How slow are zener diodes? - Page 1
Forum section: electronics / projects

author: theleakydiode
author_rank: Regular Contributor
timestamp: April 30, 2020, 05:28:11 am
content: Hi,I'm using a zener diode to clip the negative going part of a power transistors base drive waveform so it doesn't damage my transistor, but it seems to take a while to kick in. This is for a little self oscillating 20khz flyback converter I put together.Is this normal zener diode behavior? I'm using a 1N5343 7.5v 5w type, and it sees about 500mA peak zener current. Could the 5w type just be too slow to respond in time?The transistor I'm using seems to perform better with a negative turn off voltage hence why I'm using this set up, it can take up to 9v emitter-base according to the datasheet.If it is too slow then I'm thinking a TVS might be faster but they also don't clamp as tight as zeners so I'm not sure which one to pick, maybe one of the 1.3w or 3w zeners might be faster.Thanks.

author: exe
author_rank: Unknown
timestamp: April 30, 2020, 06:45:47 am
content: How slow is slow?EDIT: I believe "real" zeners are fast (those that are below ~6.2V). Avalanche "zeners" can be slower (but please check that). So, it might be beneficial to substitute a 7.5 zener with two devices. Or may be even with a 5.6/6.2V zener.One thing that limits performance of a zener is long leads (when really fast clamping is needed, such esd protection), but I don't think this is the case here.

author: theleakydiode
author_rank: Regular Contributor
timestamp: April 30, 2020, 06:48:25 am
content: Slow enough that there is a significant delay before the clamping kicks on, switching frequency is about 20khz.I've included a drawing of the waveform in the image.So I should try a lower voltage zener then? Could their wattage ratings affect their speed in clamping/clipping applications too?

author: exe
author_rank: Unknown
timestamp: April 30, 2020, 06:52:13 am
content: Quote from: theleakydiode on April 30, 2020, 06:48:25 amIt doesn't contain timebase. Anyway, I'd try to change zener. May be a diode too if changing zener doesn't help.

author: trobbins
author_rank: Frequent Contributor
timestamp: April 30, 2020, 07:13:23 am
content: Let's say it was junction charge related delay - then use the lowest power zener (ie. not 5W) and a signal diode (not the UF4004 which may well have a turn-on recovery voltage characteristic).If you are probing a very fast waveform, are you using a probe with the springy clip ground tip and the probe and scope have sufficiently high bandwidth?

author: Moshly
author_rank: Regular Contributor
timestamp: April 30, 2020, 10:01:34 am
content: Have a look at Motorola data book DL150/D. section 10 chapter 7 & 10.http://ae6pm.com/Semidata_books/Motorola/DL150-D.pdf

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 30, 2020, 01:09:17 pm
content: You can refer to this:https://www.eevblog.com/forum/projects/how-fast-does-a-zener-react/A 7.5v, 5W Zener diode will typically exhibit a pretty high junction capacitance.You can also refer to this one for alternatives:https://www.eevblog.com/forum/blog/eevblog-1157-transistor-zener-clamp-circuit/

author: exe
author_rank: Unknown
timestamp: April 30, 2020, 01:15:10 pm
content: Quote from: SiliconWizard on April 30, 2020, 01:09:17 pmShouldn't capacitance actually clamp it even more?

author: Andy Watson
author_rank: Super Contributor
timestamp: April 30, 2020, 01:17:54 pm
content: I would suggest that it is clamping - at -11.3V. After a short period the majority of the energy/current in the drive transformer is dissipated and you see the voltage fall back to -8.2V. The difference betwteen 11.3 and 8.2V being due to the slope resisstance of the zener plus the forward resistance of the diode (D2) subject to the 500mA. Either you have a poor zener with high slope resistance and/or you have have underestimated the 500mA.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 30, 2020, 03:56:29 pm
content: What's the driver? Perhaps you're seeing excess voltage due to stray inductance or resistance, at a relatively high peak current.Physically speaking, zeners respond in picoseconds, perhaps fractional nanoseconds at higher voltage ratings. Circuit strays easily dominate. Or also in this particular circuit, the forward recovery of the UF4004 (which is some nanoseconds).If the time scale of your waveform is microseconds, none of this matters. Perhaps the diode isn't rated for the voltage you thought it was?Tim

author: StillTrying
author_rank: Super Contributor
timestamp: April 30, 2020, 04:12:09 pm
content: If I go to 3A for 10ns I can get close, after the clamping the zenner supplies 0.8A back into the source.

author: duak
author_rank: Super Contributor
timestamp: April 30, 2020, 04:44:52 pm
content: Some info on Zener diodes:https://www.onsemi.cn/PowerSolutions/document/HBD854-D.PDF

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 30, 2020, 04:45:43 pm
content: Quote from: StillTrying on April 30, 2020, 04:12:09 pmWell yeah - isn't that just because of the diode's capacitance? When Vin gets back to 0V, you're actually discharging this capacitance.

