Zener Diode voltage regulation

Thread Starter

Topad

Joined Nov 30, 2021
3
Hi All, I have two designs using a zener diode for volatge regulation at 24V supply volatge. I like to know which is a better design to go with. Please see attach image. The load device is an active buzzer, that uses 15ma at 4v. Design #1: I have a shunt zener regulator using a resitor that would need to be a high power resistor at least 500mw. On design #2: which I intent to use becuase is much simpler and seem to be okay after doing some calculations. For instance, with a 20V and 1W zener diode, will allow the buzzer device to have its 4v and will consume 15ma which will be below the zener max current of I=p/v=1W/20V=50ma. Will I be okay using design#2.
 

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ericgibbs

Joined Jan 29, 2010
16,370
hi Topad,
Welcome to AAC.

How stable is the +24V.?
The right side circuit Vout will depend upon the 24V's stability.
E
 

WBahn

Joined Mar 31, 2012
27,395
As is the case whenever the question about "which is a better design" is asked, it comes down to, "What is the metric by which one circuit is deemed better than the other?" Cost? Power? Reliability? Something else?

Have you considered just using a resistor that drops 20 V when it has 15 mA flowing through it and being done with it (i.e., no zener diode at all)?

It really comes down to what the specifications of this active buzzer are. How sure are you that it will actually draw 15 mA? What is the tolerance on the 4 V input to it? How stable is that 24 V?

I'm assuming that you are activating the buzzer by switching the 24 V on and off? Or does the buzzer have its own control? If so, what current does it draw when the buzzer is on and what current when it is off?
 

DickCappels

Joined Aug 21, 2008
9,317
Being clumsy I sometimes short things out. A long time ago I use a Zener diode to drop voltage like in your example #2. You have already guessed that I shorted the Zener to ground and also probably realize the Zener acted like a fast-acting fuse. As a rule I don't use Zeners to drop voltage unless there is also a dependable current limit of some sort.

Besides, your circuit #1, as already pointed out, has other benefits over circuit #2.
 

BobTPH

Joined Jun 5, 2013
5,762
If the 24V supply is constant and the buzzer really draws 15mA continuously, then the zener is doing nothing more than a resistor would.

Edited to add: And if the power supply is variable, then the zener does worse than a resistor in circuit 2. That circuit is NOT a regulator.
 

MrSalts

Joined Apr 2, 2020
2,354
Either will work just fine. The circuit on the right is not a regulator - it just drops 20v to give the device 4v (assuming your power supply is stable and regulated at 24v, it is the most efficient and easiest to design and implement). This design on the left will generate more heat as the zener typically will take an additional 5mA to stay in compliance with the stated zener voltage so your resistor will drop 20v at 5mA for the zener and 15mA for your buzzer for a total of. 20mA. The resistor will be 20v/0.02A = 1k ohms.
 

Ramussons

Joined May 3, 2013
1,316
Hi All, I have two designs using a zener diode for volatge regulation at 24V supply volatge. I like to know which is a better design to go with. Please see attach image. The load device is an active buzzer, that uses 15ma at 4v. Design #1: I have a shunt zener regulator using a resitor that would need to be a high power resistor at least 500mw. On design #2: which I intent to use becuase is much simpler and seem to be okay after doing some calculations. For instance, with a 20V and 1W zener diode, will allow the buzzer device to have its 4v and will consume 15ma which will be below the zener max current of I=p/v=1W/20V=50ma. Will I be okay using design#2.
The left will give you 4 volts. The "24" volts can vary, but the output remains at 4 volts.
The right will give you 24 - 20 = 4 volts. The "20" is the voltage drop. If the 24 varies by +/- 1 volt, the output will also vary by +/- 1 volt.
 

Ian0

Joined Aug 7, 2020
6,294
Unless the 24V varies significantly in voltage, just select a resistor to put in series with the buzzer that provides 4v across the buzzer.
Might be worth a capacitor across the buzzer. The current may be modulated by the buzzer frequency, which would allow a much higher voltage across it when the current would be at its minima.
(Althougth, it may have some internal capacitance)
 

WBahn

Joined Mar 31, 2012
27,395
Unless the 24V varies significantly in voltage, just select a resistor to put in series with the buzzer that provides 4v across the buzzer.
I would add another "unless" -- Unless the 15 mA isn't pretty reliable and constant. Let's say that you use a 1.3 kΩ resistor and, under some conditions, it were to draw only 14 mA then the voltage across it would be 5.8 V while if it were to draw 16 mA the voltage across it would only be 3.2 V -- and that's with only a 1 mA variation from the expected current.
 

DickCappels

Joined Aug 21, 2008
9,317
With a current like that (15 ma) I would expect the "buzzer" to be a piezoelectric resonator driven by a simple oscillator rather than the crude electromagnetic buzzers that came in Erector sets and the like.
 
Regarding #2, remember that the most common Zeners have a +/-5% tolerance. A 20V zener might actually be 19 or 21V. (3 or 5V to the buzzer if the 24V is exact.) I don't know if that matters to the buzzer. (1% Zeners are available.)

