Is this mosfet good for my project?

[Xenobius]

Dear all,

This is my first post here I was looking for a good forum to settle in and I hope this is it hehe.

I am looking for a P Channel, Mosfet which will be turned ON for 10 - 100mS and off for 30 minutes. The voltage will be around 11 - 13V and current 1A (limited by a 12ohm resistor).

I am looking at this mosfet FDC6506P however I am confused.

1. I need it to turn on an electric match, what is the RDS ON value used for because many sites make a lot of fuss about it.
2. According to the datasheet, there are 2 mosfets in FDC6506P. Are they totally isolated? Meaning can I treat them as individual components in a single package? (this is what I want)
3. As for choosing P channel from N channel, the difference to me is that the N channel must have the load connected to drain while the P channel has the load connected to the source. Is this reasoning good?

Thanks a lot
Looking forward to your replies.

X

 

[(*steve*)]

1) Rds(on) is the resistance between drain and source when the device is turned on. A smaller value means less voltage is dropped by the device (when on) for a given current through it.

2) They are totally isolated

3) the load can be connected to the drain or the source, but for switching purposes typically it is connected to the drain. An N channel mosfet has it's source grounded and the gate is driven +ve to turn it on. A P channel device typically has the source tied to V+ and the gate must be pulled toward ground to turn it on.

That mosfet is in a small surface mount package and you will be pushing it's power dissipation limits, especially if you don't turn it on and off quickly. There are many mosfets in larger packages that may be a more rugged option.

edit: Oh, and welcome to Electronics Point.

 

[Xenobius]

3) the load can be connected to the drain or the source, but for switching purposes typically it is connected to the drain. An N channel mosfet has it's source grounded and the gate is driven +ve to turn it on. A P channel device typically has the source tied to V+ and the gate must be pulled toward ground to turn it on.

Hey Steve,

Thanks for you quick reply Ok so basically if I use an N channel, I will need to put a pulldown on the gate and give it 5v to turn it on WHILE the P channel will need a pullup resistor to 5V and give it ground to turn on... Hmm I think I need to show you a small schematic than because I am now confused

Will be able to upload it in3 hrs

Thanks Steve

 

[(*steve*)]

Ok so basically if I use an N channel, I will need to put a pulldown on the gate and give it 5v to turn it on WHILE the P channel will need a pullup resistor to 5V and give it ground to turn on... Hmm I think I need to show you a small schematic than because I am now confused

You seem to have that exactly correct, assuming the load is powered from 5V.

 

[Xenobius]

Yes because I am using Logic Level Mosfets.

Thanks

 

[(*steve*)]

The load can be powered from other supply rails. The term "logic level" simply refers to the voltage swing required between source and gate to turn the mosfet on.

 

[Xenobius]

Oh yes that is what I meant. I will use 5V directly from Arduino (no limiting resistors as I was reading because there is no current draw) and I will use 12V to power my electric match.

I am attaching a schematic of what I have. Should this be N channel or P channel?
Ignore the pulldown... that is ofcoarse for an N channel.

the 12V Live and 12V test are the same source... I just want to add an additional safety mechanism by switching the 12V using a relay.

 

[(*steve*)]

That circuit is (almost)correct as drawn for an N channel mosfet, however because the load is in the source, the gate may need to be pulled above the 12V rail to turn it fully on.

CHange the mosfet to a P channel device, connected essentially the same way (except the source will be at the top and the drain at the bottom. Also change the resistor connected to the gate of the mosfet to connect to the source (effectively the +12V rail).

 

[Xenobius]

Ok partially understood. So I will use a P channel, will swap the source and drain. But if I put a pullup from gate to 12v, won't that be present on the arduino? Wouldn't that burn the arduino?

I cant understand this part. Also to switch it on I will need to give it a logic 0 right?

 

[(*steve*)]

All of that is true,

Your original circuit had problems, I told you how to fix them, but now the gat needs to be pulled down from 12V -- which the arduino can't do (an extra transistor would let you do it though).

So, ask yourself... do you want another transistor, or can you figure a way to make this circuit work with the gate being pulled up to 5V rather than pulled down from 12? (what do you need to do to swap ground and 12V on your circuit?)

 

[Xenobius]

Ohh of coarse this is simple. All I need is to Pull Up the gate to 5V instead of 12V !! How silly of me!

Yes of coarse it had issues that's why I'm asking

Thanks a lot Steve.

 

[Xenobius]

I was looking for a similar mosfet which can take more power. Why is the datasheet for a P Channel all marked with Negative values? Like -30v, -4A ... this is confusing.

I am now looking at a single channel mosfet here

Thanks for your time

 

[weird_dave]

P channel fets are used 'upside-down' compared to N channel.
Take 'Vgs' for example, it's -ve because the voltage required on the gate is lower than the source, on N channel fets, it's positive.

I'd be using N channel fets.....

 

[(*steve*)]

Ohh of coarse this is simple. All I need is to Pull Up the gate to 5V instead of 12V !! How silly of me!

No, that mosfet will turn on when the gate is (say) 2.5 volts more negative that the source.

But the source is at 12V.

How can the source be at 0V and require a minimum of (say) 2.5V to turn it on?

You know that the previous configuration with an N channel mosfet was wrong, and this one is almost right except you're signal needs to be on the wrong supply rail.

How can we turn the whole thing upside-down?

Hint -- *everything* has to change...

 

[Xenobius]

Hehe thanks Dave,

Yeah but previously on this thread we have concluded that I need to use P Channel, so I have no option

Thanks (Y)

 

[Xenobius]

OMG everything means that the ignitor should be connected to 12V and the P channel ... ok im lost...

I do have a basics knowledge of electronics but unfortunately this is hard to understand. I don't know what to look for. Any hints?

 

[(*steve*)]

You're in the right track.

Draw it upside down. Point diodes in the opposite direction and change P channel devices for N channel devices and vice versa.

This assumes that there are no particular reasons why the ignitor must be connected to ground.

 

[Xenobius]

Agreed in fact I was using a schematic that I found which connected the electric match to ground. As long as current doesn't pass through this device, all should be safe.

 

[Xenobius]

Hi Steve,

I have inverted the circuit but I now figured that I cannot connect a BiLed as I had originally intended because I need a common ground and a separate voltage on the Anode. In the attached schematic I now have a common Supply and a Separate Ground which won't work for my Bi Led (which I now have lots of)...

I think I'm bound with a P Channel Mosfet no?
If so ... how should it be?

Thanks a lot for your time.

 

[Xenobius]

Ok so after a lot of reading, I now understand that for a P-Channel to be fully off, it needs to have the gate at the Source Voltage. Irrelevant if I am going to use a Logic Level Mosfet or not.

So If I intend to switch ON a P Channel using 5V, and I need the load to be working on 12V, than I need to take the gate to 12V to turn it OFF, and will need to put the gate to 0V to turn it ON. Is this case... why do I need a Logic Level Mosfet?

Also I've draw a schematic. Is this how a professional system should switch a device?

Also due to the Bi Led shown in the schematic, I cannot use an N-Channel because otherwise I wouldn't have 2x 12V and 1x Ground.. or so I think... If you know otherwise, please enlighten me!