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Patatus

Power Supply +5v with Zener

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1N5338 is a 5.1 Volt Zener rated at 5 Watt. (5 volts X 1 Amp = 5 Watts) You have to go with 5.1 or 4.7 volts. You cannot get 5 volt exactly. Also, note that this has a 5% tolerance.
You will have better current availability if you use the zener to control the base of a power transistor. (in this case you will not need one rated so high).

MP

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I have seen a power supply with the zener controlling the emitter voltage of a darlngton. The collector goes to the control. The input is also fed to the control through a resistor network. Why can't a resistor be used in place of the zener in this situation? I am thinking the signal works better on the PN junction against a constant voltage. How does this work? Has anybody seen a zener used for this purpose? Maybe the zener has a very high signal impedance.

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Kevin,
A zener diode regulates voltage well because it has a LOW dynamic impedance. Therefore if you vary the current through it, the voltage across it won't change much. See the spec sheet for a 1N5231 (I couldn't find one for a 1N5338) 5.1V zener diode here:
http://www.microsemi.com/datasheets/SA5-44.PDF
Notice that with a current of 20mA, its dynamic impedance is only 17 ohms. A resistor with 5.1V across it and 20mA through it would be 255 ohms. A good description of zener diodes is in our web-site, see Articles-General-Zener.

Patabus,
Why throw power away in a zener diode? We don't know your application but maybe a 7805 voltage regulator would be a better choice.

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Yes i know the method by IC Regulator of Voltaje, but the conflict is that the power supply is for a University Lab. were a diode zener is needed. I have simulated in Multisim 7.0 the circuit with the diode zener 5.1 Volt and an ideal transistor. What transistor would you recommend me to use?

Thanks.

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MP,
A zener regulator is always conducting current while regulating, regardless of the amount of load current. The zener conducts its wasted current to ground, not to the load. The resistor feeding the zener must supply both the zener's and the load's currents.
A series voltage regulator IC conducts only load current, without wasting any current to ground (except for its tiny operating current). So the IC will heat much less than the zener plus the resistor, above, and therefore the IC will have much less power loss.

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They both dissipate power as I stated previously. How much a regulator dissipates in dependent upon what is put into it. Yes, when there is no load, this is a different situation. But sometimes, in looking at the trade-off, a zener is more desirable to use.
....and of course, when someone posts a request for a zener, I usually give them an answer without telling them to try something else.

MP

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A comparison between zener vs series voltage regulator:
Example: 20V power source, 10V regulated voltage, 100 ma load current:

Zener S.V. regulator
Resistor voltage = 10V -
Resistor current = 200mA -
Resistor power = 2W -
Zener voltage = 10V -
Zener current = 100mA -
Zener power = 1W -
Total of resistor plus zener power = 3W -
S.V. regulator voltage = 10V
S.V. regulator current = 106mA
S.V. regulator power = 1.06W
Therefore in this case, the zener plus resistor power loss is almost TRIPLE that of the S.V. regulator.

The only time that a zener has a low power loss is when it doesn't conduct:

Resistor voltage = 10V
resistor current = 100mA
resistor power = 1W
Therefore in this case, the resistor power loss is almost the same as the S.V. regulator, except there is no regulation.

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Ante,
A zener begins losing control when it is conducting its "knee" current. A zener operating at its knee makes a lousy regulator.
Did you know that the reverse-biased base-emitter junction of an NPN silicon transistor makes a pretty good zener at about 7V. But they say that the transistor later should not be used normally, because the avalanching causes damage to its hFE spec. I think that is why there are input-clamp diodes in 5532-5534 opamps, and that protection diodes are recommended in 2-transistor multivibrators.

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sure, try connecting a 9 volt battery to a 7805 regulator and see how long your battery lasts...
Half of the voltage is converted to heat at the ground tab. It is not converted to current. It is only converted to loss. The more voltage you have to drop, the worse the problem. Regulators are not always the answer and should not be the first word spoken when someone needs to reduce the voltage. Other alternatives also exist and the function of the circuit must be considered first.
This zener vs regulator discussion is meaningless.

MP

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The circuit I was refering to is a commercial circuit. A resistor could replace the zener, but why did they use a zener. This is a simple transistor biasing arrangement with the zener at the emitter. By the way, the difference in a regulator and a zener is not much. The reverse bias zener is just another way to accomplish the same task. This is a physics debate.

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Guest

Is the circuit you are asking for?
A voltage divider R1 and R2 samples the output voltage and feeds it to the base of transistor Q1. The emitter of Q1 is held at a fixed and regulated (we hope) voltage produced by the drop across zener diode D1. This drop is produced by bias current from R3 and the emitter current of Q1. Should the output voltage drop, Q1 will tend to be turned off, drawing less current through bias resistor R4. The collector voltage will tend to rise, increasing the voltage at the base of pass transistor Q2 and therefore the emitter of Q2, which happens to be the power supply regulator output terminal. This rise in voltage will be passed to the base of Q1, compensating for the initial drop. The overall effect will be the stabilization of the output voltage.
The opposite will happen if the output voltage increases.
The above function will not happen if a zener is not there.

zener.zip

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Audioguru you got me wrong. I was mentioning about Kevin's reply.
Even i know its better to use 78XX.

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Ante,
Your experiment inspired me to try it too, and then I continued experimenting:
1) New 9V alcaline battery feeding a 7805 with no load. Current drain = 6mA. Its been going a day-and-a-half now, and its battery measures 7.8V. The regulator feels cold and its output measures 5.05V.
2) I made another one but put a 51 ohm load on it. That's 98mA into the load and the regulator was using only 6mA. The regulator was dissipating 406mW and felt slightly warm. At 4 hours, the battery measured 6.6V, and the regulator output still measured 5.05V.
3) I built a zener circuit with a 1N4733, 5.1V/1W zener, and calculated a series resistor to feed it and feed the load. With a 51 ohm 100mA load, I figured that with a 6.6V battery then a 15 ohm resistor would feed the load 5.1V. With a new 9V alcaline battery, the 15 ohm feed resistor got VERY hot since it was dissipating 1W. The Zener also got hot and was dissipating 816mW. At first, the total drain on the battery was a whopping 260mA. The battery voltage dropped to 6.6V in about 2 hours, when the zener felt cold, and the load voltage measured 4.4V. Lousy regulation.
4) With a new 9V alcaline battery, I tried the zener circuit without a load. Again the drain on the battery measured 260mA, but the zener was smoking a bit since it was dissipating more than 1.3W. The battery voltage measured 6.6V in about 3 hours, when the zener felt warm, and the output voltage measured 5.0V.
So with a new battery and 98-100mA load, the regulator dissipated 406mW, and the zener/feed resistor dissipated 1.8W.
And with a new battery and no load, the regulator dissipated 24mW and the zener/feed resistor dissipated 2.3W.
The zener regulation can be improved by feeding it even more current.
The zener/feed resistor costs more than a 7805 and eats power, so why use a zener? Maybe it's better for low current applications.

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Siddharth, you know the circuit I am talking about. The zener, because it gets it's voltage from the ouput, will drown out the signal at the emitter. The base still has it's signal. So the signal at the PN junction is really onesided. But if you use a resistor, then the function will still be the same. It's kind of arbitrary, but isn't it interesting. What is the impedance that the signal sees? I bet it's just re because of the constant voltage. But this would make for poor biasing.

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