MOSFETs - Completely confused on base voltage needed.

if your looking for a single number, you won't find it. You really should look at figure #9. Why figure #9?

It gives you an idea of Rds(ON) and junction temp as a function of Vgs

But yes, your right the max values of Vgs is where damage starts to occur. Same for some of the other maximums.

Vt is where the FET STARTS to conduct, but it isn't fully on yet. We can see from that datasheet that operating somewhere between 5 and 10 v and 100 mA is a good place for a switch.

Good numbers of Vgs are between 5 and 10V when operating as a switch.

Your actual load (Say it was a 10 mA LED) can change these numbers.

In terms of Vgs min, typ and max are really places when you want the part to behave in all cases. When designing a current limiting resistor for a LED, you should use Vf (max) when you look at the high end. So a 2.1 V led and a 0.7 V diode drop might take you to 2.8 V and suppose you have a 3.3. V +-10% supply?

Numbers don't tell all. You sometimes have to look at the graphs. The numbers HELP select.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
This is what the datasheet says:
View attachment 104175
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.
Right okay, so based at looking at this and seeing 4.5ohms you then realised that it wasn't completely open because that resistance isn't the expected resistance of the MOSFET fully open?

if your looking for a single number, you won't find it. You really should look at figure #9. Why figure #9?

It gives you an idea of Rds(ON) and junction temp as a function of Vgs

But yes, your right the max values of Vgs is where damage starts to occur. Same for some of the other maximums.

Vt is where the FET STARTS to conduct, but it isn't fully on yet. We can see from that datasheet that operating somewhere between 5 and 10 v and 100 mA is a good place for a switch.

Good numbers of Vgs are between 5 and 10V when operating as a switch.

Your actual load (Say it was a 10 mA LED) can change these numbers.

In terms of Vgs min, typ and max are really places when you want the part to behave in all cases. When designing a current limiting resistor for a LED, you should use Vf (max) when you look at the high end. So a 2.1 V led and a 0.7 V diode drop might take you to 2.8 V and suppose you have a 3.3. V +-10% supply?

Numbers don't tell all. You sometimes have to look at the graphs. The numbers HELP select.
Now this is making sense to me. So you look for that graph and when the line bottoms out its fully open and when it starts is the starting voltage when it would start to open. So you want to be somewhere relatively close to the line graph of Drain-source on-state resistance smoothing out at the bottom of the line?
 
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Alec_t

Joined Sep 17, 2013
14,280
the expected resistance of the MOSFET fully open
... is a value which may have been obtained under ideal or absolute-limit operating conditions unlikely to be met in the average circuit ;). You wouldn't really want to have Vgs at its absolute maximum allowable value just to reduce Rds(on) by a small fraction of a mV to its asymptotic minimum.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
... is a value which may have been obtained under ideal or absolute-limit operating conditions unlikely to be met in the average circuit ;). You wouldn't really want to have Vgs at its absolute maximum allowable value just to reduce Rds(on) by a small fraction of a mV to its asymptotic minimum.
Okay, just to clarify that i understand your point here lets refer back to Figure 9 and lets assume that the Tj (temp) is at 25C. Which is the normal operating temp is its opened enough, would you agree without going to far into the formula of temp calculations, would it be safe to say average working temp is going to be around the 25C mark? Now, would you say that it would be safe in terms of resistance on the Drain-Source and then safe on the temperature range to run this MOSFET gate at 5v. If i wasn't using a 5v output, would you say it may be better to provide it with 6-7v, this would be not going too overboard with the voltage to reduce the Rds(on) like you said so in your reply?

Does it seem like I understand to you?
 

BobaMosfet

Joined Jul 1, 2009
2,110
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.



Can we always assume that their test voltage is the minimum needed to turn it fully on?



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.
Note it says 'Greater Than 250uA' -- You don't know how much greater. It depends on what the OEM puts on the datasheet. You have to read the datasheet carefully and simply bear certain facts in mind. If the text is unclear, look at the graphs. Look at how the curves behave at given Vgs.

Now, if you want to talk junction thermal limits, you need to learn how to calculate that. I recommend getting yourself a copy of 'The Art of Electronics'-- they give an excellent how to for this.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Note it says 'Greater Than 250uA' -- You don't know how much greater. It depends on what the OEM puts on the datasheet. You have to read the datasheet carefully and simply bear certain facts in mind. If the text is unclear, look at the graphs. Look at how the curves behave at given Vgs.

Now, if you want to talk junction thermal limits, you need to learn how to calculate that. I recommend getting yourself a copy of 'The Art of Electronics'-- they give an excellent how to for this.
I have been reading about the thermal properties of MOSFETs and how to determine if you need a heatsink or not etc. Is this what you're talking about when you mention thermal limits? Is the thermal limit the point at wich it becomes 'unsafe/risky' to use without a heatsink?
 

ian field

Joined Oct 27, 2012
6,536
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?
Some of the MOSFETs on PC motherboards are as good as logic level types - a few will work with Vgs as low as 1.6V.

Most require a little more than that.

Ordinary power MOSFETs won't work unless Vgs reaches at least 6V.

Logic level types should work - but you need to study the MCU datasheet and see what the low and high logic levels are guaranteed to reach.
 

ian field

Joined Oct 27, 2012
6,536
This is what the datasheet says:
View attachment 104175
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.
The last MOSFETs I recovered from a scrap PC motherboard were rated for Vgs down to 1.6V.

