# Zener Diode Design

Discussion in 'General Electronics Discussion' started by Y2KEDDIE, Nov 17, 2015.

1. ### Y2KEDDIE

259
15
Sep 23, 2012
I’m trying to understand how the component values are chosen for a battery eliminator circuit by reverse engineering.
Original Design:
Vs supply +140 V
(6) 22V 5 watt Zeners In series
(5) 1N2007, (.7 X 5 +3.5V) forward biased in series
(1) 300 ohm 3 watt resistor (Rs) in series
A 70 mA load at 135V tap.
I calculated :
max Iz = 5 watt/ 22 volts = .227A - .07A (load)
Rs(min) = (Vs- Vz)/(Iz-I load = (140 – 135)/(.227- .070) = 35 / .157 = 223 ohms
Rs (wattage) = P = I (sqr) x R = (.227 X .227) X 223 = (.025) X 223 = 5.5 watts
Is there an error in my calculation?
Obviously they used a higher value Rs which lowered the wattage required. My question is how did they pick the value of 300 ohms.
Is there a rule of thumb when selecting Rs, as opposed to using the minimum value and adding resistance, trial and error, to get a reasonable wattage?

If I increase Vs (Supply) to +200V, this will also increase my wattage rating of components. Should I add more Rs resistance, or add more zeners?

2. ### Harald KappModeratorModerator

10,615
2,372
Nov 17, 2011
Show us a schematic of the circuit to help us understand it. Then we will (hopefully) be able to follow your line of thinking.

3. ### Y2KEDDIE

259
15
Sep 23, 2012
Not sure how to upload. I scanned a sketch but it says the file is too large to upload.

I've revised the supply voltage value from my first post. This is an actual working circuit, with real measured values . Hopefully I can describe the circuit:

Basically at ground I start with two (22V) zeners, (reverse biased) then 6 (.7V each, fwd bias) regular silicon diodes, and then 4 more ( 22 V) zeners, and finally the Rseries (300 ohm, 5W) limiting resistor. This entire string is across 170 VDC supply.

Up from ground the first junction is 22V, then 44v, the silicon diodes drop 3.5V , then at the cathode of the 3rd zener I have 69.7, then 91.7, 113.7. and 135.7, at the junction of the last zener and the Rseries resistor.

My load will be 70mA at the 135.7 tap.

The zeners are 1N5358, 22V, 5 Watt, Zz= 3.5 ohms
Diodes are 1N2007 silicon

IF: (my calculations):

5 Watt/ 22V=.227A + Iz

Rs = (170-135.7)/ (.227-.07)
Rs = 34.7/ .157 = 210 ohms

Rs(wattage)= I(sq)R = (.157) (.157) (210)
Rs (Wattage)= 5.176 Watts

My calculations say 210ohms @ 5.176 Watts. Obviously one wouldn't want to push all components to the limits so Rs was made bigger to reduce wattage.

My question is: How was the value of 300 ohms selected. Is there a rule of thumb?

Last edited: Nov 17, 2015
4. ### Harald KappModeratorModerator

10,615
2,372
Nov 17, 2011
Use the "Upload a file" button on the bottom. Use a compressed file format to decrease size.
Creating an image from the verbal description is difficult and the result may look different from what you intend, so we risk talking at cross purposes.

5. ### Y2KEDDIE

259
15
Sep 23, 2012
Using the formula:
Rs = (Emin-Vz)/ (Ilmaz- Iz)

I z being the maximum current allowed through the Zener 5W/ 22V,

I would think one wouldn't want to use Iz at max, but maybe minimum Iz to get breakdown current started and then add the 70mA.

???????

File size:
45.4 KB
Views:
66
6. ### Harald KappModeratorModerator

10,615
2,372
Nov 17, 2011
You have 2 conditions:
No load: Iload = 0 A, I(Rs) <= Izmax because all current flows through the zener diodes.
Max. load: Iload = ILmax, I(Rs) >= Iload + Izmin because despite current being drawn by the load still the minimum zener current needs to flow.

With two equations and one variable you get two results. The real value of Rs should be in between these two results, ideally nearer to the high value to minimize losses with no load.

