Help Reading this LL N-FET data sheet FQP27p06 and Choosing opamp for dummy load

[rwgast_lowlevellogicdesin]

Im trying to figure out the gate voltage to current ratio for this FET. I want to use it to make a digital version of daves Dummy load. This will be the first time I have ever used a FET, but i've done a lot of reading about how they work.

Anyways graph number 1, on characteristic is what I assume Im looking for. The problem is on its X and Y instead of saying like 2v, 10amps, it has like 10 to a power. Im sorry if this is really stupid bu I cant figure out how to interpret this thing and accurately figure out the gate voltage to drain current in the 1-5v region.

I need to figure out how much voltage to drive through the amp, to get 10 amps, I plan on using active cooling.

Also I have LM358s, the same as an lm324, only dual channel right? Then I have these on the way

http://ww1.microchip.com/downloads/en/DeviceDoc/21669D.pdf

will the microchip amps get a full 0 to source voltage rail to rail, unlike the lm358? Thats what im assuming by the description.

[mariush]

https://www.sparkfun.com/datasheets/Components/General/FQP27P06.pdf

You have there on the second page a graph with the relationship between the gate-source voltage and the current.

But this FQP27p06 is a P-channel mosfet, are you sure your dummy load is meant to use a p-channel fet and not a n-channel fet?


I'm using a FQP50n06 with one of those microchip opams and it works just fine.

[rwgast_lowlevellogicdesin]

Wrong data sheet, have both N and P LL FETs. FQP30N06L http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Components/General/FQP30N06L.pdf

[rwgast_lowlevellogicdesin]

The data sheets are layed out alike. I do not see any graphs on page two. They start on page 3, and Im sure graph 1 is what I am looking for, it is just formated weird as I said using some 10 to the power of for the axis labels, instead of just a list of voltage vs amps

[rwgast_lowlevellogicdesin]

This is thr graph im looking at for the N-Fet, it is what im after correct?? What I dont understand is how do you figure out where 2 volts would be, and the amount of current the fet will pass at 2v. I mean I know how to read a graph but all this 10 to the 0 power stuff is what is tripping me up, along with the multiple curves. I was excpecting a single line/curve labeld as 1v 2v 3v on the X along with 1A 2A 3A on the Y

[mariush]

The title of the page talks about FQP27P06 which is a p-channel. 
There may be an n-channel but that one has another name as you obviously noticed : FQP30N06L. 

Don't confuse them.

Since the opamp gets connected to the GATE of the n-channel, if i don't have a brain fart now, you should be looking at Figure 2, transfer characteristics.  On the bottom (horizontal) you have the voltage and on vertical you have the current.

[gibbled]

10 to the power of zero equals 1, 10 to the power of 1 equals 10.

[poodyp]

All you have to do is punch 10^0 into google and it'll tell you that's 1. Just think of it this way, it's the number of zeroes to the right. So 10^1 is 10, 10^2 is 100, 10^-1 is .1 and so forth.

For that FET, 1-3v isn't rated,. You'll have to figure it out for yourself.

[kxenos]

The reason you want to drive the mosfet with an op-amp is that you don't want to mess with I=k(Vgs-Vt)^2 because it's not a linear function.
That's why you place the gs in the feedback loop. Then it is op-amp's business to figure out the required Vgs. When I get access to a proper PC I'm going to explain better. (except if an other one does).

[rwgast_lowlevellogicdesin]

Thank you, I look forward to it! I was wondering why not just drive the FET straight from a micro controller, but the above answered that a little bit.

Im still struggling to figure out the graphs, its just that there are multiple lines/curves. So im not sure what im looking at. Other FET charts make much more sense. Basically im feeding the opamp with a DAC and and trying to figure out the correct voltage range to feed the opamp for 1ma through 10A.

[baljemmett]

I'm still struggling to figure out the graphs, its just that there are multiple lines/curves. So im not sure what im looking at.

Each curve represents the I-V relation for a given gate voltage -- see the legend at top left -- so you want to pick the one that's closest to the maximum Vgs you'll be providing.  Or alternatively pick one that indicates the current you want to pass and then choose your gate voltage accordingly.

Basically im feeding the opamp with a DAC and and trying to figure out the correct voltage range to feed the opamp for 1ma through 10A.

Well, 1mA is well off the bottom of the graph (the axes intercept at Id=1A,Vds=0.1V), and for 10A you want any of the Vgs curves apart from the lowest which doesn't quite get you there (it crosses the Id=10A gridline at Vds=9V).  So anything above 3.5V Vgs should be fine, and probably the higher the swing you can get the better.

[kxenos]

Thank you, I look forward to it!

So, as I was saying (joke by Sir Ken Robinson) in this image http://www.elin.ttu.ee/mesel/Study/Courses/Biomedel/Content/CircDesg/Fig3.gif you can see that the relation of Vgs and Id is not linear. That means that there is a region that very very small changes in Vgs have a big impact in Id. If one wanted to control Vgs directly with, say, a DAC, he would have to calibrate his values accordingly. Plus, in this non-linear region the step from one DAC value to the next would have a big difference in Id.
In this image http://www.oe.uni-duisburg.de/Reports/jb95/ni1-2.gif you can see an op-amp driving a FET. You can see that the current Id that passes through the FET passes also from R1. So the voltage drop Vr is equal to Id/R1 so it is proportional to Id. This voltage then is routed to the inverting (-) input of the op-amp. The non-inverting (+) input is biased with a constant voltage, in this example from a DAC. As a mnemonic rule of thumb you can say that an op-amp with a negative feedback will try to make it's both inputs equal. So, in this situation, the input that can be controlled by the op-amp is the inverting one. So the op-amp tries to make V- equal to V+. But V- is the voltage drop across R1 and while controlling this voltage it also controls Id. So you have Vdac = Id/R1 and thus you have a linear control of current.
Hope it helped clearing things up a little bit.