MOSFETs - Completely confused on base voltage needed.

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Good evening,

I've spent all day studying how MOSFETs work, in particular P-Channel Logic Level MOSFETs. I understand how these work even down to the internal construction and how the P-Channels use holes to flow the power from drain to source. What I just can't come to terms with it how to calculate what voltage I need to apply to the Base to toggle the MOSFET on and off. Like I have said, I have spent all day on MOSFETs and I'm really struggling and its starting to slowly eat away at me now, frustrated itsn't the best description for my emotions right now, it must be admitted.

I'm planning on powering the base through a MCU, at the moment i have it powering with 3v3 but i could change that to 5v if it means having a 5v output is more desirable. Lets take for example the NX3008PBKW P Channel MOSFET. Datasheet is here.

All that I know from here is that my circuitry needs to have a pull-up resistor on the base to make sure the base is not left floating while the MCU boots/is disconnected for some reason. I also know that the load is to be placed in between the source and ground and that the drain is connected to 5v.

This is now where I'm now lost. I'm not asking for someone to do this for me, I'm here to learn. I need to learn the process on how to calculate what is needed and why it is needed to be that value.

If you need to know the load current, its 65mA. The load is a simple PIR motion detection module, datasheet is here.

If there is any more information you require, please ask away. I'm very eager to learn this process but please bare with me as I've been studying it for the past 12 hours and i'm just struggling on this section now. I feel rather pathetic with taking this long but her, we all start somewhere...

Thanks, Sam.
 

dl324

Joined Mar 30, 2015
9,549
The MOSFET starts to turn on with Vgs=-0.6 to -1.1V.

Base is for BJT's; for MOSFETs, it's called the gate.
 

ErnieM

Joined Apr 24, 2011
8,011
That is the slimmest data sheet I have ever seen. The only hard number it offers is -4.5volts for the spec'd Rds on.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Thank you all for your responses.

Why are you using a P-channel MOSFET?
Shouldn't you be using an N-channel device instead?
Would you mind explaining why i should be? From hours and hours of reading and chatting to fellow electronic enthusiasts, i came to the P-Channel decision.

The MOSFET starts to turn on with Vgs=-0.6 to -1.1V.

Base is for BJT's; for MOSFETs, it's called the gate.
Sorry, i meant gate not base. My apologies! So it's literally a case of just giving the gate -0.6 to -1.1v no matter size of power needing to flow through the MOSFET?

That is the slimmest data sheet I have ever seen. The only hard number it offers is -4.5volts for the spec'd Rds on.
This doesn't bode well considering i was about to purchase these ones.
 

MrChips

Joined Oct 2, 2009
19,911
A P-channel device would be called a high-side switch, i.e. the switch is on the positive power rail and the load is on the low side, that is, connected to ground. This puts special requirements on the gate voltage.

An N-channel device would be on the low side, i.e. the load goes between the positive power rail and the switch.
The gate is easily controlled by logic level voltage such as that from a microcontroller.
 

ian field

Joined Oct 27, 2012
6,539
Good evening,

I've spent all day studying how MOSFETs work, in particular P-Channel Logic Level MOSFETs. I understand how these work even down to the internal construction and how the P-Channels use holes to flow the power from drain to source. What I just can't come to terms with it how to calculate what voltage I need to apply to the Base to toggle the MOSFET on and off. Like I have said, I have spent all day on MOSFETs and I'm really struggling and its starting to slowly eat away at me now, frustrated itsn't the best description for my emotions right now, it must be admitted.

I'm planning on powering the base through a MCU, at the moment i have it powering with 3v3 but i could change that to 5v if it means having a 5v output is more desirable. Lets take for example the NX3008PBKW P Channel MOSFET. Datasheet is here.

All that I know from here is that my circuitry needs to have a pull-up resistor on the base to make sure the base is not left floating while the MCU boots/is disconnected for some reason. I also know that the load is to be placed in between the source and ground and that the drain is connected to 5v.

.
If you think a MOSFET has a base and base voltage - you didn't do a very good job of reading up on it.

A MOSFET gate threshold voltage can range from about 1V for small signal and some logic level types. It may be a few volts for general purpose power types.

A logic level part should work with most types of MCU, but you need to check the specifications. Vo-high and Vo-low on the MCU don't quite go all the way to the rails.

A general purpose MOSFET will need around 8V on the gate to achieve the headline RDS-on figure quoted on the datasheet.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
A P-channel device would be called a high-side switch, i.e. the switch is on the positive power rail and the load is on the low side, that is, connected to ground. This puts special requirements on the gate voltage.

