Minimum current for zener diodes

The "minimum load 5 mA" spec on the data sheet just means that the manufacturer doesn't guarantee that the device will meet all specs below that current. It does not mean that the thing won't work acceptably below that current.

I've tested many 78xx regulators with no load other than a DVM, and have always found the output voltage to be close enough.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

Higher voltage zeners, like 12v maybe, have sharp knees and very low leakages below the zener voltage. You can do micropower regulators with a small zener and an emitter follower, running the zener at a few uA.

John

n

Oh, if you're trying to get the best efficiency from a battery, then you've gotta look at switching power supplies. (It's not a subject, I know much about.) If you want to make a 5V supply from zener's or other linear IC's then you are throwing away more than 1/2 your energy. (assuming a 12 V battery.)

George H.

here.http://www.datasheetcatalog.com/datasheets_pdf/B/Z/X/5/BZX55-C5V6.shtml

If you are on battery power you need a micropower reference. TI REF33xx is a nice series reference (like 3-terminal regulator,) LM4041xx-xx is kinda like TL431 but they work at microamps. There are others, these ones are what I personally use for battery-powered designs. Also LM4040 for a series one and others around LM404x for series and shunt (zener-like, or TL431) mode, fixed and adjustable voltage.

--
******************************************************************
*  KSI@home    KOI8 Net  < >  The impossible we do immediately.  *
*  Las Vegas   NV, USA   < >  Miracles require 24-hour notice.   *
******************************************************************

here.http://www.datasheetcatalog.com/datasheets_pdf/B/Z/X/5/BZX55-C5V6.shtml

--- You might want to try using something like an LM4040-5, which is a shunt voltage reference/regulator which has an output voltage which remains [fairly] constant over an input current range of from 75µA to

15mA.

Assuming a dead battery at 10V, 100µa into the regulator and 1mA into your panel meter load, the circuit would look like this: (View in a fixed pitch font)

+5 / +10>--[R1]--+---------+ |K | [LM4040-5] [5K] | | GND>--------+---------+

Since the drop across R1 will be 5V and the current through it will be the load current plus the shunt regulator's quiescent current,

10V - 5V 5V R1 = -------------- = --------------- ~ 4545 ohms Ireg + Iload 1e-4A + 1e-3A

The closest available 1% value is 4530 ohms, so that'll increase the current into the regulator from 100µA to about 4µA.

Assuming your battery's lead acid, fully charged it'll be at about

13.8 volts, so at that point the current into the regulator will be:

Vbat - Vreg Ireg = ------------- - Iload Rs

13.8V - 5V = ------------ - 1e-3A 4530R

= 9.4e-6A = 946 microamperes

So, fully charged, the drain on the battery would be less than 2mA and, at the recharge point, the drain would be about 1.1mA.

-- JF

here.http://www.datasheetcatalog.com/datasheets_pdf/B/Z/X/5/BZX55-C5V6.shtml

. / . Oops...104µA

. \ . Oops... 9.46e-4A

--
JF

han

ou

Your application is non-critical. A simple little regulator sufficient for your panel meter using junk box parts would be like so- uses a current starved zener under the assumption that 10% IZT gets 80% VZT: Please view in a fixed-width font such as Courier.

. . . . . PN2222A . BATT --+--------c e -----+-->

. | b | . | | | . '--[15K]---+ | . | | . c | . PN2222A b -----' . e . | . + . 5.6Vz . - . | . | . -------------+------------ . | . --- . /// . .

han

ou

I believe the most effective conservation strategy would be a momentary switch push-to-take reading...

e:

an

=A1You

s a

ike

and

xed

do immediately. =C2=A0*

r notice. =C2=A0 *

I will check that. I put the "final" solution together today, it is here:

I found a 79L05 and 78L15 and just used them as of now. However, current jumped from 4-5mA to 25. I will have to check the new meter, and try without U4, the 78L12 (which is 15 as that is what I found in my drawer). These drawer components are some 20 years old, that might affect it too. Anyway, shops are closed until monday so tomorrow I will look and think at improvements. The U4 is simply for protection as mentioned before. I have seen spikes kill my LM224 before :)

WBR Sonnich

ote:

k

than

=A1You

l

um

is a

like

I

e and

fixed

e do immediately. =C2=A0*

our notice. =C2=A0 *

I'm pretty sure that panel meter is not liking your signal common at

7V below what it thinks is power ground...

e:

an

=A1You

s a

ike

and

xed

I replied yesterday, but my posting got lost :( I will take a look at this, as of now I have an 78L15 and 79L05, found in my drawer. The meter takes 2mA, and the system with 79L05 and without the 78L15(12) consumes 7,4mA. With it all it takes a bit more than 10mA, in other words 50% is used for the regulators.

I found that the LM4040 and a 1,8K resistor with a stable 10V supply will free up 2mA. The 78L15(12) is for stopping spikes and I need to rethink that part. My charger has once fried ICs.

My schematic is here as of now:

[now I will be off for a few days]

The LM317LZ has a minimum load of 3-5mA, which will force me to fry it off.