Help with power filtering in high current circuit

[JP19]

Hello 

I have a question relating to power filtering. I'm not really sure the best way to go about it in my situation.

I've got a circuit (multiple circuits joined together actually) that draws quite a lot of current (up to a maximum of about 280mA).

I understand there isn't much I can do to filter the power from 9v as an rc filter is going to drop too much voltage.

So I thought I'd run it off a 15v power supply so I can afford to lose some volts to an RC filter.

I've currently got 2 x 6.8ohm resistors in series on the power supply giving me 13.6ohms, and 377uF of capacitors (a 330uF and 47uf) which if my calculations are correct gives me a cutoff of 31hz. The resistors are dropping the voltage a bit (down to about 12.5 volts, I can't remember exactly) but it's working well and the circuit seems to be happy running at that voltage. Happier than it was at 9v without any filtering in fact. Bonus extra headroom I guess.

The issue I have is that the resistors are getting very hot. I'm using 1W resistors but maybe I need to go bigger.

Is this the best way of filtering the power in my circuit if I want to run off the 15v supply? Can I just put some 2 or 3 watt resistors in there instead and be done with it? How hot is too hot for them to be running? Will higher rated resistors get as hot?

Or should I use a 12v regulator instead (but then how do I get my filtering?)?

Or is there a better method?


To summarise I'd like to filter the power of a circuit that draws 280mA. I have a 15v supply and would ideally like to run the circuit at about 12v. Any help much appreciated!

[antonis]

Each 6.8R resistor dissipates 0.5W so it's normal to get hot.. In case of getting very hot, probaly current draw is more than 280mA..
(you can verify it by measuring voltage drop on one of resistors..)

You can use 2W resistors, or split each of 6.8R to 3.3R + 3.3R 1W each (in case of enough space availiability..)
(solder them with a some space left between their body and board..)

IMHO, 31Hz cut-off frequency is high, although 100/120 Hz rectified ripple is well above.. (about -12/-15 dB)
Better use 2 x 330μF or even bigger value cap(s) (but don't overdo it 'cause you'll need recifier rated for high surge current) and a 100nF ceramic..

With 12V regulator (fixed or adjustable) you'll need much less filtering (if any..)
Using 1A or 1.5A item, power dissipation should be lower than 1W, so you might even use it without heatsink (maybe mounted on metal enclosure..)

[PRR]

> How hot is too hot for them to be running?

Modern "power" resistors can burn flesh or sear steak without harm. (To the resistor.)

> Will higher rated resistors get as hot?

There's two ways to get a higher rating. BIGGER part, or tougher material. Modern fashion is for small parts. So they refine the ceramic so it can run red-hot without shattering. This may burn your PCB, your adjacent parts, or you.... but the resistor is fine with that.

As said, you have maybe a fat half-Watt. When you buy, you want to at least double that. So two 1W parts "should" be ample for a half-Watt waste.

[JP19]

Ok great, thanks for getting back to me,

So what do you think would be the 'best' option?

Do I:

1) Put some 2w resistors in there, spaced a bit away from the board, and put 2 x 330uF caps?

or

2) Put a 12v regulator in there?



I'm leaning towards the regulator option as I'm guessing it might get less hot? (correct me if I'm wrong)

And if I do go for a regulator, can I just stick a big (how big?) capacitor and a 100nf to ground after it or is there more to it than that?

Thanks!

(sorry for the dumb questions!)

[PRR]

The regulator will run insignificantly warmer.

Do you actually have a power noise problem? Or are you just looking for busy-work?

[Rob Strand]

This type of circuit can provide ripple filtering without regulation.  It keeps the power dissipation low.
https://www.electronics-notes.com/articles/analogue_circuits/transistor/capacitance-multiplier-circuit.php

For large amounts of ripple it can be used but has a very undesirable characteristic that you need to choose R1 large enough and that causes the design to depend on the transistor gain.  R1 is chosen large enough so the voltage drop ensure the transistor doesn't "clip" at the bottom of the ripple.  So the way around the beta dependency is to add another resistor from the base to ground, which forms a divider.   Nonetheless you still need to decide how much voltage to drop to prevent the transistor from clipping.   The scenario also means you need more voltage drop across the transistor and it will get a bit hotter.    In short, it can be made to work but brings up issues you can usually ignore when you just want to removed a small amount of ripple.   For those reasons large amounts of ripple are  probably best handled with a regulator.

The issue I have is that the resistors are getting very hot. I'm using 1W resistors but maybe I need to go bigger.

If the input voltage is fixed, the output voltage is fixed, and the current is fixed, it doesn't matter what you use to perform the voltage drop the amount of power dissipation will be the same:   P = I*(Vin - Vout).  .   You can only use bigger parts to reduce the *temperature*.  A 5W resistor will run a bit cooler than the 1W and it will handle a higher temperature anyway.    With a transistor or regulator the temperature is reduced by choosing a larger heatsink.

A switch mode supply is not linear it reduces power dissipation by converting power.   The heat dissipated is due to circuit losses not due to physics.

[JP19]

Do you actually have a power noise problem? Or are you just looking for busy-work?

This is actually a good point!

I was having some noise issues with the 9v supply (which may have just been down to a crappy power supply) so I got the 15v one so I could afford to lose some voltage to filtering.... but I haven't actually tried running the 15v one without any resistors on the power input. Maybe I can get away running the circuit at 15v (without a resistor on the power line) rather than 12v.. 

[antonis]

You might be but series resistor doesn't only drops voltage..
It forms (with smoothing capacitor) a LPF so it attenuates rectified voltage ripple (of frequecy the double the mains, in case of full-wave rectification..)

A capacitor, by its own, forms LPF with secondary winding impedance plus rectifier diode(s) dynamic resistance, the addition of both being relatively small..
(that's one of the reasons for high ripple voltage even in high value smoothing cap power supplies..)

Another useful "advandage" of series resistor is the prevention of high surge current (inrush) when power ON.
An "empty" big electro is actually a short between its legs, so initial current is only restricted by its internal resistance (almost zero for uncharged cap) and secondary winding impedance (also almost zero..)
In case of transformer high current capability, surge current could be MUCH higher than normal (working) resulting either into stretched rectifiers or blown fuse (or both)..

Now about power: as Rob said, power dissipation is fixed for a fixed voltage drop (P=V*I) despite the way of occuring..
(series resistor, series transistor, series regulator, etc) so this fixed power has to be dissipated via equivalent amount of heat..
For 3V drop and 280mA current 840mV must dissipated either by a single item or any combination of more than one..