I've tried researching this extensively, checking existing threads, but I suppose with any uncertainties in this subject, they are hard to convey properly and therefore hard to find answers to.

So I understand static electricity (I think) and don't need to be convinced about the merits of taking precautions to protect electronic components from getting fried. But I fear there are gaps in my knowledge and/or don't understand things correctly. So I'll ask questions as I try to convey my understanding.

My basic problem is: I need the right setup for working on small electronic components. Let's say a RAM chip.

For working on internal PC components, most places I've come across suggest use of an anti static wrist strap hooked to the metal chassis of the computer and that's it.

Question 1: The internal components are all eventually connected to the conductive metal, so won't they absorb the static discharge from your body?

Then there are places which suggest connecting yourself to an ESD mat using an anti static wrist strap.

Question 2: Wouldn't this just create an open circuit to ground?

Question 3: The mat is high resistance so electrons from you would flow very slowly through it. I know electricity still moves really fast but my mind doesn't and I'm seeing discrete intervals of time. So I'm thinking that if you touch the sensitive component straight after connecting to the mat, you might still discharge to the component. So how quickly does your body take to achieve neutral charge?

Question 4: Can there be a state where you and the mat achieve equilibrium but your body still has a charge that can damage the component if touched?

Question 5: Can there be a charge in the mat that can transfer to the component once the component is rested on it?

Finally, some have suggested connecting yourself using a wrist strap to an ESD mat and then connecting the ESD mat to mains ground with the outlet switch off.

Question 6: If the switch comes on by accident or there's a fault somewhere on the circuit, will the high resistance of the mat stop the current from flowing through you freely and hence stop you from getting fried?

Question 7: Even after you've discharged, wouldn't you, the mat, and the strap be on a different (and possibly higher) potential than the component (which currently sits outside this circuit) you will now touch and therefore damage?

Now let's say with the previous setup, you are working away on the component that is resting on the mat and it's still fine. Then you rub your feet on the carpet subconsciously and gather static.

Question 8: Will there be a constant discharge through the mat while you gather charge? If so...

Question 9: Wouldn't the discharge cause current flow through the mat where the component is resting? and if so...

Question 10: Could that current damage the component?

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    \$\begingroup\$ For Q2, ESD mats are connected to ground. So if you connect yourself to the mat you also connect yourself to ground. \$\endgroup\$ Apr 26, 2017 at 16:04
  • \$\begingroup\$ With one hand, touch the GND copper of the PCB. With other hand...... \$\endgroup\$ Apr 26, 2017 at 16:38
  • 1
    \$\begingroup\$ Not gonna read all that. 10 questions for asking about ESD mats is way over the top. \$\endgroup\$ Apr 12, 2018 at 11:03
  • \$\begingroup\$ There are several questions on ESD mats with about as many questions, maybe it would be good to check those first. \$\endgroup\$
    – Voltage Spike
    Apr 12, 2018 at 22:36

2 Answers 2


Question 1: The internal components are all eventually connected to the conductive metal, so won't they absorb the static discharge from your body?

A wrist strap ensures that the potential between you and the computer is near 0V instead of thousands of volts that comes from clothing and walking across the floor. The damage comes from current flowing from you into the small gates that make up transistors. If the voltage between you and another object is 0V no current can flow.

Question 2: Wouldn't this just create an open circuit to ground?

No, when you connect the wrist strap to the mat, they really mean to ground, look at the diagram below and wrist strap two. Wrist straps should be bought with the 1MΩ resistor built in.

enter image description here

Answer to Questions 4-10:

An ideal ESD setup will lower all potentials to near 0V (which means connecting it to earth ground). There will be no current transfer into parts because there is no voltage to start with, it has been lowered to ground by a wrist strap or mat. These have a 1MΩ resistor to prevent a short circuit to ground in the event the user or other voltage should come in contact with the mat or wrist strap.

If you have a wrist strap only to the computer that's probably fine in most cases. A better setup is one with a mat and a wrist strap to ground. A true ESD controlled environment has the floor, the table surface (and no paper, or plastics that are not ESD safe), tools and lab coats (to prevent electric fields from clothing) and the humidity of the air controlled.