An N-channel device would be on the low side, i.e. the load goes between the positive power rail and the switch.
The gate is easily controlled by logic level voltage such as that from a microcontroller.
Right okay, so for starters you're advising me to change out the P-Channel and a N-Channel. Now its just to workout my thoughts on the gate voltage. I'm not too sure why I feel that its in relation to the voltage running through Drain-Source.

EXAMPLE
Lets say I opt for NX7002AK. The datasheet shows that the VGSth is MIN: 1.1v TYP: 1.6v MAX: 2.1v. Does this mean that I just need to provide 1.6v to the gate to enable the switch? Does this relate in anyway to the voltage or current needing to flow through the Drain-Source connection?
 

ian field

Joined Oct 27, 2012
6,539
Right okay, so for starters you're advising me to change out the P-Channel and a N-Channel. Now its just to workout my thoughts on the gate voltage. I'm not too sure why I feel that its in relation to the voltage running through Drain-Source.

EXAMPLE
Lets say I opt for NX7002AK. The datasheet shows that the VGSth is MIN: 1.1v TYP: 1.6v MAX: 2.1v. Does this mean that I just need to provide 1.6v to the gate to enable the switch? Does this relate in anyway to the voltage or current needing to flow through the Drain-Source connection?
Most MOSFETs can withstand upto 15V on the gate - but check the datasheet.

Within the absolute maximum rating - the more gate voltage; the more channel conduction.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Most MOSFETs can withstand upto 15V on the gate - but check the datasheet.

Within the absolute maximum rating - the more gate voltage; the more channel conduction.
The max rating of Vds is 20v so if i just give it the full 3.3v or maybe 5v if i change the MCU Voltage to 5v it will more than enough?
 

dannyf

Joined Sep 13, 2015
2,197
I'm very eager to learn this process
I still don't understand what "process" you are talking about.

You can think of a mosfet as a bjt if that helps. A p-ch mosfet is like a pnp in that if the voltage difference between the source and gate is greater than a certain level, the mosfet starts to conduct between the source and drain, just a pnp would between its emitter and collector when the emitter / base junction voltage goes sufficiently high.

The question gets more interesting if you are trying to conduct more current on a non-logic level mosfet.
 

Alec_t

Joined Sep 17, 2013
10,693
Be wary of the Vgs(th) figure. At that threshold voltage the FET is only just beginning to turn on. To turn it on fully the Vgs should be several times greater. For FETs specified as 'logic-level', 3.3V or 5V should be enough to turn them on fully; but for others a Vgs of 8V-10V is normally needed. N.b. Vgs is not the voltage on the gate (relative to circuit ground): it is the voltage difference between gate and source.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
I still don't understand what "process" you are talking about.

You can think of a mosfet as a bjt if that helps. A p-ch mosfet is like a pnp in that if the voltage difference between the source and gate is greater than a certain level, the mosfet starts to conduct between the source and drain, just a pnp would between its emitter and collector when the emitter / base junction voltage goes sufficiently high.

The question gets more interesting if you are trying to conduct more current on a non-logic level mosfet.
By "process" i meant the theory of how much voltage the gate needs to allow the drain-source conductivity. Thank you for that information. So is the VGSth typical of 1.6v the lowest voltage point the gate will open and the absolute max rating of 20v the highest the gate can go before damage on the example mosfet linked above?
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Be wary of the Vgs(th) figure. At that threshold voltage the FET is only just beginning to turn on. To turn it on fully the Vgs should be several times greater. For FETs specified as 'logic-level', 3.3V or 5V should be enough to turn them on fully; but for others a Vgs of 8V-10V is normally needed. N.b. Vgs is not the voltage on the gate (relative to circuit ground): it is the voltage difference between gate and source.
So, on the datasheet i posted for an example, could you explain to me how you would work out or know what voltage you would need to apply to the gate to have it fully open and therefor not create too much resistance on the mosfet itself.
 

shortbus

Joined Sep 30, 2009
7,327
The datasheet shows that the VGSth is MIN: 1.1v TYP: 1.6v MAX: 2.1v. Does this mean that I just need to provide 1.6v to the gate to enable the switch? Does this relate in anyway to the voltage or current needing to flow through the Drain-Source connection?
If you look at the chart you found that set of numbers in, table #7, it also says under that heading, ""> = 250 µA; VDS = VGS; Tj = 25 °C". So it is only passing 250mirco amps at that gate voltage. Unless your building an audio amplifier you should just/always ignore the Vgs th.

Table #5 is where you want to look in that data sheet. Under symbol "ID", that shows thad amperage being passed at the correct Vgs or 10V.

