a little trick for regulated or not regulated supplies

[dschwartz]

one proble i allways had with the pedals i sell was the power supply requirements..the dual rectal (wich is every day farer away from dr boogey) or other high gain pedals really don´t like unregulated supplies, or even switching regulated supplies, so everytime i sell one i had the trouble of getting clean 9V from any DC supply the buyer had..first i tried adding a 7809 regulator inside the box, wich worked very well for unregulated cheap supplies, but if the guy buyed a regulated supply, it worked badly..

so i gave that a little thought and ended up with a nice solution..using a 9.1V zener diode at the dc jack..that way, when used with a regulated 9V supply, the zener doesn´t affect the voltage..and if the supply is regulated, it cuts the voltage to a clean, steady 9.1V..it works great!!

probably many people use this trick..but i hope it helps someone who has the same problem and doesn´t know what to do..

greetings from chile!

[brett]

Daniel - greetings from Australia.
That's a cool idea.  But I've always wondered how much resistance you should put in series with the Zener.  If the input voltage is 11V and the Zener is 9.1V, that is putting 2V across the Zener with minimal resistance.  Won't that burn it unless you put at least 10 ohms in series? (2V at 10 ohms = 0.2A, and 2V x 0.2A = 0.4W That's heavy load on both the Zener and the supply.)
cheers
PS it's been 3 months since I was in Chile.  Man I miss it!  Andes, Arica and the Attacama !

[Sir H C]

Have you looked at low drop-out regulators?  Often they will work down to 50mV across them then gracefully go down with the input voltage. 

[dschwartz]

Daniel - greetings from Australia.
That's a cool idea.  But I've always wondered how much resistance you should put in series with the Zener.  If the input voltage is 11V and the Zener is 9.1V, that is putting 2V across the Zener with minimal resistance.  Won't that burn it unless you put at least 10 ohms in series? (2V at 10 ohms = 0.2A, and 2V x 0.2A = 0.4W That's heavy load on both the Zener and the supply.)
cheers
PS it's been 3 months since I was in Chile.  Man I miss it!  Andes, Arica and the Attacama !

haha you liked the desert? i lived for 10 years in antofagasta..ended up hating the desert...but it surely has some mystique..

about the resistor..i used a 270 Ohm in series to protect the zener..

[dschwartz]

Have you looked at low drop-out regulators?  Often they will work down to 50mV across them then gracefully go down with the input voltage. 

no? what are those? any part code to check the datasheet?

[stm]

Check the LM2941CT, available at Victronics (a chilean electronics store).  Datasheet indicates around 100mV of voltage drop for output currents of 100mA. 

Datasheet:
http://www.victronics.cl/inf_tecnica/Nationas%20Semiconductors/Datasheets/LM2941.PDF

[dschwartz]

ok i checked that out..but what if the input voltage is smaller than the requested output? i assume that it won´t work properly, just as a 7809..
the data sheet say that you need Vin= Vout +2  son it doesnt solve the problem..

i needed something that will work with 9v regulated or those 17V that unregulated Wall warts give, and still have nice regulation..if i use LM2941 with 9 volts input, it wont work right.

[Ripdivot]

Are you using the zener with the resistor in series with the supply and then the zener to ground?

[stm]

If you look at the datasheet on page 4, the "Dropout Voltage" graph indicates you need a 100mV Input-Ouput differential voltage for operation at 100mA in case of the LM2941.  At lower currents this is even less.  On the other hand, a 7809 requires full 2V of input output differential to operate!

Anyway, there is no free lunch and you must accept to trade something in order to have regulation.  A practical use of the low dropout regulator would be to set its output for 8.5V.  Then, any input voltage above 8.6V would be properly regulated to 8.5V.

This isn't much different than using a series diode to prevent reverse polarity.  You have to live with 0.6V drop, or 0.3V drop at best if you use a schottky diode.

[dschwartz]

but.. what´s wrong with the zener?
for me, it looks easier and cheaper (1 zener, 1 resistor, 4 solder connections, and it´s ready to go)....and with the right resistor you can set the current to protect the zener in case of shortcut or too much over voltage..
i´m gonna measure the voltages to calculate how much power dissipates the zener when using unregulated DC...

i like the regulator..but that means an auxiliar board with the hassle it involves and you have to set it for less than 9V...

If i´m missing something wrong with the zener? they explode? burn? are expensive? hard to get? sucks current? look ugly?