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    \$\begingroup\$ A concise answer to an insanely broad question. +1 \$\endgroup\$
    – uint128_t
    Apr 26, 2017 at 16:11
  • \$\begingroup\$ Yes agreed, it did turn out to be an insanely broad question but it was a case of get all these suspicions cleared together at once or understand nothing :) \$\endgroup\$
    – Ash
    Apr 26, 2017 at 16:49
  • \$\begingroup\$ @laptop2d Thanks, this does clear up a few things. In the case where you don't use a mat, I get that there will be 0V between the chassis and the contact point where the strap meets your wrist thanks to the high resistance of the strap. But what if your palm touches the chassis. Won't that produce a circuit where the built up charge can escape down your arm and through your fingertips and not via the high resistance path from wrist through the strap? Apologies if that doesn't make much sense. \$\endgroup\$
    – Ash
    Apr 26, 2017 at 16:56
  • \$\begingroup\$ If you have the strap on your arm before you touch the chassis, they will be the same potential, even though it is high resistance, the charge is small. \$\endgroup\$
    – Voltage Spike
    Apr 26, 2017 at 18:45
  • \$\begingroup\$ Are we assuming in this case that the PC is connected to mains ground? \$\endgroup\$
    – Ash
    Apr 27, 2017 at 1:30

As an answer to question 3, which is very interesting, I think a look at this webpage or this webpage, which discusses the importance of discharge rates when designing electronics, would help clear many things up.

EDIT: Some explanation from the above link (this is going to be a bit long, but very extensively descriptive, so bear with me please):

Take into account three things:

(A) Consider the Human Body Model (HBM). The ESD Association models it as a 100pF capacitor that is discharged through a switching component, along with a resistor of 1.5kΩ connected in series (ESDA's Fundamentals). The website above amends that model a bit to take a worst-case scenario, so the values used are C = 200pF and R=10kΩ.

(B) When engineering for safety (both equipment and personnel safety), we are bound to make a significant "concession": There is no such thing as total annihilation of an ESD event. Yes, you read that right, things, equipment, and even standards about precautionary measures, have to take into account that an ESD event will happen...multiple times, no less. Remember that notion, it is more important than it reads at this point.

(C) Consider, also, that the ESD mat circuit you mention is, technically, a circuit with resistors in series. Typically, everything you touch is "connected" to the mat and from there to ground through a (its) resistor. The "connected to the mat" part is of paramount importance. If you lift it off, it ceases to be in connection to ground through the mat's resistor. Remember that point too, whatever you do with devices on the mat, the ESD events you will cause, drain your charge to ground through the mat, in a controlled (i.e. slowed down) manner, through its resistor and to ground. If you lift your device off, the excess (or deficit) of charge on you is "handled" by your device, and no slowing-down occurs anymore because there is no resistor. So the mat's resistor controls the ESD event.

So, when taking precautions, two time durations are considered. These time durations are the results of two corresponding concerns. (Remember, it is a given that you are grounded and the objects you touch are also grounded, so you are all at the same potential, i.e. you are NOT expected to touch anything "floating", or outside your "ESD mat circuit", although you are prone to make it "flow" afterwards, i.e. literally lift it off).

These two concerns are:

1) When causing an ESD event to a sensitive device on a mat connected to the ground, your charge will travel through the device, then the mat, all the way to ground level. This will happen and you cannot avoid it. However, when you manipulate the device, you may suddenly disconnect it, e.g. grab it, lift it off to relocate it, drop it again, etc. The moment you lift it off, you disconnect it from ground. If the "controlled" ESD event has not "finished", i.e. charges have not been totaly equalized with ground level between all ESD event participants (you and the device), whatever potential (charge) difference remains will be handled solely by the device as a floating device. So, the first time duration is Tmax, i.e. the maximum time it should take for ESD caused charges to drain to ground through the mat before you "disconnect" the device from it (lift it off), because from then on, it's "on its own" to handle the rest of the undrained charge. This time depends on how fast you "do things" with the devices and tools and is delineated in the link above. What is concluded in the webpage is that, from various measurements, while tampering with devices, a worker may disconnect a device from the mat in as little as 153ms. Engineering a safety factor of x2 (double the safety), this is made to be 76.5ms, just to be sure. So, the device you touch and cause an ESD event must participate in such a connection to ground (i.e. through such a resistor), which will drain it in no more than 76.5ms, because if it delays a bit more, a remainder of charge will have to be handled from the device itself with no help from the resistor to slow things down.

