Is it mandatory to use series resistors for zener diodes? - Page 1
Forum section: electronics / beginners

author: syntax333
author_rank: Regular Contributor
timestamp: May 24, 2020, 06:18:03 pm
content: Hi, I have a question. Is it mandatory to use series resistors for zener diodes?For example I want to switch a power supply using a microcontroller as shown in attachment.I want to protect the P-channel MOSFET's gate to source voltage. Vcc is 28V, R1 and R2 is 20kohms, C1 is 10nF.When power is switched on the Vgs becomes 14V due to Voltage divider. Thus voltage across 12V zener diode is 14V. Lets assume zener resistance is 9ohms.Does this mean zener diode will consume (14-12)/9 = 222mA without series resistor? Or Am I missing something?I cannot use series resistance since it will also limit the current that is drawn by Load.

author: TimFox
author_rank: Super Contributor
timestamp: May 24, 2020, 06:25:54 pm
content: When dealing with a constant-voltage device, where the voltage is roughly independent of the current through it, it is mandatory to limit the current (for example, with an external series resistance) through the device.It is unsafe to apply a voltage from a constant-voltage source (where, again, the voltage is roughly independent of the current) directly to a Zener; both your external voltage source and the Zener itself have tolerances, and there are two possibilities: 1. The voltage source is less than the Zener knee voltage: nothing happens. 2. The voltage source is greater than the Zener knee voltage: prepare for smoke. The Zener resistance is always small, or it would not be a useful device, and the voltage tolerance is usually +/- 5% or 10% (measured at a specification-point current).

author: syntax333
author_rank: Regular Contributor
timestamp: May 24, 2020, 06:36:29 pm
content: If it is a bad idea to use zeners for my case, how can I protect my MOSFET?Using only voltage divider?

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 06:37:00 pm
content: Quote from: syntax333 on May 24, 2020, 06:18:03 pmNo resistor automatically means you are pulling gate directly to GND. How do you expect zener protecting anything when not limiting current means zener won't be able to keep the voltage? It's like attaching voltage limiter and not wanting voltage to be limited at the same time. And in what miraculous way series resistor is supposed to limit load current to begin with?

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 06:39:16 pm
content: And why do you need C1 and C2 when they make switching time worse?

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 24, 2020, 06:43:01 pm
content: Quote from: syntax333 on May 24, 2020, 06:36:29 pmUse 16V or 18V Zener.

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 06:48:44 pm
content: Quote from: Vovk_Z on May 24, 2020, 06:43:01 pmThat does not change anything fundamentally since Vcc is 28V. Also even if it wasn't, using zener without series resistor means something will explode once voltage is exceeded. Though it's better to use zener rated for more than 12V since MOSFET gate usually is rated to 20V, so around 15-16V zener will result in less power loss and still will be safe.

author: mikerj
author_rank: Super Contributor
timestamp: May 24, 2020, 06:53:07 pm
content: You do have a series resistor, it's R1. R1 will see 28-12v across it, giving a current of I(R1) = (28-12)/20k = 0.8mA.R2 will shunt some of the current from the zener: I(R2) = 12/20k = 0.6mA. This means the zener current is just 0.8-0.6=0.2mA. At such a low current a 12v zener will have less than 12v across it, so these numbers will be a little different in practice.

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 06:56:50 pm
content: Quote from: mikerj on May 24, 2020, 06:53:07 pmOh, I thought he talks about R1 as series resistor. Now I see that most likely he thinks that resistive divider miraculously has zero output resistance.

author: syntax333
author_rank: Regular Contributor
timestamp: May 24, 2020, 07:10:12 pm
content: Quote from: wraper on May 24, 2020, 06:39:16 pmI asked the question inhttps://www.eevblog.com/forum/beginners/purpose-of-external-capacitor-between-gate-and-source-of-a-mosfet/msg3069326/#msg3069326this topic.Slowing down the transistor does not matter in my application since I don't need speed.

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 07:11:22 pm
content: Quote from: syntax333 on May 24, 2020, 07:10:12 pmIf that does not matter, what positive effect they are supposed to do?Edit: explanation by Peabody makes little sense since gate has it's own capacitance and even without that, there will be no GS voltage upon applying Vcc.

author: syntax333
author_rank: Regular Contributor
timestamp: May 24, 2020, 07:19:37 pm
content: Quote from: wraper on May 24, 2020, 07:11:22 pmWell I have never used any circuit like this before. I just searched soft-latch circuit on google images, and found the circuit that I linked in the thread. I didn't understand either thats why I created a post about it in the first place.

author: Vovk_Z
author_rank: Super Contributor
timestamp: May 24, 2020, 08:19:16 pm
content: Quote from: wraper on May 24, 2020, 06:48:44 pmOh, yes, I was inattentive (I thought power supply is 14 VDC).But still, with R1=R2 and 28VDC supply a 15-18V Zener will consume a little less power.To TS:There is nothing wrong with a circuit. There is R1 = 20 k. It won't (can't) work without this resistor. It is mandatory, of cause, otherwise, Zener will die because of large current.

author: syntax333
author_rank: Regular Contributor
timestamp: May 24, 2020, 08:37:01 pm
content: Quote from: wraper on May 24, 2020, 07:11:22 pmSome sources say that it prevents the MOSFET to turn on immediately after VCC. Dave's soft-latch power switch circuit is at power-on state when VCC is applied.I found this post on different forum if you want to check it out.https://electronics.stackexchange.com/questions/304133/need-a-soft-latching-power-on-off-that-starts-off

author: wraper
author_rank: Unknown
timestamp: May 24, 2020, 09:00:38 pm
content: Quote from: syntax333 on May 24, 2020, 08:37:01 pmI don't see the cap in circuit presented there, nor anything written about it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 25, 2020, 03:47:04 am
content: What you've missed is not that the zener's own resistance is important, but the total circuit resistance. Add them up, and then calculate.Here, the zener's resistance is in series with the Thevenin equivalent resistance of R1 and R2. Which is the resistance R1 || R2 = R1*R2 / (R1 + R2).It would be very strange, after all, to discover more current flowing in the zener than is being delivered by R1.Another way to look at it is a current divider, where R1 is delivering current to D1 || R2. We start by assuming the zener is off, then assuming it's on, and checking that the currents are sensible. That is, if R2 has 12V across it, will it draw more current than R1 is delivering? If not, then D1 is sinking the difference; else, D1 is off (and the actual drop will be less than 12V). Either criteria is sufficient (V > Vz, Iz > 0).Tim

author: David Hess
author_rank: Super Contributor
timestamp: May 25, 2020, 09:31:17 am
content: Quote from: syntax333 on May 24, 2020, 06:18:03 pmA series resistor is not required however the current through the zener diode needs to be limited in some way. This can also be accomplished with a transistor acting as a current source or some other ballast type of circuit using capacitors and inductors.

author: not1xor1
author_rank: Frequent Contributor
timestamp: May 25, 2020, 05:28:57 pm
content: Quote from: mikerj on May 24, 2020, 06:53:07 pmit is easier to consider that a voltage divider is just a voltage source with a series resistance equal to the two resistors in parallel. So here you have 28V /2 = 14V and a series resistance of 20k || 20k = 10k. That means a current of 2V / 10k = 0.2mA