Large voltage drop in MOSFET

[zhiling0229]

Hi guys,

I'm rather new in power switching circuits.

I'm using a IRF 640 N-channel mosfet with the drain connected to approximately 9 V and the source to the load. I'm using the MOSFET as a switch to turn on my power supply using a microcontroller this power supply will be fed into a 5V regulator and a 3.3 V regulator.

I have a few problems.

The voltage drop I measure is approximately 1.6 V which drops to about 7.4 V. I'm worried that it is not enough to regulate my regulator at 5V.

Is there any other way to do it instead of using IGBT cause IGBT is really expensive.

I look in the datasheet the Vgs that was written was min 2V to max 4V does this means that I only can send a pulse to the gate in this range? Or can I send a 5V logic pulse to it?

Thanks a lot

 

[v_c]

What kind of Vgs do you have on your MOSFET? Is it fully turned on? To turn on the MOSFET fully and have low Rdson, you need to have Vgs be something like 10V or so.
Since you have your MOSFET on the high-side, you need to make sure that the voltage that you see at the gate is about 20V with respect to ground. This is assuming that you have low voltage drop across the MOSFET and the voltage at the source is around 9V, then the 20V at the gate would give you 11V of Vgs. You might need a high-side MOSFET driver to give you the boost needed for the gate. Something like an IR21xx should do.

Do a search on "high-side MOSFET" and "high-side MOSFET gate drive"

Best regards,
v_c

 

[dfullmer]

Hi zhiling0229,


v_c is correct. Don't forget that the gate voltage is in reference to the MosFet's source leg not to ground. So if you have 9 volts feeding the drain of the MOSFET and 8 volts coming out of the source you must have a voltage potential 2-4 volts higher than the source output. In other words a gate voltage of about 12 volts. The High side driver that v_c mentioned would solve your problem. It pumps the voltage up to a good gate drive level based on the source voltage.

Good luck

dfullmer

 

[Kral]

zhiling0229,
If all you need to do is turn on the power to your regulator, A PNP power transistor might be a better choice, Connect the emitter to your raw supply, the colllector to your load. Connect a resistor from the raw supply to the PNP base. Use an NPN transistor or N channel Fet to turn on the PNP transistor. The NPN base or FET gate would be driven by the microcontroller. Connect the NPN collector or FET drain to the PNP transistor base through a resistor. The value of the resistor would be chosen to provide enough base current to the PNP to ensure that it is saturated when turned on.
Regards,
Kral

 

[IanP]

P-Channel MOSFET seems to be much better option ..
See:
"P-Channel MOSFETs, the Best Choice for High-Side Switching"
**broken link removed**

Regards,
IanP

 

[zhiling0229]

Thanks guys for such a wonderful suggestion.

But all your suggestion suggest of driving with a much higher or lower voltage. Such as 15V etc.

However the only power supply I have is approximately 9 V battery which is a combination of 6 AA batteries.

Is there any other way without require a higher voltage potential or with less voltage drop or to boost the voltage above 9V?

Can anyone show me the schematics how Karl suggest to use a power transistor PNP and a NPN connected to a microcontroller to drive it. If I'm not mistaken a BJT has a smaller voltage drop compare to MOSFET. Is it true?

Thanks

 

[IanP]

Below is a circuit for switching positive voltage with PNP/NPN pair ..
Depending on the load's current you will have to select suitable PNP ..

MOSFETS will also work with switching voltages of 9V and less, but this will depend on the availability of suitable MOSFETS at your place ..
For example, 2N7000 as driving N-channel and IRF5305 as P-channel load switch .. see circuit from my previous post ..

Regards,
IanP

 

[nguyennam]

Plus, IGBT is not designed and made for low voltage circuits.

 

[zhiling0229]

I just develop the circuit of PNP/NPN pair using BC327 and BC337. It gives an excellent low voltage drop of about 0.2 V only. I have yet to test it on my load the only load i use was a volt meter.

The best part is it is cheap just cost me less than a dollar to construct