2) The second concern is, and always will be, that ESD events can be lightning-fast (pun not intended). Because the speed/rate of an ESD event determines how much power/current the device is forced to take, the speed of the ESD event must NOT be too fast for the device. This is Tmin and is a little trickier (though not the trickiest) to engineer for safety. Determining this value is at the heart of ESD event rates (speeds).

From then on, a mat's resistance for connecting to ground is the main factor of controlling an ESD event's charge dissipation rate. The resistor must be such, that connection to ground drains the charge slower than Tmin but not slower than Tmax. Then you have maximum safety for your device, when working not properly grounded yourself, i.e. causing significant ESD events all the time. If you ground yourself too, then the safety is a lot higher because the ESD events you cause will always have a negligible charge difference (unless you are Flash, that is).

Now, how do you calculate Tmin... what follows is the approach of the link considered above.

i) You (the human) cause the ESD events, so you need to know what you cause. C = 200pF is your capacitance in the worst case (100pF is the model, double it for safety).

ii) Determine what sensitivity level your device belongs to. Take, for example, a very sensitive device, characterized as Class 0, with upper limit of 250V.

iii) Consider the discharge to be almost complete at 99% equalization of voltages. Remember the RC time constant? This means that V(Tmin)/V0 = 1%, i.e. after a Tmin amount of time, you are at 1% of charge difference with the device you touch. In other words, in the time duration to consider, a full ESD event is taken to be 99% of the discharge. Since \$t = -R*C*\ln(\frac{V(t)}{V_0})\$, it is clear that the time depends on the resistance and the capacitance (but you know the capacitance from (i)).

iv) \$I = \frac{V}{R}\$, so the current we will cause when discharging to a 0V device from 249V on ourselves (worst-case), based on our sole resistance of 10kΩ, is: I = 249/10000 = 24.9mA. Engineering a factor of x2 for safety, we must not cause more than ~12.5mA of current to run through the device!

v) We can find the Tmin from this current by considering that we have 12.5mA = 12.5 mC / sec (\$Q = I*t\$). The full charge to flow is expected to be \$Q = C*V\$ = 200pF*249V = 49.8nC This is taken to be constant, so, from the equation Q=I*t, if t gets any smaller (i.e. faster discharge), then I becomes larger!

vi) Tmin = Q/I = 49.8 nC / (12.5mC/sec) = 3.984ms. This is Tmin. Anything faster than that for a discharge will risk running more than ~12.5mA of current through the device, which is about half the maximum amount it can handle because it has been designed to belong to Class 0 of sensitivity.

vii) Either with Tmin and Tmax, or with Imax and Tmax, it is possible to select the range of resistances the Mat circuit must have to dissipate charges of ESD events brought forth on a Class 0 ESD-sensitive device, i.e. based on a maximum discharge from a voltage potential difference of 249V. This can be done by using R = V/Imax to find the MINIMUM resistance of the mat circuit (i.e. the resistance that slows the discharge enough to avoid higher currents), this being Rmin = 249V / 0.0125A = ~ 20 kΩ.

viii) Considering Tmax and a full discharge being at 99% of charge dissipation, we have Tmax = -RmaxCln(V/V0) <=> \$0.0765s = -Rmax*200*10^{-12}*ln(0.01)\$ <=> Rmax = -83058819.7 = ~83MΩ.

So.. provided that voltage differences are going to be at 249V max, a mat circuit with a resistance less than 20kΩ will drain charges too quickly, risking damage from high current. A circuit with a resistance more than 83MΩ, will drain them too slow, risking an incomplete drainage prior to someone disconnecting a device from the circuit (i.e. lifting it off, moving it around, etc.), forcing it to handle the rest of the discharge by itself.

Of course, things are not so simple in real world, so this still is a simplification, but it highlights in an interesting way, the theory behind such safety precautions for ESD.

That about sums it up. I hope I didn't miscalculate or misunderstand anything!

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    \$\begingroup\$ Try to include the main points of "link" answers in the answer. We have no guarantee how long the links will last. \$\endgroup\$
    – RoyC
    Apr 9, 2018 at 16:11

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