Re: Newbie: Zerner diode voltage regulator ?

I'm self teaching myself electronics

Hmm... this is quite different then what I had envisioned happening
with the no current passing thru the zener until breakdown. I'm still
trying to get a grasp on what's happening on all the interactions
between components here. I wish there was an educational animation type
program that showed you in SLOW motion exactly how the current flowed,
voltage, reaction of components etc. Now that would be a great learning
tool. Anyway, I'll go pound my head against the wall a few more
times and I'm sure I will finally pound it in there. :)

Thanks everyone for their response.

J



Chris wrote:
hdjim69 wrote:
Hi, I'm self teaching myself electronics as a hobby so I don't have
an instructor to ask simple questions of so please forgive me if this
has been explained before.

Regarding using Zerner diodes in reverse bias configuration as a
voltage regulator, I just want to make sure I got this right. Here is
a link to HyperPhysics website displaying a simple voltage regulator
circuit using a Zerner diode:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/zenereg.html

So, let's say the load needs 12v to operate properly and the
Zener's breakdown voltage is 13v. So when the unregulated power
supply's voltage spikes at 13v, are we saying the Zener will fail
"shorted" ? That is, conducting full current flow just like a
piece of wire ? Kinda like a dam failing when too much water pressure
is applied to it ? If so, then when this happens the Zerner is now in
parallel to the load and we know that adding a parallel branch does not
effect voltage or current in the other branches since all branches are
connected directly to the voltage source and the current through that
branch is determined by it's resistance, so the new "closed"
Zener branch has no effect on the load branch. But the key to all this
is the dropping resistor connected in series with both the Zener & load
branch. Now that there's extra current flowing through the Zener
branch, this current will now have to flow through the dropping
resistor too which will now drop more voltage (Vd = I * R thru the
component). So for all practical purposes the circuit is really the
load in series with the dropping resistor. And the Zener just acts
like a control mechanism to the dropping resistor.

Do I have this right ? Is this how Zerner diodes are used in this
configuration ?

J

As mentioned by others, the zener diode is a shunt regulator -- that
is, it shares the current coming from R with R(l). As load current
decreases, the voltage at the node will rise. But as the voltage
rises, the zener will instantly take more current, keeping the voltage
fairly constant.

And if the load current increases, the voltage at that node will start
to go down. But the zener will then instantly take les current, and
the voltage again stays fairly constant. This effect is called shunt
regulation -- current is shunted away from the load in such a way to
keep voltage constant.

Without getting into details, zener regulators are holdbacks from the
1960s. A shunt regulator wastes the most power when there's minimum
load current. If there's variation in load current or input voltage,
you end up having to waste a lot of power to keep the shunt regulator
within regulation. And their ability to regulate a changing load isn't
anywhere near as good as a standard 3-terminal regulator like the
LM7812, which uses an internal diode fed by a constant current source
for its internal voltage reference. In short, I don't believe there
are any advantages to using a 1N4742 with a power resistor over using
an LM7812, except possibly saving a few cents for an application with a
very steady load current and input voltage (and if that's the case, why
not just use a resistive voltage divider?)

http://www.fairchildsemi.com/ds/LM%2FLM7812.pdf

Learn it, learn how to design one, and then forget it -- it's doubtful
you'll ever need it.

Good luck
Chris

.

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