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I am using mosfet (IRHJ9A7234) as a load switch and Id current is about 0.7A and maximum Rds from datasheet is 110mohm. So I thought based on power calculation P= I^2*R = 77mW which is way below than rated power of 75W so I thought it was safe to use for my application. Now I am looking at the SOA curve as shown below. Vds in my case is Id * Rds = 77mV. However I don't see this range in Vds and not able to align with the conclusion I made with power calculation. So which one (SOA or power calculation) do I need to refer and make decision whether this Mosfet is good to use for my application? Same thing for Id vs Vds charateristic curve.. I do not see Vds below 0.1V enter image description here

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  • \$\begingroup\$ You forgot to square: V = I*R = 0.077V, times 0.7A again is 54mW. \$\endgroup\$ Commented Mar 1 at 3:04
  • \$\begingroup\$ Thanks. But it dose not affect my original question. I will rectify the number tho \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 3:48

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If the MOSFET is as "on" as it's possible to be, then \$R_{DS}=110m\Omega\$. In this condition the graph of \$I_D\$ vs. \$V_{DS}\$ will be a straight line:

$$ I_D = \frac{V_{DS}}{R_{DS}} = \frac{V_{DS}}{0.11\Omega} $$

This condition is represented by the diagonal portion of line at the upper left, shown in red here:

enter image description here

Since it is technically impossible to obtain (DC) currents above this line, due to the minimum attainable \$R_{DS}\$ of 110mΩ, you need not concern yourself with that region.

In other words, it doesn't make much sense to plot values above that line in a graph of SOA, since you can't even get there in normal operation.

Your application requiring \$V_{DS}=77mV\$ and \$I_D=0.7A\$ will also necessarily lie on this line, and not above it, and is therefore "safe".

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  • \$\begingroup\$ Hey Simon, thanks for your explanation. How dose this relate to maximum power rating from data sheet? It’s 75W. By looking at the graph my condition is near or on the red line. As I mentioned in the post, my calculation of power loss is way smaller than maximum power. However, what you are telling me seems to be close to the max SOA.. I don’t understand. What if I want to drive 2A on this mosfet as a switch(ohmic region) is it possible to achieve ? Also do you also agree that VDC line in SOA from its datahseet is somehow off? It dose not match max rating. \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 1:57
  • \$\begingroup\$ @AliaSana Your condition, in the Ohmic region, is smack-bang ON the red line, for the reasons I just explained. It's fine. Anything to the left of the DC line is fine, since you can never get above that red line anyway. I calculate maximum power to be \$P=I^2R=(0.7A)^2\times 0.11\Omega=54mW\$, which is a thousand times less than the maximum of 75W. The thing will get slightly warm, that's all.... \$\endgroup\$ Commented Mar 1 at 2:11
  • \$\begingroup\$ I see. Thanks Simon. Last question, if you look at DC line in SOA, for example, 100V and 0.2A, shouldn’t this power (100x0.2) be the same number as maximum power stated from datasheet? So 75W.. why is it off so much ? Unless SOA graph is drawn somehow wrong \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 2:15
  • \$\begingroup\$ @AliaSana The DC line will account for any rise in \$V_{GS(TH)}\$ with temperature, which, for any given fixed \$V_{GS}\$ would effectively increase \$R_{DS}\$, possibly leading to thermal runaway. This graph shows you where you need to operate in order to guarantee no thermal runaway, which, for long-term, DC operation, is necessarily less than 75W. \$\endgroup\$ Commented Mar 1 at 2:20
  • \$\begingroup\$ I understand that. What I’m trying to say is that 75W is based on temp range between 25 and 150 Celsius as shown in the graph and in datasheet (Tc and Tj). My point is that DC curve in SOA, its Vds x Id on the DC line should match with same power as 75W or close to it. I remember discussing with someone from Infinenon long time ago that their graph is sometimes drawn wrong so I just wanted to make sure that it is the same case. \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 2:30
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In this case use your power calculation, you would have to extend that graph a long way to the left to get your case.

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  • \$\begingroup\$ That wouldn’t make any sense tho. If graph is extrapolated such that I can find operating point, it wouldn’t match my load current (0.7A) and Vds (77mV). Like it won’t be in SOA dc area. Also are you sure that we are allowed to extrapolate the graph ? \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 1:14
  • \$\begingroup\$ Even If extrapolated, max power is approximately 1A * 0.1V = 100mW which is close to my operating point (77W) and this contradicts to the margin I calculated above. 77mW vs power rating 75W whereas 77mW vs 100mW as you suggested. I think there seems to be something wrong on the graph itself.. DC line should yield maximum power rating stated from datasheet but it dose not yield close to 75W. For example, at Vds = 10V and Id = 4A gives 40W not 75W. \$\endgroup\$
    – Alia Sana
    Commented Mar 1 at 1:31
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    \$\begingroup\$ @AliaSana The power dissipation rating applies when no other limits do; transistors also generally exhibit 2nd breakdown, limiting power dissipation as voltage rises. This is reflected by the slope being less than 1:1. Notice ~70W applies at low voltage (20A, 3-4V). In any case, you are well below this curve. \$\endgroup\$ Commented Mar 1 at 5:20
  • \$\begingroup\$ The sloped line does not indicate Ids being limited by power dissipation but by Rds you already calculated that which would put you on that line. \$\endgroup\$
    – RoyC
    Commented Mar 1 at 8:29

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