Newbie: Zerner diode voltage regulator ?

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

  1. hdjim69

    hdjim69 Guest

    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
    #1
    1. Advertising

  2. hdjim69

    john jardine Guest

    "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
    #2
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  3. hdjim69

    Bill Bowden Guest

    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
    #3
  4. 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
    #4
  5. hdjim69

    Chris Guest

    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
     
    Chris, Oct 26, 2006
    #5
  6. hdjim69

    hdjim69 Guest

    >> 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
    #6
  7. hdjim69

    Chris Guest

    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
    #7
  8. 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
    #8
  9. hdjim69

    john jardine Guest

    "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
    #9
  10. hdjim69

    Rich Grise Guest

    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
    #10
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    Mar 25, 2014
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