author: graybeard
author_rank: Frequent Contributor
timestamp: April 30, 2020, 04:52:02 pm
content: The spike at the beginning is the charge emptying out of the transistor base. It is probably a lower impedance path than the diodes. Once the charge has emptied out of the BJT the diodes turn fully on.

author: free_electron
author_rank: Super Contributor
timestamp: April 30, 2020, 05:21:30 pm
content: You are reverse biassing the b-e junction. This will "poison" the transistor over time and trap electrons in the junction. The gain will collapse and the transistor will die. this process is irreversible ( unless you cook the transistor at high temperatures ( like 200 degrees + ) to knock the trapped electrons loose.NEVER-ever let a b-e junction see a reverse voltage larger than about 6 volts.Your zener construction will not work. zeners are slow. They can have a capacitance of 1 to 10 nanofarads. ( as reverse voltage increases , the 'plates' get pushed closer , so capacitance increases )use a stack of silicon diodes. Two or three fast silicon diodes will work wonders and the reverse b-e will not go above 2 volts.

author: exe
author_rank: Unknown
timestamp: April 30, 2020, 05:29:33 pm
content: Quote from: graybeard on April 30, 2020, 04:52:02 pmShouldn't this create a low-impedance path for the current? I think the base is charged to the opposite polarity, so I'd expect it to counter-act. And may be it actually does. If the switching speed of the circuit is 20kHz, then the time-base on the plot is hundreds of microseconds. So, I'm more in favor of theory that inductor stores a lot of energy and dumps a lot of current for extended period of time, and zener or diode cannot clamp it well.Quote from: free_electron on April 30, 2020, 05:21:30 pmCan you please explain how capacitance makes clamping slow? I'm also not sure if capacitance really grows with voltage, for regular pn-junction it's the opposite.

author: graybeard
author_rank: Frequent Contributor
timestamp: April 30, 2020, 05:53:05 pm
content: Try moving the 22 ohm resistor in the base lead in-between the diodes and the base. If the issue is the base charge, it should help.You could also split it and put part of the resistance between the diodes and the base and leave part of it where it is.

author: TimFox
author_rank: Super Contributor
timestamp: April 30, 2020, 09:30:10 pm
content: To get good clamping action with a Zener to ground, I suggest biasing the Zener to a current above it's knee specification, and connecting the clamped node to the Zener through a fast diode. Any large current through the fast diode will flow through the Zener, raising its voltage slightly (according to the Zener resistance).

author: David Hess
author_rank: Super Contributor
timestamp: April 30, 2020, 10:24:35 pm
content: The Motorola/On application notes linked earlier show a different configuration where a capacitor placed in parallel with the zener diode provides low AC impedance and the dynamic performance only relies on a forward biased rectifier diode. Then the turn-on time and dynamic impedance of the zener diode are much less important.I know you gave the frequency but it would have been helpful to have your drawn graph of the waveform include time.Quote from: free_electron on April 30, 2020, 05:21:30 pm"Hot carriers" will cause dislocations in the silicon lattice which act as recombination points lowering minority carrier lifetime and hfe. This is a major problem for high gain transistors but of little consequence for power transistors and gold doped transistors which operate at low hfe anyway.

author: graybeard
author_rank: Frequent Contributor
timestamp: April 30, 2020, 11:14:50 pm
content: Quote from: David Hess on April 30, 2020, 10:24:35 pmThe beta degradation mechanism has nothing to do with dislocations. It is a surface effect at the SiO2-Si interface between the base and emitter. It is calledMcDonald Effectafter the guy who first described it.Hot carriers due to the reverse bias of the base-emitter junction get enough energy to occupy traps at the SiO2-Si interface. The charge that is then stored in the traps at the interface bends the bands at the interface and changes the relative energy level of the traps and they become recombination/generation centers that both allow conduction in the Si at the interface. This provides a path for current from base to emitter that bypasses the active area of the transistor.What McDonald did to show this was the mechanism was add at gate electrode over the junction that allowed him to modulate the surface conduction with applied bias.The rate at which the degradation occurs is exponentially dependent on the applied reverse bias and the construction of the transistor.Degradation shows up first at low forward current current levels (~nA) but grows with time and shows up at higher levels as the current through the leakage path increases.The reverse bias at which it occurs depends on the construction of the transistor. In high frequency Si BJTs and SiGe HBTs it can be as low as 3.5V and much less than the breakdown. In low frequency power devices it can be higher. However it always starts before the base-emitter junction goes into hard breakdown.In this case the specified breakdown or "maximum reverse bias limit" of the devicetheleakydiodeis using is specified at 9V, which is very high. I suspect if he limited the reverse voltage to 7.5-8V there would be no issue.