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Motors and MOSFETs

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solis365

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This is likely a really simple question but I seem to have confused myself.

Say I am doing a very simple thing: Driving a motor with a low-side N channel MOSFET. Thus VS = 0. Lets say the supply voltage for the motor is 30V. so the circuit is: 30V to +motor. Mosfet drain to -motor. Mosfet source to gnd. Thats it, lets assume the gate is driven by an ideal voltage source.

I want the MOSFET to be in saturation for maximum efficiency. Therefore I must set the gate to some VGS such that VGS-Vth < VDS.

How would I calculate VDS in this situation? The way I see it VDS is going to be determined by the motor current and the RDS(on). So if RDS(on) = .02Ω, and the motor requires 5A, then VDS = 0.1V. Therefore I will want VGS to be very close to Vth such that VGS-Vth < 0.1V.

This makes NO sense to me as you want higher gate voltages to gather more carriers in the channel and thus make it easier to pass current.
I am thinking that in the case of the motor driver, the MOSFET sees 30V and therefore VDS = 30V so your VGS can be something like 10 for saturation... which is how I know the mosfets will work. The motor is a DC short, right?


im looking at the Id/Vd curves on the IRFZ24N datasheet just for a reference.
 

BrownOut

Banned
I think you're getting the term "saturation" in mosfet confused with that term as it applies to bjt's. It means something completely different in mosfet. What you really want is to operate the mosfet in the "linear" or "ohmic" region, in which case, you want

vgs - vt > vds.
 
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solis365

New Member
Forward Active mode in the BJT is where you get current amplification. You want saturation mode for switching: Vc > Vb > Ve. Saturation in a mosfet is sort of analogous to forward active mode in bjts, but...

In MOSFETs, I thought that the best mode for loads would be in saturation, or the point where the most carriers are available in the channel and the channel resistance is the lowest. In the linear/ohmic/triode region of the mosfet, the lower your gate voltage, the higher the channel resistance. So you now are changing the resistance based on your gate voltage... for high loads, wouldnt you want this resistance to be as low as possible, so that it is closest to a closed switch? and the lowest resistance would be found in the saturation mode, would it not?
 
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indulis

New Member
Correct!!! Look at the data sheet... there will be a graph of Vgs vs. Rds on. You don't want the MOSFET to be a variable resistor (ohmic region).
 

BrownOut

Banned
NO! When you're using a fet for switching, you want to be in the ohmic region. The channel resistance will be a function of Vgs in that mode. The higher Vgs, the lower channel resistance. The actual resistance is the inverse of the line on the Id/Vgs curves.
 

solis365

New Member
in the datasheet I have there is no such graph, but I thought that my thinking was right. (http://www.electro-tech-online.com/custompdfs/2009/07/irfz24n.pdf)

But how do you make this consistent with its operation as a switch?

What voltage does the mosfet see (VDS)? if it sees 30V, then VGS = 10V will give VGS-Vth ~= 6V. VDS = 30V so VGS-Vth < VDS ---> saturation

if it sees .1V as I thought (see above post), then I am still wrong. but it must see the full voltage when acting as a switch for the motor.

i.e. motors take a rated voltage, not a rated fixed current, and you just need to make sure you can pass this much current to the motor safely.

If I wanted to control the speed of the motor I would use the ohmic region as this would change the resistance and hence lower the voltage to the motor. (I realize this is not an efficient way to control motor speed and that in reality you would use PWM)
 
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smanches

New Member
I think you are confusing yourself.

According to the datasheet, 10V Vgs (Id 10A) will turn it on with 70m resistance. It's not the drain voltage that affects the Rds_on, but the drain-source current. The higher the current, the harder you have to drive it. There is a graph for this.

Strange that datasheet does not have a Vgs to Rds graph.
 
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BrownOut

Banned
RDS is only vaild for the ohmic region ( hence, the value of resistance ) It can be approximated as the inverse of the IDS/VDS slope in the device's output characteristic graph.
 

MikeMl

Well-Known Member
Most Helpful Member
Here is a plot of FET power dissipation vs gate voltage for a specific example. Note that as the gate is driven harder, the dissipation comes down, but not dramatically.
 

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smanches

New Member
I think I'm starting to understand your question better, but I think you are trying to hard to figure out the ohmic boundry curve.

Is there a reason you're trying to find the exact voltage that puts it into saturation mode, instead of just driving it at the 10V as specified on the datasheet?

Mikes sim shows the various regions. Just left of 1.1V is the cut-off region. Between the 1.1v and about 1.8v is the ohmic region (for his particular current flow) and everything to the right of that is the saturation region.
 
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BrownOut

Banned
Here is a plot of FET power dissipation vs gate voltage for a specific example. Note that as the gate is driven harder, the dissipation comes down, but not dramatically.

Notice however, that dissipation comes down dramatically when you enter the ohmic region. BTW, how do you get the plot for the product of quantities. Help out a LTSpice noob. :)
 

MikeMl

Well-Known Member
Most Helpful Member
If you have already simulated,

to get POWER in a component, place the cursor over the component in the schematic, and hit the Alt key. LTSpice computes the power by generating the expression automatically.

