# Newbie: Zerner diode voltage regulator ?

Discussion in 'Electronic Basics' started by hdjim69, Oct 25, 2006.

1. ### hdjim69Guest

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

hdjim69, Oct 25, 2006

2. ### john jardineGuest

"hdjim69" <> wrote in message
news:...
> 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
>

("Zener" diodes)

You've nearly got it right but coming in from the wrong angle.
It is the Zener itself that controls the voltage to the load. A 13V Zener
will supply 13V to the load. If you've a 10V load then you'll need a 10V
Zener.
Basic idea is that the Zener is set to -always- be conducting a small
holding current (say 5ma) and it's kind of hovering on the cliff edge in a
situation where it's running at it's Zener voltage but really wants to pass
lots and lots more amps ("on the knee"). If allowed, yes it would quickly go
short circuit but the dropper resistor will not allow this.
The dropper is sized so that the load can take the maximum current it needs
and also supply a small additional Zener current. Zener current is usually
set at (a non critical value) 10% of the maximum load current.

Say you need a 5V 100ma load and only have a rough 12V power supply.
1] Get hold of a 5V zener.
2] Unpack the calculator.
3] Allow 10ma zener current. (10% of 100ma). That's a total of 110ma coming
out of the 12V supply.
4] Dropper resistor value, is incoming supply voltage minus the zener volts,
then divided by the total current. Which is (12V-5V) / 110ma. Which is
63.6ohms. Use nearest standard value of 68ohms.
That's it. You've a working, Zener controlled, 5V supply!.

For good design, afterwards always check that the dropper and Zener are
sized for the correct power rating.
Zener power is at a minimum under normal running conditions and is 5V * 10ma
=50mW (zilch).
Zener power is maximum when the load becomes disconnected and all the design
current runs through the Zener. It's then 5V * 110ma = 0.55W. A 1W rated
Zener would be fine.
Dropper resistor rating is (12V-5V) * 110ma = 0.77W. A 1W resistor would be
OK.
john

--
Posted via a free Usenet account from http://www.teranews.com

john jardine, Oct 25, 2006

3. ### Bill BowdenGuest

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

Yes, that's about right. The zener just passes any extra current needed
to keep the load voltage constant.

So, if the input is 12, and the output is 6, and the load varies from
100 to 200mA, the series resistor is 6/.2 =30 ohms. But you probably
want to use something a little smaller so the zener is always
conducting. Zeners need a minimum current to regulate well. You can
look up the curve for the zener to determine the voltage at various
currents. So, if you add a minimum zener current of say 10mA, the
resistor is 6/.21 = 28.6 ohms. Round it off to 27 which is a standard
value.

Then you need to make sure the zener doesn't overheat at minimum load.
At that point, the zener will conduct 110mA at 6 volts which is 660mW
so the zener should be rated above that, or around 1 watt. The series
resistor power will be 36/27 = about 1 watt. A 2 watt size would give a
little extra margin.

Now, if the input voltage also changes, the problem gets more
complicated, since you need to consider the minimum and maximum input
voltages, and also the min and max output currents.

-Bill

Bill Bowden, Oct 25, 2006
4. ### John PopelishGuest

hdjim69 wrote:
(snip)
> Is this how Zerner diodes are used in this
> configuration ?

Not quite. Think of the zener as a pressure relief valve
that leaks current to keep the voltage from rising above its
opening point. It can only act as a continuous regulator
if its current is not allowed to go all the way to zero,
because then it looses control of the voltage. Ordinarily,
the upstream limiting resistor is chosen to make sure that
at minimum input voltage and maximum load current, the zener
current is just approaching zero. The zener wattage is
chosen by evaluating how much power it has to dump with
maximum input voltage and minimum load current, while the
zener is passing all the excess current (at its breakdown
voltage) that gets through the input limiting resistor and
has no place to go except through the zener.

John Popelish, Oct 25, 2006
5. ### ChrisGuest

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
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/LM7812.pdf

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

Good luck
Chris

Chris, Oct 26, 2006
6. ### hdjim69Guest

>> 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/LM7812.pdf
>
> Learn it, learn how to design one, and then forget it -- it's doubtful
> you'll ever need it.
>
> Good luck
> Chris

hdjim69, Oct 26, 2006
7. ### ChrisGuest

hdjim69 wrote:
> >> 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/LM7812.pdf
> >
> > Learn it, learn how to design one, and then forget it -- it's doubtful
> > you'll ever need it.
> >
> > Good luck
> > Chris

Hi, J. A couple of things. First , take a look at the data sheet of a
zener diode:

http://www.onsemi.com/pub/Collateral/1N5333B-D.PDF

For this one, look at the representation/graph sketch "Zener Voltage
Regulator" on the top right corner of p.2. That, along with the data
sheet chart showing the parameters, is all you really need.

When the zener is forward biased, it acts pretty much like any diode.
When it's reverse biased, it acts something like a regular diode until
the reverse voltage approaches the Zener Knee Voltage (Vzk). As it
gets closer to that zener knee voltage, the resistance of the zener
diode starts to decrease in such a way that the current passing through
the zener will imicrease dramatically as the voltage is slowly
increased. This characteristic continues past the
manufacturer-designated test current (at which the device is
characterized), up to the point where the power being dissipated by the
zener exceeds the package/die rating, when it is destroyed by
overheating.

If you're designing with zener diodes, you want to set things so the
zener current is between the test current (Izt) and the maximum
current. If you're daring, you might go less than the test current,
but definitely not low enough to get close to the zener knee -- you'll
lose any voltage regulation there.

You don't need a .mov file, you just need a v-i curve.

Good luck
Chris

Chris, Oct 26, 2006
8. ### John PopelishGuest

hdjim69 wrote:

>... 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.

(snip)

LTspice is a free circuit simulator that lets you see the
current through and voltage across every component, as well
as lots of other things.

It could easily simulate this regulator with a varying
voltage input and/or a varying current load.

http://www.linear.com/designtools/softwareRegistration.jsp

There is also a very active Yahoo discussion group dedicated
to the use of this simulator where you can get some hand
holding as you come up to speed on its use.

http://tech.groups.yahoo.com/group/LTspice/

John Popelish, Oct 26, 2006
9. ### john jardineGuest

"hdjim69" <> wrote in message
news:...
> >> 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

Forget the animations and sims. Way, way, way better to just connect a
resistor and Zener across a power supply and look at the output voltage,
while you vary the PSU. Feel the Zener body temperature, watch for the
smoke.

--
Posted via a free Usenet account from http://www.teranews.com

john jardine, Oct 26, 2006
10. ### Rich GriseGuest

On Wed, 25 Oct 2006 09:00:01 -0700, 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 ?

No, not at all. It's more like a calibrated spillway at the top of the
dam that lets different amounts of water over to keep the level about
constant.

In this case, the water level represents voltage, the water pressure
at the bottom of the dam is the source for your load (say a turbine), and
what falls over the spillway represents what gets dissipated in the series
resistor. When your turbine needs more flow (called 'current'), less water
falls over the spillway, and so on.

Hope This Helps!
Rich

Rich Grise, Oct 31, 2006