What does the gradient of the Ids vs. Vds graph represent?

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imperza

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A simple question , but what does the gradient Ids vs Vds represent?
 

Above 0.8V supply V, the device is in a 'good range of operation'. It turns on to a normal degree. Closer to the ideal device. Current is constant at all supply volt levels.

Below 0.8V, the device shows greater resistance than in normal operation. It turns on less than expected. Less like an ideal device and more like a plain resistor.
 

It is correct to say that the gradient represent the transistor began to saturate?
 

imperza said:
It is correct to say that the gradient represent the transistor began to saturate?
Click to expand...

No, the graph depicts the smallest limit where the transistor just starts to operate. Current is very low, less than 70 uA. I think the bias is very low too.

Saturation occurs when the transistor is biased strongly so that C-E current is maximum. In other words, C-E is down to its lowest 'On' resistance. Further increase in bias current will have no effect. This is typically at a bias of several mA.

If we're talking about a mosfet (as I believe the graph portrays) then the bias is in volts, nevertheless the same concepts apply.

- - - Updated - - -

Your graph would be the area inside the green region below.

 

imperza said:
A simple question , but what does the gradient Ids vs Vds represent?
View attachment 84444
Click to expand...

Coming back to the original question regarding the slope (gradient) of the Ids-Vs graph: The slope of this curve dI/dV is identical to the dynamical output conductance of the device, which is rather small (equivalent to a
large output resistance). This property is the reason to treat transistors as (non-ideal) current sources.
 

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