Other tricks:
You can generate a trace on the plot by doing arbitrary arithmetic using things like I(R1)*V(node2,node1) [i.e. power]. Use the ADD TRACE menu pick. See the list of AVAILABLE DATA, voltages and currents in parts of the circuit. You can create algebraic expression of great complexity. See "Waveform Arithmetic" the HELP FILE

Note how to get the "average value" of a trace.
 

BrownOut

Banned
Awesome! I really wanted to create arbitrary expressions. I tried "Add TRACE" That's pretty cool!
 

audioguru

Well-Known Member
Most Helpful Member
The graphs are for "typical" parts. You don't know if your Mosfet is low, typical or high unless you measure it.

So just give the gate 10V as shown on the datasheet so that the on-resistance is guaranteed to be equal to the spec or less.
Note that the on-resistance gets higher (about 1.5 times) when the Mosfet gets hot.
 

solis365

New Member
Is there a reason you're trying to find the exact voltage that puts it into saturation mode, instead of just driving it at the 10V as specified on the datasheet?
I am not, I am trying to understand why 10V. Though finding the exact point would probably help with my understanding of why.

I think this is more of a question about motors, as I know how MOSFETs work and their regions of operation, etc. Imagine Mike's circuit and replace the resistor with a motor. What happens?

The motor looks like a small resistance if I am correct. Thus, when in cutoff, there is ~0 current in the resistor. The transistor sees a VDS of 30V. Thus you need a fairly sizeable VGS to get it to saturation mode. If more current is required and you want to keep power dissipation the same, you need a higher VGS to get more carriers to the channel so they can conduct.

If the above paragraph is correct then my question is answered.


Also, this should be correct:
__________________
For BJTs:
Amplification: Forward-Active mode desired
Switching (digital, power): bjt Saturation mode desired

For MOSFETs:
Amplification: mosfet Saturation mode desired
Switching (digital, power): mosfet Saturation mode desired, minimizing the time the transistor spends in the ohmic region.
__________________


EDIT: I also meant to put this entire thread in general electronics chat rather than projects. I am not working with that specific mosfet, though it seemed like a good example one.
 
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BrownOut

Banned
For MOSFETs:
Amplification: mosfet Saturation mode desired
Switching (digital, power): mosfet Saturation mode desired, minimizing the time the transistor spends in the ohmic region
No, you want ohmic region for switching. The many responses on this thread should make that clear.

For your motor, using 10V on the gate should be sufficient to drive your fet into the desired ohmic region. Once current starts flowing, Drain voltage will drop quickly to a much lower voltage. You will no longer have 30V across the motor, as there will be a very small drop across the fet channel's resistance, but the motor voltage should be very close to 30V, and the fet's collector voltage will be less than 1V.
 

MikeMl

Well-Known Member
Most Helpful Member
...
I think this is more of a question about motors, as I know how MOSFETs work and their regions of operation, etc. Imagine Mike's circuit and replace the resistor with a motor. What happens?

The motor looks like a small resistance if I am correct. ...
A STALLED motor looks like a small resistance in series with an inductance; a RUNNING motor is much more complex, because of the back EMF it generates.

From the standpoint the FET, I was trying to show that once you get to Vgs>than about 2Vt, there is not a huge reduction in power dissipation.
 

smanches

New Member
The motor looks like a small resistance if I am correct. Thus, when in cutoff, there is ~0 current in the resistor. The transistor sees a VDS of 30V. Thus you need a fairly sizeable VGS to get it to saturation mode. If more current is required and you want to keep power dissipation the same, you need a higher VGS to get more carriers to the channel so they can conduct.
You are correct in that it would take an extremely high voltage to go from cut-off directly to saturation. But you never go directly there; you have to go through the ohmic region first. While this transition is happening, you are starting to get some Id flowing, which will reduce the Vds voltage drop. This is then going to require less of an increase in Vgs to reduce Rds even further, and also why the Vgs to Rds graphs are curves.

BrownOut. I think you might have the ohmic and saturation regions in reverse. I've never heard of using the ohmic region for switching at all, and it doesn't quite make sense to me. The lowest Rds_on is in saturation which is what you want when used in switching applications.

I've attached a great paper on mosfets by Fairchild. I keep returning to it every time and it keeps answering my questions.
 

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BrownOut

Banned
No, you have backwards. Read your own paper that you posted. Ohmic is on the left side of the vgs-vt=vds curve. (see figure 6) At turn on, the fet goes through saturation on it's way to ohmic. That's where the big buldge in power is in Mike's simulation.

The confusion is understandable because saturation in bjt's means the complete opposite of what it means in mosfet.

Maybe instead of saturation and ohmic, we should use pinch-off and linear. RDS is only vaild in the linear region (VGS-VT>VDS) Otherwise, ID is constant current, and device resistance is in megohm range, as any good constant current device would be.
 
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smanches

New Member
Maybe instead of saturation and ohmic, we should use pinch-off and linear.
Yes, that's where I was getting confused. I did not realize that saturation started at the same place the pinch-off effect did.

I was thinking it was more like Cut-off, Ohmic, Saturation, pinch-off. But after reading just about everything I could find on it for the past couple of hours, I'm starting to understand that pinch-off and saturation are exactly the same region.
 
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