MOSFET threshold voltage from characteristic curveby pastro Tags: characteristic, curve, mosfet, threshold, voltage 

#1
Sep1509, 10:54 AM

P: 15

Hi,
I have Ids vs Vds characteristic curves for a MOSFET at Vgs = 2V and Vgs = 2.5V in both the linear (ohmic) and saturated regions. I wanted to try to figure out the threshold voltage for the MOSFET from this information only. I had two ideas of how to do this, but they differ by a factor of 2. I want to know why the estimates are so different from each other. Idea 1: From the characteristic curves, I know that in the saturation region, Ids' = 1.4mA @ Vgs' = 2.5V and Ids = 0.65mA @ Vgs = 2V. Use, Ids'/Ids = (Vgs'Vt)^2/(VgsVt)^2 => 2 = (2.5Vt)^2/(2Vt)^2 Solving for Vt, I get Vt = 0.8V, 2.2V. I neglect 2.2V because Vds > Vgs  Vt in the saturation region Idea 2: Use the Vds of the "knee" (where the curves turn over in transitioning between the linear region and the saturation region) as the the saturation voltage of Vds, and use the condition Vds_sat = Vgs  Vt. On the Vgs = 2V curve, this knee occurs near Vds = 0.5V, which means Vt = 1.5V, fairly different from the 0.8 volt guess I got above. Similarly, on the Vgs = 2.5V curve, the knee occurs near Vds = 0.7V, which means Vt = 1.8V The 1.5V and 1.8V threshold voltage estimates from method 2 are pretty similar (especially since I am eyeballing the knee for the curve) but these differ a lot from 0.8V from method 1. Why? 



#2
Sep1509, 01:57 PM

Sci Advisor
P: 2,751

Exactly how are you "eyeballing the knee of the curve". I wouldn't describe point where Vds = VgsVt as exactly the "knee" of the curve because it's actually the point where the curve (ideally) becomes horizontal. Of course in practice the curve mightn't quite become horizontal (due to channel length reduction with increasing Vds), so this may make judging the boundary between triode and saturation regions a bit more difficult.
Anyway my money would be on the following for the most likely sources of error. 1. Under estimating Vds_sat by taking the "knee of the curve" at a point before Ids gets horizontal. 2. The model you're working with is only an approximation to the actual characteristics. BTW with your first calculation you're a bit careless with the rounding, as in 1.4/0.65 is not equal to 2. Using the exact values you get Vt=0.93 volts. It's ok to round off after the calculation but best not to be too cavalier at the beginning. 



#3
Sep1509, 02:02 PM

P: 480

You know if it's a short channel MOSFET the quadratic dependency on gate voltage is not even close.. That might be the source of error. Because of extreme electric fields the carriers quickly reach the velocity saturation limit and do not give extra current after some point in these short devices... 



#4
Sep1609, 01:00 AM

Sci Advisor
P: 4,003

MOSFET threshold voltage from characteristic curve
The "gate threshold voltage" is the voltage from gate to source when the drain current just starts to flow. Is this the "threshold voltage" you are after?
If you have a look at the atatched curves for a MTP3055 mosfet, it is clear that this voltage for this device is 4 volts from the RH curve. On the left curve, you can say that it is less than 5 volts because this is the lowest Vgs curve shown. That is the bottom curve on the graph. If these curves were shown at closer intervals, you could estimate it closer than "less than 5 volts", but with this graph, you can't do better than that. But you don't need to. It is clearly shown on the RH graph. It has nothing to do with the saturation knee on the left of the curves on the left graph. 



#5
Sep1609, 05:25 AM

Sci Advisor
P: 2,751

For example, take the first of your two graphs and the Vgs=5 trace (btw the x axis on that graph should be labelled as Vds not Vgs). The point at which this curve just enters the constant current region (which for a fet is also referred to as the saturation region) is approx Vds=1 volt. From this we can predict that Vt = 5  1 = 4 volts. This follows directly from the (ideal) equations for a FET. Look them up. The main problem here is that the FET may not accurately follow the ideal quadratic relationship too well for various reasons, see for example sokrates reply above. BTW, the above shown FET clearly does depart quite significantly from the ideal quadratic relationship at higher current levels. 



#6
Sep1609, 07:12 AM

P: 15

Thanks for all the great responses. After reading you replies, I think the problem is that I was using the "knee" instead of the point where the curves get horizontal.
Cheers! 



#7
Sep1609, 07:26 AM

Sci Advisor
P: 4,003

See attached diagram.
If you have a look at the attached curves for a MTP3055 mosfet, it is clear that this voltage for this device is 4 volts from the RH curve. On the left curve, you can say that it is less than 5 volts because this is the lowest Vgs curve shown. That is the bottom curve on the graph. If these curves were shown at closer intervals, you could estimate it closer than "less than 5 volts", but with this graph, you can't do better than that. But you don't need to. It is clearly shown on the RH graph. It has nothing to do with the saturation knee on the left of the curves on the left graph. 


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