While I'm not as experienced as Ian Field, any Nmosfet I've seen usually gives a Vgs 10V as the "ideal" gate voltage. Unless it is a logic level device.
 

BobaMosfet

Joined Jul 1, 2009
927
So, on the datasheet i posted for an example, could you explain to me how you would work out or know what voltage you would need to apply to the gate to have it fully open and therefor not create too much resistance on the mosfet itself.
Looking at the Datasheet for 'NX3008PBKW', Vgs is clearly stated as anything between -8VDC and +8VDC. So -8. They were kind enough to state it. If you look further at Vgs(th), you'll see that is between -0.6 to -1.1VDC. RDSon is about 4.1Ohms. Looks like this FET is used in negative logic. More important is-- junction temperature. You need to determine if your desired numbers will exceed the thermal junction temperature, because this is what actually determines how much energy you can put into the device (and whether you need countermeasures with heatsinks, etc).

Looking at the Datasheet for 'NX7002AK', the Vgs range is -20 to 20VDC. NOTE, their 'TEST' condition: 10VDC. That, therefore, is the full 'ON' minimum (and why they tested at 10VDC). Vgs(th) is 1.1 to 2.1VDC (they give a range because of differences in silicon) for beginning transconductance. RDSon = 4.1 Ohms.
 
Last edited:

dannyf

Joined Sep 13, 2015
2,197
you look further at Vgs(th), you'll see that is between -0.6 to -1.1VDC. RDSon is about 4.1Ohms. Looks like this FET is used in negative logic.
Vgs(th) is typically specificed at a very low current level -> it is the Vgs needed to turn OFF the mosfet.

it would be rare for a mosfet to be of depletion type.
 
A MOSFET is a transistor that uses the special effects of an electrical field to switch the flow of power. To turn a p channel mosfet on you are apply a negative voltage to entrance. In circuit u connect mosfet’s basis terminal to a positive power supply and drain to regulator linked to ground and in N channel mosfet turns on whenever u apply a positive voltage as its entrance station. The power is better than the positive high voltage source supply at the drain terminal.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
If you look at the chart you found that set of numbers in, table #7, it also says under that heading, ""> = 250 µA; VDS = VGS; Tj = 25 °C". So it is only passing 250mirco amps at that gate voltage. Unless your building an audio amplifier you should just/always ignore the Vgs th.

Table #5 is where you want to look in that data sheet. Under symbol "ID", that shows thad amperage being passed at the correct Vgs or 10V.
Maybe i'm reading the datasheet incorrectly then, because i thought that "> = 250 µA; VDS = VGS; Tj = 25 °C" means that it will let 250uA through when VGS is the same as VDS.

Looking at the Datasheet for 'NX3008PBKW', Vgs is clearly stated as anything between -8VDC and +8VDC. So -8. They were kind enough to state it. If you look further at Vgs(th), you'll see that is between -0.6 to -1.1VDC. RDSon is about 4.1Ohms. Looks like this FET is used in negative logic. More important is-- junction temperature. You need to determine if your desired numbers will exceed the thermal junction temperature, because this is what actually determines how much energy you can put into the device (and whether you need countermeasures with heatsinks, etc).

Looking at the Datasheet for 'NX7002AK', the Vgs range is -20 to 20VDC. NOTE, their 'TEST' condition: 10VDC. That, therefore, is the full 'ON' minimum (and why they tested at 10VDC). Vgs(th) is 1.1 to 2.1VDC (they give a range because of differences in silicon) for beginning transconductance. RDSon = 4.1 Ohms.
Can we always assume that their test voltage is the minimum needed to turn it fully on?

A MOSFET is a transistor that uses the special effects of an electrical field to switch the flow of power. To turn a p channel mosfet on you are apply a negative voltage to entrance. In circuit u connect mosfet’s basis terminal to a positive power supply and drain to regulator linked to ground and in N channel mosfet turns on whenever u apply a positive voltage as its entrance station. The power is better than the positive high voltage source supply at the drain terminal.
I understand the workings and method of the MOSFET (i think, not sure about that regulator you mentioned linking to ground though). Its how to work out the voltage needed to turn the MOSFET on and off that i'm not understanding.
 

Alec_t

Joined Sep 17, 2013
10,693
So, on the datasheet i posted for an example, could you explain to me how you would work out or know what voltage you would need to apply to the gate to have it fully open and therefor not create too much resistance on the mosfet itself.
This is what the datasheet says:
Rds(on).PNG
As you can see, when Vgs = 5V the Rds is 5.2Ω. Clearly the FET is still not quite fully open, because when Vgs = 10V, Rds drops to 4.5Ω.
 
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