# Completed ProjectMOSFET Source to GND voltage

#### anishkgt

Joined Mar 21, 2017
549
Hello All,

This is a spot-welder project that i plan on starting, now in the design phase. After a spot weld is done there would be some voltage returned to the ground plane so should it be connected to the common GND or via a 10k resistor. Its obvious that the MOSFET would need to be conneted to the GND without a doubt to complete the circuit but just curious to know if there is any advantage of adding a resistor from a N-MOSFET Source to GND. The voltage from drain-to-source is 8.1(MAX) from ultracapacitors in 3s2p configuration.

#### ZCochran98

Joined Jul 24, 2018
192
Frequently (in small-signal amplifiers, anyway), source-to-ground resistors are there for two primary reasons: to allow finer control of the drain-source voltage (which controls operational point), and to mitigate against thermal runaway. If there is a resistor at the source to the ground, it tends to consume some power, rather than the MOSFET taking the full brunt of whatever power is being drawn. This in turn, theoretically anyway, prevents the MOSFET from overheating and hitting that "thermal-runaway" point, where the MOSFET will start heating up more quickly as its conductivity rises (MOSFETs are weird in that, for a while, as temperature increases, their effective resistance decreases), which causes higher current to be drawn at the same drain-source voltage, and eventually burn itself out.

So, tl;dr: it's not always necessary to have that source-to-ground resistor, but in some cases it's helpful in protecting your MOSFET and controlling its operational point.

#### anishkgt

Joined Mar 21, 2017
549
Thanks man. I appreciate it.

So in a way its good to have it in my design. I always thought it as to clean the 'return-voltage' (if thats the term for it) of any voltage when it enters the GND plane so that other components would not be affected. So would this only make sense in inductive loads i suppose ? So if were to have one how would i determine the value and wattage ? i was thinking 10k 1/4W ( I = V/R)

to allow finer control of the drain-source voltage
Can you elaborate on this a bit more. Finer control in the sense, the rise and fall of the Gate voltage ?

cheers

#### ZCochran98

Joined Jul 24, 2018
192
Prepare for a lot of reading (apologies)....

By "to allow finer control of the drain-source voltage," I mean this: If you just have a resistor at the drain-supply connection, you can, after doing some rather tedious math, control the voltage across the MOSFET's drain and source pins, simultaneously controlling the current through it (and the resistor). The drain-source voltage controls the operational point, which, in turn (combined with the gate-source voltage), determines what the current through the MOSFET will be. However, only having one resistor only gives you one parameter to tweak in the case you need to adjust something later on (if you have a resistor from drain to supply to begin with). The second resistor gives a second parameter to allow you to adjust the operational point further, if desired. Now, in your circuit, you don't have a drain-to-supply resistor (and those aren't always necessary), so having a source-to-ground resistor will allow you to control the operational point of your MOSFETs, if necessary.

To determine the wattage you need to know how much current is going to pass through those MOSFETs, as you already know. Now, you have three MOSFETs connected to that one resistor, so you may want to consider giving each of those MOSFETs its own resistor, rather than one for all three (it's more parts, but it makes the calculations easier and allows for lower-power resistors). If you know what kind of MOSFETs you're going to use, the gate-source voltage you plan on using, and the current you ideally want your MOSFETs each to draw, you can use the datasheet to figure out the drain-source voltage and from there determine the size of the resistor you need (if you want to be super-precise about it). If you know the drain-source voltage you want, you can determine the current the MOSFET will draw, also based on the datasheet, and use that value to determine the resistance you need: (Vsupply - VDS)/IDS will give that value of R, which, in turn will also let you calculate your required power (via P = IV = I^2*R = V^2/R, as you had). I'm guessing 1/4W will probably be sufficient, but without knowing MOSFET type you're planning on using it's only a guess at the moment. There is a more precise way to determine the values, but it's also much more tedious.

Now, I do have a couple questions for you: first, did you intend to ground your Electrode- line? Because the way you have your circuit drawn at this moment, your Electrode- line goes straight to ground, which will result in no voltage across those MOSFETs, making them effectively useless. Second, what is the purpose of the Q1n_Gs lines? As far as I can tell, they're not going to control the MOSFETs' behavior at all. According to the datasheet for your gate drivers you've selected, the SOURCE and SINK pins are typically connected together. Then, a (very large - typically on the order of hundreds of kilohms to even, in some cases, megaohms) resistor connects the gate pin to the ground. If you haven't already, take a look at the datasheet here; the last page has a good application circuit example. Your 100 ohm resistor would be sufficient for R1 in it's example, and R2 would be the massive resistor. You don't have an RLOAD, but your source-to-ground resistor would function equivalently (just in a different place).

