Why resistor affect voltage curve from Zener diode?

[hurricanehenry]

Hi - from one of Dave's videos (on Zener diodes) I don't understand why substituting a 1K ohm resistor for a smaller, 100 ohm resistor, changes the curve?

Actually I am not sure why the curve is not as sharp in the first place.

It looks to me like a phase shift with the larger resistor, and less of a phase shift with the smaller resistor?

[Gyro]

I would think it's a thermal effect. The lower the resistor value, the higher the current, so the faster the Zener reaches its final temperature.

[TheUnnamedNewbie]

I don't really know what timescales those are, but a diode junction has a capacitance. With the resistor, you are making an R-C filter circuit. Bigger resistor in series means larger timeconstant, and thus more time to charge.

There is also the fact that the current through the zener changes (a zener does not have a perfect knee, it's voltage is still dependent on the current through it. It's just a far sharper curve than a resistor. That's why in circuits with zeners as references, you will almost always see a constant-current source driving it)

[lordvader88]

they have a lot of equations for how and what changes as temperature also changes

So far I have not needed any of those types equations for stuff I've made/modeled

[danadak]

Probe has 10 pF of C, diode has reverse biased junction C, so change of source Z
(lower)causes increased I to charge caps causing the reduced risetime.

Q = C x V
I = C x dV/dT or dV/dT = I / C


Regards, Dana.

[MrAl]

Hi - from one of Dave's videos (on Zener diodes) I don't understand why substituting a 1K ohm resistor for a smaller, 100 ohm resistor, changes the curve?

Actually I am not sure why the curve is not as sharp in the first place.

It looks to me like a phase shift with the larger resistor, and less of a phase shift with the smaller resistor?

Hi there,

One thing you dont show is the TIME scale, so we dont have all the information we need to make a good guess.

I tend to agree with the equivalent capacitance ideas posted elsewhere in this thread, but seeing the time scale for both waves would help a lot.

The equation for a resistor charging a capacitor when the capacitor has zero volts across it to begin with is:
Vc=Vcc*(1-e^(-t/RC))
where RC=R*C.

Using that equation you could try different capacitances and try to mimic the waves you are seeing by changing only the resistor by 10 fold.

If you do that you will see the waves look like in the attachment.  Note you need to keep the time scale set the same for both waves on the scope to see them compared like this.  This is what you get when you change the resistor value 10 fold.  Ignore the axis scales as this is a normalized comparison.  You will see slightly different waves because there is a limiting zener too, but the comparison should be about the same.