but it is certainly registering a varying AC voltage on my Fluke (especially noticeable at higher RPMs, thus more AC leakage [side note-this is a brand-new alternator with a new voltage regulator]). I don't think it's a large enough ripple to call for any particularly specific inductor, so any recommendation/ballpark size inductor would be helpful (I don't understand milli-Henrys nor inductor theory, not for lack of trying).

The 14.1 incoming voltage is from the charging current on my vehicle (at engine idle). Engine-off voltage is 13.2.

Ideas?
 
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Reloadron

Well-Known Member
Most Helpful Member
Given a choice for two amps I would use a simple 3 amp LM323 5 Volt regulator. Fixed regulator mounted on a heat sink with an input cap of .1 uF and maybe 1,000 uF and the same on the output. This is the LM323 regulator. The problem with trying to do this with a zener diode is for your 2 amp current requirement you will need at least a 10 watt with greater preferred stud mount diode that will really need heat sinking. The series resistor would be dropping 14 - 5 volts = 9 volts or so at 2 amps so the resistor should be about 20 watts min and in reality about a 40 watt 4.5 Ohm resistor which is not easy to find off the shelf. It will be large and run hot.

I am not saying that a zener diode solution won't work, I am saying I do not see it as a practical or viable solution to get where you want to go.

As to the capacitors you want a higher working voltage cap than the voltage you plan to run at. So for example at 14 VDC I would look for a cap rated at 25 or better 50 VDC. I would build a circuit around the examples in the data sheet I linked to. I have used 1 amp LM7805 regulators in automotive applications with sensitive devices and they worked fine. The LM323 should work fine providing more current than you need with less cooling problems.

Just My Take
Ron
 

bradosaurus

New Member
This makes a lot of sense. I could use this voltage regulator in place of the zener diode. What about the inductor? Could a capacitor on either side of the regulator suffice? I know it's a tough question to answer since it is completely dependent on unknown parameters; but I'm not looking for perfection in the voltage supply, just no AC and a fairly smooth voltage supply.
 

Boncuk

New Member
Hi bradosaurus,

there are 78S05 voltage regulators on the market. They deliver 5V at 2A at their outputs.

I do not recommend to clean the dirty DC using "normal" diodes, but use TRANSIL diodes rated for the necessary input voltage of the regulator which is 14.4V at max.

A TRANSIL diode "kills" any voltage beyond it's rated break through value taking spikes of 82A without any problem.

I recommend an 1,5KE-13CA, a bidirectional TRANSIL diode with a break down voltage of 12.4 - 13.7V (min and max) with a peak pulse current of 82A, clamping voltage 18.2V.

That way you'll have an almost clean DC on the input of the voltage regulator. Using a 2.200µF electrolytic cap rated 40V at the regulator input nothing can go wrong.

Here is the datasheet of 1,5KEXX diodes.

Boncuk
 

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Diver300

Well-Known Member
Most Helpful Member
As for the capacitor, I mis-wrote my units. I can't seem to find a capacitor greater than 250 mF for a 12V system. Can I use a capacitor rated higher than 12V? Say, for example, 100V? (I've read up on capacitor theory, I just don't understand all of the technicals).

You should use a capacitor rated higher than 12 V. The rating is the maximum that it should ever be subjected to and 25 V rating or more would be my choice for a car system.

Capacitors are usually rated in µF, not mF, so do you mean 250,000 µF? That is very large and I don't know why you would need that. It might be needed in a mains powered system where the mains drops to zero twice each mains cycle and the capacitor has to keep the DC going. However in a car system there may be noise and spikes, but there are no gaps in the power.

Of course, when the engine is starting, the voltage can drop to 6 V or so for 5 seconds, and no capacitor that you could fit inside the car is going to ride that.
 

Diver300

Well-Known Member
Most Helpful Member
There are several automotive rated voltage regulators. They are designed to withstand the big voltage surges that you can get if a "load dump" happens.

If you are running a linear regulator with 5 V output, 14 V input and a 2 A load it will produce 18 W of heat, so a large heatsink is required. However, that large heatsink can be part of the car that is already there. There are often metal brackets behind the dashboard that are large enough to act as as good heatsink.
 

MrDEB

Well-Known Member
Where does on obtain a 7805 that outputs 2A!!
For automotive use a LDO regulator is the better choice. An autos electrical system has LOTS of noise.There are regulators made specifically for the automotive use. Have seen posts on this forum that have used 7805's and after a month or so it's history.
Not sure but look for a 3-5A LDO regulator?
 