[Sir H C]

ok i checked that out..but what if the input voltage is smaller than the requested output? i assume that it won´t work properly, just as a 7809..
the data sheet say that you need Vin= Vout +2  son it doesnt solve the problem..

i needed something that will work with 9v regulated or those 17V that unregulated Wall warts give, and still have nice regulation..if i use LM2941 with 9 volts input, it wont work right.

It acts like a small resistor in series for low input voltages, it will pretty much track the input voltage down pretty far below the desired output voltage.  It cannot pump the output higher than the input, but there is not much difference between the input and output.

[Sir H C]

but.. what´s wrong with the zener?
for me, it looks easier and cheaper (1 zener, 1 resistor, 4 solder connections, and it´s ready to go)....and with the right resistor you can set the current to protect the zener in case of shortcut or too much over voltage..
i´m gonna measure the voltages to calculate how much power dissipates the zener when using unregulated DC...

i like the regulator..but that means an auxiliar board with the hassle it involves and you have to set it for less than 9V...

If i´m missing something wrong with the zener? they explode? burn? are expensive? hard to get? sucks current? look ugly?

You should also have a cap in there to keep the impedance low so you don't get bouncing on the supply.  To current limit with a relatively large drop (say 3 volts) on the zener, you will get problems with the low voltage operation.  These regulators are cheap, often sub dollar, just need two caps and you are set.

[stm]

but.. what´s wrong with the zener?

Almost everything...  Let's workout a practical design and see what happens:

1) Design Goals:
a. Develop a circuit that limits and regulates the supply voltage for a pedal at 9V.
b. Voltage drop when the input is close to or lower than 9V should be minimal.
c. Range of input voltage from 9V up to 17V.
d. Use a shunt zener as a regulator.

2) Solution:
a. Let's consider using an inexpensive and easy to find 1N4739A zener diode (9.1V, 1W maximum dissipation).
b. It must be noticed that the 1W dissipation for this zener is in open air at 25ºC or so, and the junction is at the verge of destruction by heat.  A proper design should establish a security factor of 100% or more, so we'll consider for our purposes that this zener cannot dissipate more than 0.5W safely.
c. Let's assume our pedal needs 20mA (0.02A) of current at 9V.
d. Let's assume that a 0.2V drop at 9V input is acceptable for our purposes.
e. From (c) and (d) we can calculate the series resistance as: Rs = 0.2V / 0.02A = 10 ohms.
f. When the supply voltage is 9V, there is virtually no conduction in the zener, thus the supply delivers 0.02A or 9V*0.02A = 0.18W.
g. Power dissipation in the resistor will be 10ohms*(0.02A)^2=4mW (negligible).
h. Let's see what happens when the supply voltage is 17V (the other extreme).  In this case the zener is conducting and keeping a near 9V voltage between anode and cathode.
i. The current through Rs is now given by: (17V - 9V) / 10 ohms = 0.8A (!!!)
j. The power dissipation in Rs is: (17V - 9V)*0.8A = 6.4W (!!!), thus, good design practice dictates using a resistor rated for not less than 15W or so!
k. The power supply is delivering: 17V * 0.8A = 13.6W (!!!)  The power supply may be greatly overloaded by the extra zener current.
l. The zener is conducting a current of: 0.8A - 0.02A = 0.78A (!!!)
m. The zener power dissipation is then: 9V * 0.78A = 7.02W (!!!) or 14 times more than the maximum reasonable power dissipation.

3) Another approach to the solution:
a. Let's start the other way round.  This time let's calculate Rs so it limits the current in the zener to a safe value when the input voltage is maximum.  The maximum allowable current in the zener can be expressed as: 0.5W / 9V = 0.056A
b. Rs can be calculated considering it will pass the current across the zener and the current needed by the pedal, thus: Rs = (17V - 9V) / (0.056A + 0.02A) = 105 ohms.
c. Under this condition the power dissipated by Rs will be: (17V - 9V) * (0.056A + 0.02A) = 0.608W.  Thus, a 100 ohm, 2W resistor would be more than adequate for this purpose.
d. Let's see what will happen when the input voltage is at the lower limit of 9V.  Under this condition the zener is essentially off and only the pedal current flows through Rs.  This produces a voltage drop of: 100 ohms * 0.02A = 2V (!!!)  In other words, the pedal is being fed by 7V only.

4) Conclusions:
A simple resistor-zener shunt regulator is not practical for large input voltage variation without sacrificing power efficiency and voltage drop.  Even if this circuit is redesigned for lower input voltage variations, like 9 to 12V only, the circuit is prone to suffer permanent damage if the input voltage exceeds a certain value.

[dschwartz]

Oh..  thats pretty interesting.

..thanks Sebastian!