zener diode inuition machanism for regulation

[yefj]

Hello,I have the following circuit of a shunt regulator.I kow that the zenner diode enforces certain voltage threw it.
In case of sudden increase in R_L the V_l wil rise so Vbe will be smaller.
So how the zenner diode willinteract with this process and make the V_L voltage smaller beack to the original value?
Thanks.

 

[rjenkinsgb]

So how the zenner diode willinteract with this process and make the V_L voltage smaller beack to the original value?

It will not!

In fact as the base current has decreased, more of the current through Rs will have to pass through the zener, so the voltage at the base will also increase slightly.

(And that does not account for the input supply increasing when the load current decreases).

That simple emitter follower is a simple, non-exact circuit, which is why more complex voltage regulators are far more common.

 

[yefj]

Why the base current would be decreased? what is the analog logic of that?
Thanks.

 

[yefj]

Update:
given that the base current will decrease what will happen inside the zenner current mechanism?
Thanks.

 

[Galgso]

The circuit you have there is a simple zener diode voltage reference with a BJT buffer.

A simplified explanation of how a zener diode works is that it does not allow current to flow at voltages lower than its reverse breakdown voltage or zener voltage (Vz). At its zener voltage, it becomes much more conductive, far more conductive than resistor Rs. So if any voltage higher than Vz is put across the input the resistor will drop the excess voltage and the zener will maintain a constant voltage drop. This constant voltage is fed into the base of the NPN BJT which acts as a simple buffer in emitter follower configuration. It is needed to reduce loading on the zener diode which would make the circuit no longer regulate voltage properly unless extremely high currents are used through the zener which would reduce efficiency.

I suggest allaboutcircuits.com/textbook if you want to learn more about BJTs. Also you can try simulating the circuit using falstad.com/circuit

 

[Galgso]

yefj try this: https://tinyurl.com/2dz2y9um (link to falstad simulation)

 

[Tony Stewart]

Zener . Try not to mispell again. We say Zee' ner

Increase Load R with thumbwheel and memorize the behaviour.
https://tinyurl.com/2c9o66gm If Zener is using 350 mW, then it must be rated for twice this power to not be running at 150'C as using full power means 125'C above ambient.

 

[danadak]

 

[Nigel Goodwin]

Zener . Try not to mispell again. We say Zee' ner

In the proper English speaking world we say Zen' er

 

[yefj]

Hello Galgco,Yes I understaood the if Vin is higher then Vz then the voltage drop will be divided betwee the Rs and the zenner in reversed biased.
given the manual below we see the If RL rises Vo rises which will decrease Vbe by the formula below it will decrease the base current of the PNP.
Is=Iz+ib
So what will happen nextcurrent wise in so base current will rise again to the previos value and restor the previos situation?

 

[danadak]

As Vo tries to rise, base current does drop dropping Iload which forces Vo
to correct and drop. Small drops in Ib result in large drops in Ic,
Ic = Beta x Ib

 

[yefj]

Hello Dana, Yes that I understood.
But we have the Is=Iz+ib
when Ib dropswhat happens to Is an Iz?
Thanks.

 

[danadak]

Look at the graphs .....

What is Is ?

 

[Galgso]

Hello Dana, Yes that I understood.
But we have the Is=Iz+ib
when Ib dropswhat happens to Is an Iz?
Thanks.

Is and Iz in your original schematic should remain fairly constant if the circuit is well designed across the entire required output current range. The base current should be negligible compared to the current through the zener diode to make the output voltage less load-dependent.

If the output voltage is greater than Vz minus Vbe, the base current will drop severely as there will no longer be voltage across the base-to-emitter junctionl. This will cause the collector current to also drop and the voltage will go back down.
When the voltage on the emitter which is also the output voltage drops there will once more be a potential difference across the base-to-emitter junction so the transistor will conduct again.

 

[yefj]

Hello Dana,according to you graphs, when Ibase drops Ir2 rises,why is that happening?
Thanks.

 

[yefj]

Hello Galgso, Could you give mathemtical intuition regarding why in case of Zenner current>> base current then output voltage will be much stable?
Thanks.
"The base current should be negligible compared to the current through the zener diode to make the output voltage less load-dependent"

 

[danadak]

when Ibase drops Ir2 rises,why is that happening?

Ir2 - Ibase = Iz
Ir2 = Iz + Ibase

I had the I probe at wrong end of R2 -

 

[Galgso]

Hello Galgso, Could you give mathemtical intuition regarding why in case of Zenner current>> base current then output voltage will be much stable?
Thanks.
"The base current should be negligible compared to the current through the zener diode to make the output voltage less load-dependent"

If the base current is not much smaller than the zener current, the zener voltage reference will change with load current because base current = load current ÷ beta. By making the zener current much larger than the largest possible value for base current you are making sure that the base will not steal current away from the zener diode and change the reference voltage.

The required base current can be calculated by dividing the load current by beta for your BJT. A rule of thumb is that the zener diode current should be the base current * 10 using the lowest expected beta value and highest expected load current.

Lower current through a zener diode will cause the zener voltage to drop because real zener diodes are not ideal and exhibits current dependent reverse voltages. If you try to use a 2W 12v zener diode with 10uA flowing through it in reverse you will see a much lower voltage than Vz even though your supply voltage may be high enough in theory to break down the diode junction in reverse. This is why it is necessary that the base not load down the zener diode and cause the reference voltage to change.

Try my falstad simulation link or Tony Stewart's link which is nicer than mine and see what happens if you increase the value of the resistor in series with the zener diode.

 

[danadak]

From 1 uA to 1 mA, 11.75V to 12.05V (not 2 W though)

 

[Tony Stewart]

In the proper English speaking world we say Zen' er

Hence the improper spellings.

Research indicates:
The name’s origin likely ties to Clarence Melvin Zener’s family background, which may have German or Dutch roots—common for many American surnames. In German, a name like "Zener" might be pronounced with a "ts" sound for the "Z" (e.g., "TSAY-ner"), but Zener, born in Indianapolis in 1905, would have used the Anglicized form, as was typical for immigrants or their descendants in early 20th-century America.

For example, in educational videos and lectures on electronics, the name is consistently pronounced as "ZEE-ner," reflecting its use in American English.