Because you can use it in the same way that you use pentode curves to determine the required resistance for a particular current.
Vertical axis: Ids
Horizontal axis: Vds
Suppose that you wanted to pass 9mA. Looking at that current, you find that the second curve down is pretty well 9mA, and it achieves this with Vgs = -1.6V. If you put a resistor in the source circuit of 1.6/0.009 = 178R, and connect the gate to the bottom of that resistor, you have created a 9mA two-terminal CCS.
Additionally, the curves show the minimum voltage required across the device (Vds) for it to achieve a constant (high) output resistance.
Vertical axis: Ids
Horizontal axis: Vds
Suppose that you wanted to pass 9mA. Looking at that current, you find that the second curve down is pretty well 9mA, and it achieves this with Vgs = -1.6V. If you put a resistor in the source circuit of 1.6/0.009 = 178R, and connect the gate to the bottom of that resistor, you have created a 9mA two-terminal CCS.
Additionally, the curves show the minimum voltage required across the device (Vds) for it to achieve a constant (high) output resistance.
9mA achieved with -1,6v.. so where does that put you on the graph..i.e. where is the operating point?
Where is resistance on the chart? I mean..how does one see that? Left of the dotted line?
Additionally, the curves show the minimum voltage required across the device (Vds) for it to achieve a constant (high) output resistance.
Where is resistance on the chart? I mean..how does one see that? Left of the dotted line?
Attachments
Resistance
Resistance is inversely proportional to the slope of the curves. A perfectly horizontal curve has slope = 0, and implies infinite resistance. For best CCS operation, you want to avoid operating where the curve deviates from horizontal, so at higher currents, you need a higher Vds to stay on the horizontal part.
Resistance is inversely proportional to the slope of the curves. A perfectly horizontal curve has slope = 0, and implies infinite resistance. For best CCS operation, you want to avoid operating where the curve deviates from horizontal, so at higher currents, you need a higher Vds to stay on the horizontal part.
Operating point
Your operating point is somewhere along the 1.6V line. If you were using the CCS in the cathode circuit of a valve, and the grid was grounded, then the cathode might be at 8V, so that would place the operating point at 8V less the 1.6V drop across the source resistor = 6.4V, which is well to the right of the curved region.
Your operating point is somewhere along the 1.6V line. If you were using the CCS in the cathode circuit of a valve, and the grid was grounded, then the cathode might be at 8V, so that would place the operating point at 8V less the 1.6V drop across the source resistor = 6.4V, which is well to the right of the curved region.
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