Re: Ripple filter

On Fri, 13 Jan 2006 07:09:00 GMT, "John Doe" <xxx@xxxxxxx> wrote:

>I would like to use a brushless 12v (0.07a) computer fan to vent a small
>area in my RV. I tried this, but the battery charger cooked the fan after a
>while due to ripple- it could not handle the peak voltage I guess. I
>replaced the fan but placed a 2 ohm (5w) resistor in series with it, and a
>0.1mfd cap and 15v (3w) zener in parallel with it to filter and hopefully
>protect the fan- but now the 15v zener gets too hot as it "clamps off the
>ripple" (I only want it to conduct for transient protection). Fan still
>going ok though.
>
>A transistor regulator would not be good as- correct me if wrong- it would
>show a 0.7 Vce drop. This would cut the fan performance and wasting current
>for regulation is not desirable when on battery power. Would a bigger
>capacitor provide enough filtering to prevent the zener from conducting
>(0.1mfd was a guess)? The resistor needs to be as smaller (ohms) if
>possible for faster fan. Can anyone improve on this? I am pushing my
>design abilities already. Thanks for any ideas!

---
For your application, here's what I'd do: (View in Courier)

Vin Vout
/ /
CHARGER+>---[1N4001]--+--[7812]---+
+| | +|
[ ] | [FAN]
| | |
GND>------------------+----+------+

Assuming that your charger output is unsmoothed and outputs 16V
peaks, then the cap will charge up to about 15.3V.

Since the 7812 will output 12VDC for your fan, and the cap will be
charged up to 16V, the differential voltage across the regulator
will be:


dV = Vin - Vout = 16V - 12V = 4V


with a fan current of 0.7A, the power dissipated by the regulator
will be:


P = dVI = 4V * 0.07A = 0.28 watts


With a dropout voltage of 2.5V for the 7812, a 70mA load, and a
full-wave rectified, unsmoothed output from the charger at 120Hz,
the value of the capacitor will be:

I dt
C = ------
dV

where C is the capacitance of the capacitor in farads
I is the load current in amperes
dt is the period of the ripple
dV is the allowable ripple in volts.


Now, since we have 16V across the cap and the output of the
regulator is at 12V, that gives us 4V of headroom. Since the
regulator requires that its input never fall more than 2.5 above its
output, that means that we must make sure that the voltage across
the cap never go below:


Vin(min) = Vout + Vdo = 12V + 2.5V = 14.5V


Since we'll have 16V across the cap when it charges, somewhere on
the rising edge of the charger's output peaks, we have to make sure
that it never falls below 14.5V when the cap is dicharging into the
load and the charger's output isn't charging the cap and driving the
load. Since the difference between 16V and 14.5V is 1.5V, that
means that the allowable ripple voltage out of the cap is 1.5V

If we lower the stress on the cap and lower the ripple to 1V we can
solve for the capacitance like this:


I dt 0.07A * 8.3ms
C = ------ = --------------- = 581µF
dV 1V

Most common small aluminum electrolytics have a capacitance
tolerance of +/- 20%, so to make sure you get at least 581µF you
need to increase its value to 697µF.

The closest commonly available value is 820µF, which would be fine
for full-wave rectified 60Hz.

However, on the chance that the output of your charger is half-wave
rectified 60Hz, the capacitance would need to be doubled for the
same ripple, bringing the value to 697µF * 2 = 1394µF.

A 1500µF/25V Panasonic EEU-FC1E152 would work, as would anything you
might have lying around with at least that capacitance and a working
voltage somewhat greater than the peak output voltage from your
charger.

There's not much reason to worry about transients since the cap will
soak them up, but if you want to you could put a Zener in parallel
with the cap with a voltage rating equal to greater than the peak
output of the charger but less than the working voltage of the cap.

--
John Fields
Professional Circuit Designer
.

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