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,419
2,790
Jan 21, 2010
you also need to maintain a reasonable current through the zener diodes because the voltage is somewhat dependent on the current. This is more an issue with low voltage zeners though.

ideally, the voltage across the resistor remains constant, thus so does the current through it and the power dissipated by it, unless you attempt to draw too much current. At this point the voltage falls, the zeners no longer draw current, and the resistor dissipation rises. You would not typically operate in this mode.

The resistor is normally chosen so that at full rated load, there is still sufficient Zener current to ensure good regulation.

The details will vary depending on the nature of the load, but at minimum current you'd typically see the zeners operating close to their max power (considering any necessary derating), and at max load at a current required to achieve the required degree of regulation -- and here you'll need to read the data sheet(s).

8. ### Y2KEDDIE

259
15
Sep 23, 2012
If I use Zero Amps:

Rs= (170-135.5)/ (0+.07) = 34.5 / .07 = 493 ohms, so somewhere between 152 and 493 ? So 300 would be a little more than half way between. Would this be the rule of thumb? Why not use use the Iz( min) from the data sheet and add my load current to determine Rs.

If you use 300 Ohms then ..047A goes thru the Zener, and .07A through the load:, and .117 through the Resistor.

9. ### Y2KEDDIE

259
15
Sep 23, 2012
The data sheet for 1N3558 shows the nominal Izt, is 50 mA, (test current).

I've been confused by the formulas given in different texts for Rs (series resistor) . Why not just say " use the minimum Zener test current, maybe a little more." It makes sense as you need to have a specific threshold current flowing to get the specified Zener output .

10. ### Harald KappModeratorModerator

10,615
2,372
Nov 17, 2011
It is not effiient to use a resistor-zener configuration at highly varying load currents. A simple transistor to decouple zener current from load current is what I recommend.
You can read up a bit on this topic in this document.

(*steve*) likes this.
11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,419
2,790
Jan 21, 2010
That, plus the maximum load current. Remember that you need to allow for the load as well!

And ensure that the zener is not dissipating too much power if the load current is 0 (or minimum if it cannot be zero).

A lower series resistor will give better regulation because the current through the zener varies less. Harald's suggestion really requires all the same calculations, but the load current is lower (it is lower by a factor of the transistor's gain) so the dissipation in the zener is far less. Using a transistor changes the regulator from a shunt regulator to a series regulator, which is typically far more efficient.

12. ### Y2KEDDIE

259
15
Sep 23, 2012
Thank you for the Student - Hobbyist link.

So if my calculations are correct:
If I use 50 mA as Iz, and 70 mA I load, Izt=120 mA, using a 287.5 ohm resistor .72 Watts will be created, well within the 5 Watt maximum.

From the Data sheet: 50ma is Izt minimum and 216 mA is Izm (Max Reg Current)

If I use max Iz: 216 mA(from spec sheet),and an I load: 70mA, Rs=236, Pd, Iz = 5 watts, provided the Vs never goes higher, or the R-load never goes higher.

The actual circuit uses the values shown. I still question why they chose Rs to be 300 ohms. Safety margin?

Last edited: Nov 19, 2015
13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,419
2,790
Jan 21, 2010
it looks like they allow a lower minimum zener current. Have you called what it is?

you may also like to calculate figures for the input voltage varying a little (assuming it's not regulated.

14. ### Y2KEDDIE

259
15
Sep 23, 2012
This is a partial circuit , the one used in my ABCE1 Battery Eliminator, (the B Supply). They rectify the 120VAC transformer secondary which gives 170 VDC across the series resistor and diodes to ground. They claim that it will provide 70mA to a load.

I'm basing my calculations from these values and the Zener spec for Iz min (50 mA).

I calculated Rs= 291 ohms, they used 300 ohms, a standard value.

My calculated Pz= Vz X Iz max, 135V X .120 A = 16.2 Watts. 16.2W / 11 diodes = 1.47 Watts,.70mA load
Pz= Vz X Iz min, 135V X .050A = 6.75 Watts 6.75W/11 diodes= .60 watts, no load