Let me suggest a third approach, although it would add an NPN transistor. See attached figure - an emitter follower. The Zener would be biased by about 5.8mA. The buzzer current would flow through the transistor. Note: The base-emitter junction will drop about 0.7V. My sketch uses a 4.7V Zener, so the voltage going to the buzzer will be closer to 4.0V. The transistor will dissipate about 0.6W, so I suggest a 1W or higher transistor.
4V Regulator.jpg
 

dcbingaman

Joined Jun 30, 2021
720
A standard LM317 linear regulator is just as cheap but also offers built in short circuit current protection using just one 3 pin device and two external resistors:

1662323695067.png

The resistor values are determined from the datasheet as follows:

1662323801147.png

Because IADJ is 50 microamps, you may be able to get this to work with just one Resistor by removing R1 and making R2=(4V-1.25V)/50uA = 55Kohms
 
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Thread Starter

Topad

Joined Nov 30, 2021
3
1. 24V stability? The 24v is very stable the device uses an AC to DC converter AC to DC converter 24v with load regulation of +-0.2% and load regulation of +-0.5% which when I measure it is about 23.84V and stable.

2. WBahn questions "which is a better design"?. I like to make the circuit as simple as possible yet effective using these components. I know could use an LM317 linear regulator or an emitter follower using transistors, but for external reasons, I can't use them. But I can implement the circuits I suggested, preferable design number #2, but I need help to find the right appropriate values for the Cricut to work under my given values such as the power supply, buzzer and Zener diode specifications.

3. Buzzer datasheet BUZZER and current see pictures below at 4V using my power supply the buzzer draws about 16.87ma, and
at 5v about 20.66 and at 8v it draws 30ma.

4. I have tried to use a 1k resistor 1W, this will be great if I can make work, but I see something strange on my measurements maybe because the buzzer is an active self-driven magnetic buzzer, but I can't figure out why as see that behavior. For example, using a 1K resistor in series I only get about 3-4ma and the buzzer voltage drop is about 20V. The buzzer starts to sound around 650 ohms, and the buzzer voltage drop increases. I don't seem to understand why I see this behavior, because I was expecting to see 5v drop on the buzzer at 19ma 24-5v =19/1K =19ma

5. I forgot to include an extra NPN transistor that the design will use, which will add about 0.7. I calculated the values one more time and this seems to be a good option, see the image below, I use this Zener diode BZT52-B18X . The buzzer only sounds when the capacitive touch is used so the buzzer will not sound continuously but only when someone touches it. The Zener diode is an 18v with 590mw power and 2.22%. I calculated the values by taking into account how the given values might vary. Please see the image below if the design is good enough to work. Also, I could use a high-power resistor 1W, but I don't know why I see the behavior on the voltage drop of the buzzer I describe above. Thank you, all for your help!

1662340985818.png

1662330414952.png1662330353208.png
 

WBahn

Joined Mar 31, 2012
27,395
I haven't done much with magnetic buzzers, but my guess is that the current draw is FAR from DC -- it is likely all-but turning on and off and the average value is what you are measuring. If so, that means that the voltage using a resistor is flailing around like crazy and the buzzer has a hard time coping with it.

Since the buzzer can operate over a range from 4 V to 8 V and since you seem to have a pretty solid supply voltage, try using a Zener that will drop about 18 V and see how that works. That should park it nominally in the middle of your allowed voltage range to give you plenty of wiggle room. The fact that you will have no power-supply rejection and aren't actually regulating anything shouldn't matter for this application.
 

MisterBill2

Joined Jan 23, 2018
13,128
Post #1 states an active buzzer, which is some sort of electronic device. So a resistor selected to drop 20 volts at 15 mA in series will make the most sense, and be the simplest. But measure the voltage with that circuit and see how close the buzzer voltage actually is. Changing the resistor may be required.
 

WBahn

Joined Mar 31, 2012
27,395
Post #1 states an active buzzer, which is some sort of electronic device. So a resistor selected to drop 20 volts at 15 mA in series will make the most sense, and be the simplest. But measure the voltage with that circuit and see how close the buzzer voltage actually is. Changing the resistor may be required.
Yet, when he used a resistor to do that, it behaved erratically. He later gave the datasheet (such as it is) for the part he is actually using. It's a magnetic buzzer. The data sheet gives no details of the internals, but there is some kind of oscillator circuit in there that is essentially switching it on and off (or otherwise modulating it considerably). That likely means that the instantaneous current is fluctuating significantly and a resistor that is there to drop the voltage will then have a voltage drop across it that will be fluctuating wildly, which will likely interact with the internal drive circuitry in a way that is not desirable. It was worth a try, but his experience is that it doesn't work adequately.
 

MisterBill2

Joined Jan 23, 2018
13,128
Yet, when he used a resistor to do that, it behaved erratically. He later gave the datasheet (such as it is) for the part he is actually using. It's a magnetic buzzer. The data sheet gives no details of the internals, but there is some kind of oscillator circuit in there that is essentially switching it on and off (or otherwise modulating it considerably). That likely means that the instantaneous current is fluctuating significantly and a resistor that is there to drop the voltage will then have a voltage drop across it that will be fluctuating wildly, which will likely interact with the internal drive circuitry in a way that is not desirable. It was worth a try, but his experience is that it doesn't work adequately.
OK, the normal approach to holding the supply voltage to an active component with varying current requirements is a shunt capacitor. That has been standard for many decades, and is still common. Capacitors with low internal resistance provide the best results. That is usualy mentioned in the second semester electronics courses where active devices are introduced.
 

WBahn

Joined Mar 31, 2012
27,395
A shunt capacitor might well do the trick. Certainly worth a try. It will also delay the on and off action by a bit, but if that's tolerable and it can stabilize the voltage sufficiently, then that's good enough.
 
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