Does this MOSFET datasheet contain missing data?

Question

This is the datasheet of a mosfet and I have two issues:

http://www.ram-e-shop.com/ds/tr/IRFP150.pdf

First, I can not find the pinout diagram. I don't know where drain, source and gate are. Which pin is gate and which is drain?

Second, Drain-to-Source Breakdown Voltage (VDSS) = 100 V.

When I hear the word "Breakdown", I feel like it is OFF characteristic. I think if I applied more than 100V while the transistor is OFF, A damage will occur. What I mean by "OFF" is when the gate voltage is zero (no gate signal) or when the MOSFET is reverse-biased (i.e. Drain voltage is 0V and source voltage is +100V).

But I don't know the normal drain-source voltage when the MOSFET is on. What is the maximum supply voltage or the maximum operating voltage? Is It 100V too?

Thank you very much,

Answer 1

It might help if you looked at the safe operating area for the MOSFET in question: -

The X axis is voltage and you can see it is limited to 100 volts. The Y axis is current and this is limited to 140 amps i.e. the "Pulsed Drain Current" specified in the "Absolute Maximum Ratings" table on page 1.

What this tells you is that if you could turn on and off the MOSFET in 10 us you could take 140 amps but with a limited voltage of about 60 volts. That's an instantaneous power of 8.4 kW.

Alternatively you could withstand 100 volts but at a reduced current of about 90 amps. That's an instantaneous power of 9 kW.

This is very much defined as a "single pulse" (bottom left corner of graph) i.e. it is a single event of 10 us.

If your pulse was 10 ms then you could withstand 100 volts whilst taking no more than about 2.5 amps. Notice now that the power taken during this much longer pulse is only 250 watts.

Eventually (if the graph showed this detail) you would find that the continuous power rating of 160 watts means a continuous current of 1.6 amps whilst withstanding 100 volts. Or it could mean 10 amps at 16 volts or 16 amps at 10 volts etc..

What I mean by "OFF" is when the gate voltage is zero (no gate signal) or when the MOSFET is reverse-biased (i.e. Drain voltage is 0V and source voltage is +100V).

You can't apply anything more than about a volt in reverse. The body diode in the MOSFET will conduct and, if the reverse supply could provide more than 42 amps, the body diode would break and so would the MOSFET. See "Source-Drain Ratings and Characteristics" for numbers on page 2.

See also this on the last page: -

Answer 2

When the FET is fully on, it behaves more or less like a resistor of value RdsON, which is 36 mOhm in this case, so there would be no way to have 100 volts across it.

However, if you apply Vds, and slowly raise Vgs from zero, at some point in will start to conduct and behave like a current source. See datasheet characteristic. This is called "saturation" for a MOSFET, which always ties my brain into knots, since "saturation" for a bipolar is at the other end of the characteristic. Some prefer to call this "linear mode", since this is the mode to use when you want to make a linear amplifier or a source follower with the FET.

When Vgs is high enough, then it will enter ohmic mode and meet its RdsON spec.

Breakdown maximum Vds applies in all modes. For example, if you set Vgs so that Id=1 mA, then you cannot have more than 100V Vds. It is determined by the physical construction of the FET.

Also, both instant and average dissipation are limited (see SOA graphs) which will give you a time- and Vds-dependent current limit if you don't want to fry it.

when the MOSFET is reverse-biased (i.e. Drain voltage is 0V and source voltage is +100V).

Internal diode would conduct and prevent this.

As for the pinout, you can bet on the standard one, ie GDS.

Answer 3

But I don't know the normal drain-source voltage when the MOSFET is on. What is the maximum supply voltage or the maximum operating voltage? Is It 100V too?

No. You have to consider allowed power dissipation and maximum current too. What will happen if you apply 100V on 0.036 ohm (which is Rdson)? The current would be 2777 amperes. And power dissipation 277 kW. From datasheet you can see, that maximal allowed current is 42A, and power dissipation 160W. You would be orders of magnitude somewhere else = MOSFET would die immediately.

Answer 4

Pin assignment is buried within the mechanical drawings. The most important rule about any data sheet is this: Never miss the small print, let alone the footnotes. Data sheets are marketing. The pin assignment is, of course, nothing that the manufacturer wants to hide from you, but there are many examples where something is important - it just doesn't look good in the advertisement (= data sheet).

ON-state voltage: For (fully-on) MOSFETs, you get a drain-to-source ON resistance, which is specified in the data sheets. Using the resistance and the current you pass through the MOSFET, you can calculate the ON state voltage using V_ON=R_DS,on*I_Drain. The Safe Operating Area will tell you what the MOSFET can take under any worst case scenario you are using it in.

Answer 5

Yes, it is complete.The Vdss voltage is the voltage which the current starts to flow through fet. It is very high since it is a Rectifier that which is used for rectifying AC voltage levels at 120VAC or 220VAC.