Mosfet as load switch

[kevinwinder]

I am using mosfet as load switch application. anybodies can help to calculate if value of resistor R1, R2, RB are correct if AO3401 and MMBT3904 was used?

 

[ronsimpson]

Looks like you chose a MOSFET that turns on at low voltage. Good job!
The resistors look good to me.

 

[kevinwinder]

Looks like you chose a MOSFET that turns on at low voltage. Good job!
The resistors look good to me.

Can you help to verify those values by showing the calculation?

 

[crutschow]

What is the switching frequency?

 

[ronsimpson]

Can you help to verify those values by showing the calculation?

NPN transistor: 100k pull down resistor. Probably not needed. The micro that drives this pin might have all of its pins in 3-state at power up. There might be some noise on this line.
NPN transistor: 10k on base. You will have about 1 volt across this resistor. 1.8V-0.65V= 100uA Approx.
NPN transistor: When NPN is just turned on, but the MOSFET's gate capacitor has not charged up yet, there could be 5V across the 1k resistor so about 5mA of current in the collector. 5mA/0.1mA= current gain of 50. That transistor can do that. Even if it could not that is OK for this very short time.
Then the MOSFET's gate has 4.8 volts across it the NPN collector current drops to 5V/470k ohms. I did not check to see what the NPN's collector leakage current is when the base is at 0V and the temperature is high. (50nA??) So what is the collector voltage (gate voltage) with 50nA and 470k ohms? = not much

What is the switching frequency?

This is to turn on off the power to a circuit. I assume it gets turned on at power up, or once/hour. So driving the gate with a 1K resistor is fine. The MOSFET will turn on slow which is good in this case.

 

[kevinwinder]

NPN transistor: 100k pull down resistor. Probably not needed. The micro that drives this pin might have all of its pins in 3-state at power up. There might be some noise on this line.
NPN transistor: 10k on base. You will have about 1 volt across this resistor. 1.8V-0.65V= 100uA Approx.
NPN transistor: When NPN is just turned on, but the MOSFET's gate capacitor has not charged up yet, there could be 5V across the 1k resistor so about 5mA of current in the collector. 5mA/0.1mA= current gain of 50. That transistor can do that. Even if it could not that is OK for this very short time.
Then the MOSFET's gate has 4.8 volts across it the NPN collector current drops to 5V/470k ohms. I did not check to see what the NPN's collector leakage current is when the base is at 0V and the temperature is high. (50nA??) So what is the collector voltage (gate voltage) with 50nA and 470k ohms? = not much

Do you mean R2=470K is too big?
Have checked with some application notes in internet, some application note for mosfet as a load switch mentioned that the value of R2 should be in ranged 1K to 10K, how to determine the correct value of R1 and R2?

 

[spec]

Hello Kevin

You know I have looked and looked at your schematic and felt that something was wrong but couldn't put my finger on it. Then the penny dropped: Q28 is a PMOSFET so the arrow on its schematic symbol should point outwards not inwards. This makes no difference to the circuit function though.

Can you explain the purpose of L20. I asume it's just high frequency filtering.

Data sheets for reference:
AO3401 PMOSFET data sheet: http://www.aosmd.com/pdfs/datasheet/AO3401.pdf
MMBT3904 bipolar NPN transistor data sheet: **broken link removed**

PS: you have put your post inside the closing quote tag so it looks like it's part of ronsimpson's post. Just Lclick on 'edit'. Lclick on 'More Options' and move your text down and outsite the closing 'quote' tag. Lclick 'Save Changes'

 

[kevinwinder]

Hello Kevin

You know I have looked and looked at your schematic and felt that something was wrong but couldn't put my finger on it. Then the penny dropped: Q28 is a PMOSFET so the arrow on its schematic symbol should point outwards not inwards. This makes no difference to the circuit function though.

Can you explain the purpose of L20. I asume it's just high frequency filtering.

Data sheets for reference:
AO3401 PMOSFET data sheet: http://www.aosmd.com/pdfs/datasheet/AO3401.pdf
MMBT3904 bipolar NPN transistor data sheet: **broken link removed**

PS: you have put your post inside the closing quote tag so it looks like it's part of ronsimpson's post. Just Lclick on 'edit'. Lclick on 'More Options' and move your text down and outsite the closing 'quote' tag. Lclick 'Save Changes'

 

[kevinwinder]

The arrow is already pointed outward for pmosfet.
yes, L20 is used for high frequent filtering.

 

[spec]

Appologies Kevin- my error.

Often it's easier to analyse a circuit if it is rearranged; I had a go at yours:

As ronsimpson says, the circuit is fine and should work well. It will be fairly slow to switch on and take an age to turn off, but not in human terms.

What currents are you switching into the load, AMP_DVDD?

Could you define the drive circuit that provides the 1V8 VIN?

What is the 5V source?

In case you don't know, the Shottky diode shown within the Q28 schematic symbol is there as a consiquence of making a power MOSFET. Normally you can ignore it, as in this case. Other times it is useful and some times it is a nusance.

 

[AnalogKid]

R381 is pretty large, and could cause slow turn-off time for Q28. I would reduce it to 10 K, and eliminate RR383.

ak

 

[spec]

Hi AnalogKid,

Thats what I thought. Also, with Q29 turned off there is a very high impedence node at the Q28 gate wich may cause problems with pick up etc, in theory anyway. R383 doesn't do anything in terms of the basic function, but it does provide a bit of protection for Q29, but more importantly protection for the diodes in Q28 gate in the event of any spikes and nasties on the supply lines.

The other thing is that with Q29 off, Q28 may be liable to hoot, especially with that inductor in the source feed. Its always best to put a 10 Ohm gate stopper in, physically close to the gate. They are great things MOSFETS but they have a very high frequecy respons and quite large parasitic capacitances, a combination guaranteed to ecourage parasitic oscillation.

If Kevin posts the data about the input circuitry to Q29, the 5V power source, and the maxiumum output load curent we will get a better picture. That MOSFET is quite a beast, 20A at 2.5V Vgs, which may not be optimum if he only wants a 100mA load say.

Great fun!

 

[spec]

Just one more thing- there always is.

There is no short circuit protection which may be fine if everything is in a case, including the load, but imagine this circuit being fed from NiMH batteries with the output provided by an exposed connector which could short to 0V. Not only would it potentially destroy the MOSFET but the batteries could explode. The answer is to put a fuse in the supply which would be a wise precaution for a development circuit. Just make sure you have a big box of them!

Of, course if the 5V input is from a current limited source, like a normal power supply, all would be well.

But even so, if the current limit from the input power source were 10A say, that would mean 50W disipation in the MOSFET under short circuit conditions. So, without a massive heat sink, the MOSFET would be destroyed anyway.

 

[spec]

Kevin,

Don't let any of the above posts put you off; it's just the way that engineers analyse circuits- always looking for problems. Your circuit will be fine especially with a few adjustments here and there. But we do need to know the information that we have requested to arrive at a good design. Having examined the data sheet for the MOSFET it looks like a great device.

 

[simonbramble]

Put a capacitor on the Source of the FET (1uF electrolytic should do). When the FET switches off you have a residual current in the inductor which will probably put a positive spike on the Source of the FET (it is acting like a boost converter). A cap will soak up the spike.

 

[ronsimpson]

you have a residual current in the inductor

The inductor is probably a bead and will have very little energy stored.
The 470k turn off resistor will turn off the MOSFET v e r y s l o w and so the inductor will not kick up. (MOSFET is slower than the time to reset the inductor)