Hope that helps!

#### anishkgt

Joined Mar 21, 2017
549
Thanks that did help clear out some things. My bad about the schematic. Got it wrong. Hope the attached schematic clears things. The electrodes are CAP+ and Electrode- The Gate driver Source connects to the Gate of the MOSFET and the SINK pin is connected with a schotky diode in series for quicker discharge.

#### ZCochran98

Joined Jul 24, 2018
192
Ah, I see now - makes sense. Good luck on the rest of your project! Glad to have been of some assistance.

#### anishkgt

Joined Mar 21, 2017
549
The Gate voltage would be Vgs = 10v so based on the datasheet of the MOSFET it would be ON allowing current to flow.
you can use the datasheet to figure out the drain-source voltage
Isn't it something we determine ? The CAPS are charged to 8..1. A voltage of 1.2 to 2 is generally seen during spot-weld. So based on that
(Vsupply - VDS)/IDS if am correct it would be
ie average current during a spot weld is 1100A so 1100A/number of mosfet is 366A each mosfet would handle.
(3v - 10v)/366A = 2.97R
So would that be 2.97Ohm in series to each MOSFETs source to GND. Did not get how you did the Power calculation. P=IV i can understand, i lost the remaing part of it. Finally its
$$P=V^2 R$$ so that would $$P=3^2 2.97=3W$$

#### ZCochran98

Joined Jul 24, 2018
192
For power, the derivations are:
$P = IV = \left(\frac{V}{R}\right)V = \frac{V^2}{R}$
$P = IV = I\left(IR\right) = I^2R$

You sort of determine the drain-source voltage - the current going through the MOSFET will pass through the drain and source (so IDS) to the resistor. So the 8.1 volts will be dropped by IDS*R (whatever value you choose for R), leaving VDS = 8.1 - IDS*R (Kirchhoff's voltage law). This value of VDS combined with the value of VGS (VGS = VG - IDS*R) will, based on the datasheet, "select" the current you're running at.

However, in the case of your design, you have three MOSFETs each pulling around 366 A (you're going to need some serious heat sinks on those). I hadn't thought of it when initially answering your question, but with that number in mind, you'll probably not want a resistor there at all - 366A * R for almost any resistor you get will far exceed the 8.1V supply, making the MOSFETs effectively useless (8.1 - 366*R = VDS < 0, which is not useful, especially with N-channel MOSFETs). With that in mind, you can get rid of that resistor entirely; the MOSFETs will be dissipating so much power anyway. And as I mentioned, it's not necessary to have that resistor, just "nice to have," especially for low-current or small-signal applications. Upon realizing your circuit is anything but low current, you'll probably be fine without the resistor there. Just, again, you'll need a massive heat sink and possibly some kind of airflow to keep the device within its appropriate temperature range (typically between -50 and 175 Celsius). I would recommend, if you haven't selected a MOSFET for it yet, to go with a MOSFET that can handle 400 A continuous current (for example, IXFH400N075T2, which I found on Mouser - it's not cheap, but it can handle 400 A continuously. Datasheet link seems to be broken, though, so you'd need to go digging to find its datasheet. There's another that costs 4x as much, but has a datasheet directly on Mouser).

Hope this helps/clears things up!

#### Alec_t

Joined Sep 17, 2013
12,441
Where did the relation (3V-10V)/366A come from? It resolves to a small negative fraction, not 2.97Ω!
A source resistance of 2.97Ω would limit the MOSFET drain current to about 8.1V/2.97Ω = 2.7A per MOSFET (assuming an ideal MOSFET with a negligible Rds(on)). That's a long way from 366A.
Get rid of the source resistors, unless they are being used to promote load sharing (in which case their value would be only a few milliOhms).
Btw, P=V²/R.
In which component is the power being measured?

#### ZCochran98

Joined Jul 24, 2018
192
Get rid of the source resistors, unless they are being used to promote load sharing (in which case their value would be only a few milliOhms).
Basically what I said, in way fewer words.

But yes: also the (3-10)/366 [=-19.1 mOhm] - I'm not entirely sure where that came from myself (I think a backwards version of (Vsupply - VDS)/IDS), but it would come down to only a few milliohms. I'm also not sure where the choice of VDS (I'm guessing the 3V?) comes from - maybe from a datasheet?

So yeah: it is recommended you not have those resistors, thus greatly simplifying some of the work required to design it. Especially for as much current as you're wanting to draw.