N-channel depletion mosfet for power switching
- Thread starter ashuppie
- Start date
Sorry, I don't know much about it. It requires Vgs of -2 to -4v. Does that mean I need to have Vg @ -9vdc?
Please post a schematic as 800x600 *.jpg, *.gif.
How much current draw?
You should choose another N MOSFET, because the Vdsx of IXTP3N50D2 is too high and that kind of type will be have a high Rds(1.5Ω), and the high Rds will causes a high voltage drop (Vds=Id * Rds), and that does not suit for your low voltage application.
N, P type MOSFET Components Simple Data.
How much current draw?
You should choose another N MOSFET, because the Vdsx of IXTP3N50D2 is too high and that kind of type will be have a high Rds(1.5Ω), and the high Rds will causes a high voltage drop (Vds=Id * Rds), and that does not suit for your low voltage application.
N, P type MOSFET Components Simple Data.
Depends upon the source voltage.
Source voltage is 12.7vdc.
According to your schematic, the source of M1 and M2 are connected to ground.
And you need to show the rest of the circuit.
Since others are already on this, I'll just make some comments on depletion mode FETs.
Depletion mode power MOSFETs are rather rare and somewhat expensive, especially when compared with enhancement mode types of similar voltage and current rating. They are rarely used as simple load switches but are very useful for some special applications, often in common-drain configurations, where the fact that they are "normally ON" and turned off "actively" simplifies circuitry.
example of use: The controller for a switch mode power supply operating from the AC line requires a power supply itself to get started. Often the rectified line voltage is around 400 volts, but a supply of perhaps 12 volts at (let's say) 10 mA, is required for the controller. If you try to do that with a resistor it will dissipate almost 4 W (388 V x 10 mA). Once the main supply is going it powers the controller so you no longer need current through the startup resistor. A depletion mode power MOSFET is perfect for the job of switching off current through the resistor - it is ON by default and turned off actively, making a very power-efficient startup circuit. It is because of aps like this that many of them have a high voltage and low current ratings.
Depletion mode power MOSFETs are rather rare and somewhat expensive, especially when compared with enhancement mode types of similar voltage and current rating. They are rarely used as simple load switches but are very useful for some special applications, often in common-drain configurations, where the fact that they are "normally ON" and turned off "actively" simplifies circuitry.
example of use: The controller for a switch mode power supply operating from the AC line requires a power supply itself to get started. Often the rectified line voltage is around 400 volts, but a supply of perhaps 12 volts at (let's say) 10 mA, is required for the controller. If you try to do that with a resistor it will dissipate almost 4 W (388 V x 10 mA). Once the main supply is going it powers the controller so you no longer need current through the startup resistor. A depletion mode power MOSFET is perfect for the job of switching off current through the resistor - it is ON by default and turned off actively, making a very power-efficient startup circuit. It is because of aps like this that many of them have a high voltage and low current ratings.
Here is what I am trying to do, what should I do differently?
Use enhancement-mode N-Mosfets. That circuit won't work with depletion-mode devices.
Reverse the direction of the LEDs.
Recalculate the resistor values for the LM317 (which is a voltage regulator, not a transistor as you show). R3 should be no larger than 120Ω.
Attached is a more readable copy of your circuit:
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Good catch.
I didn't notice they were different.
So for that, there's a number of options.
On is to use two enhancement-mode N-MOSFETs with an inverter circuit in between.
Another is to use one enhancement-mode N-MOSFET and one enhancement-mode P-MOSFET.
OK, Thanks!
Inverter Circuit? Sorry, I don't know a lot about that.
It can be a common-emitter transistor circuit with a couple resistors, or a logic chip inverter, such as a CD4049.
Ok, thanks for your help
There is a sneaky and devious thing you can do with the CMOS version of the 555 that could be advantageous here.
An astable multivibrator can be built in the usual way except for the resistors. A single resistor is used between the output and the timing capacitor. When the output is low, the capacitor is discharged to the trigger level, setting the output high. The cap now charges to the threshold level, sending the output back low. The duty cycle, except for the first cycle, is very close to 50%. If you put a pullup resistor on the discharge pin, it becomes the logical complement of the output pin, so you could drive one N-FET with the discharge pin and another with the output pin.
[Edit] You can do this with the bipolar (ordinary) version of the 555, but the duty cycle won't be 50% because the output HIGH voltage is less than the supply voltage. The higher the supply voltage, the closer the duty cycle will be to 50% because the difference between the output HIGH voltage and the supply voltage is more or less constant.
see Figure 17
http://www.ti.com/lit/ds/symlink/lmc555.pdf
An astable multivibrator can be built in the usual way except for the resistors. A single resistor is used between the output and the timing capacitor. When the output is low, the capacitor is discharged to the trigger level, setting the output high. The cap now charges to the threshold level, sending the output back low. The duty cycle, except for the first cycle, is very close to 50%. If you put a pullup resistor on the discharge pin, it becomes the logical complement of the output pin, so you could drive one N-FET with the discharge pin and another with the output pin.
[Edit] You can do this with the bipolar (ordinary) version of the 555, but the duty cycle won't be 50% because the output HIGH voltage is less than the supply voltage. The higher the supply voltage, the closer the duty cycle will be to 50% because the difference between the output HIGH voltage and the supply voltage is more or less constant.
see Figure 17
http://www.ti.com/lit/ds/symlink/lmc555.pdf
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