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Old 11th September 2003, 01:29 AM   #1
MIKET is offline MIKET  United States
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Default How's about a Depletion Mode / Enhancement Mode CCCS

I have some DN2540's and IRF 510's and have been considering using the DN2540 on the supply side and the IRF510 on the tube side. Any thoughts?
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Old 11th September 2003, 03:14 AM   #2
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Hey-Hey!!!,
If you use the enhancement mode, you will need bateries. Use the depletion mode and you cna get away without, though it is beter to use them on the bottom ones anyway...
regards,
Douglas
ps. don't let me stop you from trying something
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Old 11th September 2003, 11:58 AM   #3
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Default Show us your curves!

This might be useful for the DN2450. Top curve is -1.5V, and the curves go progressively more negative at 0.1V per curve.
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Old 11th September 2003, 01:07 PM   #4
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Hi EC8010,

How is this graph useful? Because I would like it to be!


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Bas
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Old 11th September 2003, 01:20 PM   #5
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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.
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Old 11th September 2003, 02:06 PM   #6
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9mA achieved with -1,6v.. so where does that put you on the graph..i.e. where is the operating point?


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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?
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Old 11th September 2003, 02:41 PM   #7
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Default 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.
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Old 11th September 2003, 02:57 PM   #8
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Default 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.
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Old 11th September 2003, 03:12 PM   #9
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Thanks
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Old 11th September 2003, 03:44 PM   #10
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You're welcome.
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