Reloadron

Well-Known Member
Most Helpful Member
The 78S series of regulators are rated for 2 amps. Example can be found here. The problem is that it is not a LDO type regulator. Most of the 5 volt LDO regulators do not deliver the wanted current of 2 amps. The LM2940 is a good LDO 5 volt regulator well suited for automotive applications but only good for about 1 amp. There are some higher end regulators and DC / DC converters out there that would work but as to keeping things simple and inexpensive with a low parts count I would just go with a LM123 / LM323 type.

As to load dump? The possibility always exist. Just a matter of wanting to design around it or not. There are chips like the LT3692 that deliver 3.5 amps but the parts count and complexity of the circuit seem to grow exponentially, an outline can be found here.

I have used as I mentioned LM7805 chips in automotive and they worked fine for me. I had one in an old '78 Pontiac Bonneville for a decade. Then too, that was on the accessory side and not subject to cranking voltage drops.

Ron
 

crutschow

Well-Known Member
Most Helpful Member
............................
For automotive use a LDO regulator is the better choice.
.................
For 5V output from a 12V battery an LDO regulator has no advantage. The 2-3V dropout voltage of a standard regulator is still much less than the 7 plus volts difference between 5V and 12V.

Perhaps you were thinking of trying to provide regulated 12V from a car which would indeed require an LDO regulator.
 

bradosaurus

New Member
I'm trying to keep up with the discussion in between running around for the holiday. Thanks everyone for the great input.

This is what I've come up with so far:

uqMsR-1.jpg


The general diode is 1N5402. How does this design fare in the discussion? Will a simple heat sink suffice?

So far I've been able to target all of these parts for less than $5.
 
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Reloadron

Well-Known Member
Most Helpful Member
You want to get a .33 uF and a .1 uF cap on the input and output respectively as shown in the data sheet for the 78S series regulators. Leave the existing but add those small caps for the HF noise.

Ron
 

Diver300

Well-Known Member
Most Helpful Member
You will need to get rid of around 15 W of heat from the regulator when running at the full 2 A. You need a heatsink rated to around 4 °C per W or less, unless you can use part of the car.

I would suggest that you add a 2 Ω resistor, rated to 10 W or more, on the input. That will reduce the heatsink needed and it will reduce the current that the Transil has to absorb if there are any spikes.
 

Boncuk

New Member
Hi Diver300,

the TRANSIL will not serve its purpose well when connected via a resistor. It must be the first part in an automotive circuit to eliminate spikes efficiently.

The 1,5KE-13CA can stand pulse currents of 82A, more than the car alternator is able to produce.

If the 1,5KE-13CA's rating is too low for the specific alternator he might use a 1,5KE-15CA, which should be perfect for the dirtiest DC an alternator can supply.

Using a heat sink rated 0.7K/W the OP will have a relatively small, but dependable power supply for the car.

Your calculation for heat dissipation is certainly the worst case. When taking all effects into consideration I come up with a heat dissipation of 8 to 10W.

Boncuk
 

Diver300

Well-Known Member
Most Helpful Member
I meant to put the resistor before the 1,5KE-13CA, but leave the 1,5KE-13CA protecting the rest of the circuit. The resistor would be in series with the top left wire in the circuit.

That will stop the 1,5KE-13CA from protecting anything else in the car, but will provide better protection for the 5 V regulator.

Load dump, if it happens, can continue for some time. It can also be repeated if there is a bad connection that is connecting and disconnecting as things move. The specification sheet for the 1,5KE-13CA shows that it can absorb 2 kW for 10 ms, or 5 W continuously. Well a load dump is around 1 kW and will decay to zero over 100 - 500 ms, and that is a lot of energy to absorb, and I think that it is above what the 1,5KE-13CA can manage, but the graph doesn't extend beyond 10 ms.

Suppressors like that are excellent for absorbing inductive spikes that are of short duration, but load dump goes on much longer.

I once repaired a car where the engine to body earth wire had broken, but had not fallen away, so it kept making and breaking the connection. The car finally stopped when the protection diode in the engine management computer shorted out, and the 5A fuse feeding it blew. It made me realise that load dump protection is very hard to do, and adding resistance means that a supressor has a lot less current to absorb.
 

bradosaurus

New Member
The sizing of the power supply (2A) is based on a 500mA continuous load system, with 2A peak load capacity. The supply will go to 2 USB ports, one of which will power my power-hungry phone/GPS, the other for general use electronics (girlfriend's iPod, etc.).

I like the idea of load-dump protection. Integrating that into the system, as well as the small caps for high-frequency noise reduction, this is what I have:

36T6d-1.jpg


EDIT: the image shows the caps as mF. The units should be uF.
 
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