| ??? 08/07/03 17:01 Read: times |
#52250 - ESD calculation Responding to: ???'s previous message |
Hallo Mahmood, hallo Lee,
advantage of zenerdiode in preventing damage is not only the existence of threshold voltage but also the rather high junction capacitance. Standard zenerdiode needs about some tens of nanoseconds for fully turning on. In the first nanoseconds, standard zener diode has no voltage limiting property, then only junction capacitance can do the job. Standard 4V7 zener diode has a junction capacitance of about 200pF at reverse voltage of about 2V. This true for a 500mW type. Bigger zeners have higher junction capacitance. Doubling power rating is normaly achieved by doubling junction area, which also causes doubling of junction capacitance, as rule of thumb. Keep in mind, that many people do not always mean pure zener diode, when they are talking about zeners. In such applications as can be seen here, often transzorbs are used instead of pure zeners. That's important, because transzorbs are not only extraordinary faster than standard zeners, but also show drastically increased junction capacitance, which is of extreme importance, when dealing with ESD events. 1.5kW transzorbs are turning on in only a few 100psec! If longer turn-on times are observed, then this results only from unavoidable inductivity of terminals. But the pure transzorb element is extremely fast. For a 4V7 transzorb junction capacitance is about 10nF! Although in some situations very disturbing and annoying, here in our application, this high junction capacitance is very helpful, as we will see in the following rough calculation. Standard 1A rectifier diode shows about 20pF at reverse voltage of 1V. It's interesting to note, that this is true not only for standard diodes, but also for fast recovery types. This, at least, I could observe, when carefully anaylizing databook of Motorola. Let's have a look at the following schematics: ![]() When dealing with ESD events, high frequency equivalent circuit has to be taken into account. So, 11k resistor mutates to a complex circuit, consisting of parasitic inductance, caused by wire wounding, and parasitic capacitance (C2 and C5), caused by interwinding capacitance. It was assumed, that 11k resitor is a high power, wire wound type. Estimation of parasitic inductance and capacitance is not so easy. It highly depends on used resistor type. For the following calculations we want assume a parasitic capacitance of about 2pF. C3 and C6 were discussed, already. Consequently, also for 100R resistor a parasitic capacitance must be assumed. But here a metall film or carbon film type is used normally, which shows only very small parasitic capacitance, less than about 0.5pF. Associated time constant is in the psec range and can be neglected. If we now connect a 200pF capacitor, charged with 8kV, to our circuitry, acccording to well known 'machine model', at the same time nearly 8kV is dropping across our circuitry. This true, because series capacitance of C2 and C3, or similar C5 and C6 is smaller than 2pF, and this is much smaller than the 200pF of C1 or C4, means, C1 or C4 does not loose much charge. And, not to forget, there is no charging resistance which could introduce any slowing down time constant!! Now, both circuits become different looking: In Michaels circuit C3 is charged to about 80V, while at the modified circuit C6 is charged to about 800V! While in the modified circuit there's NO current limitation, opto-LED will be destroyed! But in Michael's circuit current through opto-LED is limited to about 80V / 100R = 800mA. Datasheet of 4N35 tells, that up to 3000mA are allowed for a duration of 1µsec. So, when duration of this 800mA current flow is limited to less 1µsec, then opto-LED is protected! How long will the current flow? There are two time constants, which are involved in the decaying of applied high voltage: C2 x 11k = 22nsec and C3 x 100R = 20nsec. So, this 800mA current flow through opto-LED, will not last longer than about 100nsec, which means, that opto-LED is protected. As you can see from this rough calculation, not only the current limiting 100R resistor is needed for protecting, but also the rather high junction capacitance of zener diode. When using a transzorb instead of zener diode, C3 becomes about 10nF and voltage drop is about 1.6V, when applying 8kV to circuitry! So, in contrast to zener diode, threshold voltage is NOT EVEN reached during an ESD event, which makes the circuit almost undestroyable. Question arises, whether 4V7 zener diode will withstand 80V. Well, some manufacturers say 'yes' and recommend them as direct ESD protectors. But, because I'm paranoid, I would only use a transzorb. And I would NOT omit this 100R current limiting resistor, either! Some of you might ask yourself, whether Michael has really made this calculation. I don't even think so. A really experienced person like Michael, with decades of being in business, JUST KNOWS IT! Hard to believe, but true.... Regards, Kai |




