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Electronics => Beginners => Topic started by: Nicals on April 20, 2017, 04:53:41 pm

Title: [Art of Electronics|solved] Zener power supply exercice
Post by: Nicals on April 20, 2017, 04:53:41 pm

Hello,

I'm currently reading the Art of Electronics (http://artofelectronics.net/) book, and I am having difficulties with an exercice relating to a simple zener diode regulated power supply.

The exercice says this (ex 2.6, p 82, 3d edition):

Quote

Desing a +10V regulated power supply for load currents from 0 to 100mA; the input voltage is +20 to +25V.
Allow at least 10 mA zener current under all (worst-case) conditions.
What power rating must the zener have ?

This is all about resistance with a Zener diode in serie.
The next exercice is the same thing, but with a transistor in emitter follower configuration to show the difference in required zener power.

(https://i.imgur.com/IAhj56D.png)

And then added some computation to it.

We want at least 100mA load, so I just assume that I had a 100mA current sink through the right wire.
I then computed the R1 resistance value so that 100mA flows through it.

Code: [Select]

R = Umin / Imax = 20 / 0.1 = 200

Without any load connected, I still have my 100mA flowing. This is at least the 10mA asked. So we are good.

At this point, I know I did something wrong. I should have computed the current through the Zener diode.
But there is no indication on the Zener internal voltage.

I also don't understand why we would have a Zener current of at least 10mA.
What does this relates to ?
Shouldn't we want the zener current to be minimized to prevent power loss ? Or is their a good reason to have a minimum current though the zener diode ?

I'm really lost at this point. It seems easy, but I can't manage to find from where I should start to solve this problem.

Thanks for helping.

Title: Re: [Art of Electronics] Zener power supply exercice
Post by: rstofer on April 20, 2017, 05:13:22 pm

The zener current keeps the diode in a mode where it regulates the voltage drop.

When you calculate the resistor, you need to consider both the load current plus the zener current at the same time.  You need to consider this at Vmin which you did.  So, Vmin / (10 + 100) mA...

When Vin is higher than Vmin, there will be more current flow through the zener and the maximum zener current will occur when there is no load and the voltage is at Vmax.  Once you know the maximum zener current (based on the resistor selected above), you just multiply by the zener voltage (10V) to get the power dissipated.  As a practical matter, the actual device will have a rating somewhat higher than the calculated rating.

Title: Re: [Art of Electronics] Zener power supply exercice
Post by: djnz on April 21, 2017, 04:04:57 am

If you look at a typical zener's I-V curve, for very low currents, you are still in the "rounded" region of the curve, but beyond a certain current, the curve becomes a more vertical straight line. Look at Fig 1.15 on page 12 of your book. For good regulation, you want to be in the stiff straight line part of the curve, not the rounded part. Mathematically, you want dV/dI to be as close to zero (the definition of regulation!) or, dI/dV to be very large in magnitude. Thus, to be safe, the authors suggest that current through the zener should not drop below 10mA.

However, this 10mA is not the output current!  There is some input current coming in; a portion of it goes through the zener and the rest is delivered through the load.

Title: Re: [Art of Electronics] Zener power supply exercice
Post by: Nicals on April 22, 2017, 02:48:46 pm

Thanks, it is now much more clear. It really helped me.