Not all are that good - but with Id rated somewhere in the direction of 80A, the Rds-on was low enough with the voltage I could give it.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
I think the issue is that i'm expecting there to be some magical equation that gives me the voltage needed on gate when realistically its not just as easy as a simple single voltage.
 
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.[/QUOTE]





When the input voltage to the gate of the transistor is 0, the MOSFET conducts almost no current and the output voltage is equal to the supply voltage then the MOSFET is OFF. Minimum ON gate voltage essential to ensure that the MOSFET remains ON when carrying the selected drain current can be determined.
 
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.



Can we always assume that their test voltage is the minimum needed to turn it fully on?



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.

When the input voltage to the gate of the transistor is 0, the MOSFET conducts almost no current and the output voltage is equal to the supply voltage then the MOSFET is OFF. Minimum ON gate voltage essential to ensure that the MOSFET remains ON when carrying the selected drain current can be determined.
 

Alec_t

Joined Sep 17, 2013
14,280
Its how to work out the voltage needed to turn the MOSFET on and off that i'm not understanding.
You don't need to be too fussy working it out, since even nominally-identical FETs vary from sample to sample. As a rule of thumb, just stick as much voltage as you've got (providing it doesn't exceed the rated maximum) on the gate ;). That will be the best you can do to minimise the Rds(on). If you've got only 5V available, then the FET needs to be one specified as a 'logic-level' type. If you've got 10V available then almost any FET that can handle the required load current will do.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
I was just looking for a mosfet to use, and came to the decision of using http://uk.farnell.com/fairchild-semiconductor/bss123/mosfet-n-100v-sot-23/dp/9845321RL.

From its datasheet, i see the graph for "On-Resistance Variation with Gate-to-Source Voltage". Does this graph show how much resistance there will be in the MOSFET for the difference in voltage from the Gate to the source? So if there is 5v on the drain pin and then obviously going through to the source pin, does this mean that if i apply 5v on the Gate there would be no resistance what so ever because there is no difference in voltages from Gate and Source?
 

Alec_t

Joined Sep 17, 2013
14,280
Does this graph show how much resistance there will be in the MOSFET for the difference in voltage from the Gate to the source?
Yes. It's the resistance between drain and source.
So if there is 5v on the drain pin and then obviously going through to the source pin
No. It wouldn't 'go through to the source pin' unless the FET were fully switched on with Rds(on) =0Ω (which it never is; though it can get down to a few mΩ). So Vds is always >0V.
A hypothetical example. Consider the source is our 0V reference. If Vgate = 3V, say, then Vgs=3V and the FET may be partly turned on with Rds = 5Ω. If Vd = 5V then Vds=5V and the drain current = 5V/5Ω = 1A. If Vg is then increased to 10V, Vgs=10V, Rds goes down to, say, 20mΩ and if Vd is still 5V the drain current would rise to 5V/20mΩ = 250A.
 
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dl324

Joined Mar 30, 2015
16,846
So if there is 5v on the drain pin and then obviously going through to the source pin, does this mean that if i apply 5v on the Gate there would be no resistance what so ever because there is no difference in voltages from Gate and Source?
When an N channel MOSFET is operated as a switch, it's source is usually grounded. When you apply sufficient Vgs, the MOSFET is fully on and is operating at it's minimum Rds. So the correct way to view this is the drain is near ground.

If the MOSFET was operated as you're implying (load on the source), voltage developed across the load would decrease Vgs and cause it to start turning off.
 

eetech00

Joined Jun 8, 2013
3,859
The first thing to understand is that basically a mosfet is a voltage controlled device. Current thru the load is not really dependent on the current thru the gate (for a BJT transistor, the load current is dependent on current thru the base). It is controlled by the voltage level applied at the gate, that in turn, varies RDS(on). You can think of Rds(on) reflecting the value of a voltage controlled resistor that is controlled by the voltage applied at the gate pin. For an NMOS, the gate voltage level needs to be high enough (relative to GND) to turn the mosfet on, after that, rds(on) will decrease as the gate voltage is increased. A PMOS device works the same way except the polarities are reversed.

What are the voltage levels available that will be applied to Mosfet gate? 0v=low +5=high?
What voltage/current is required for the load (I couldn't find this info in the PIR doc you posted)?
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
The first thing to understand is that basically a mosfet is a voltage controlled device. Current thru the load is not really dependent on the current thru the gate (for a BJT transistor, the load current is dependent on current thru the base). It is controlled by the voltage level applied at the gate, that in turn, varies RDS(on). You can think of Rds(on) reflecting the value of a voltage controlled resistor that is controlled by the voltage applied at the gate pin. For an NMOS, the gate voltage level needs to be high enough (relative to GND) to turn the mosfet on, after that, rds(on) will decrease as the gate voltage is increased. A PMOS device works the same way except the polarities are reversed.

What are the voltage levels available that will be applied to Mosfet gate? 0v=low +5=high?
What voltage/current is required for the load (I couldn't find this info in the PIR doc you posted)?
This is exactly what i've learnt and you have just made me realise that i've learnt it correctly, i think. However like i keep saying i'm struggling to know at what gate voltage it will start to turn the mosfet on and at what point it will be on 'enough'. I've been looking at the resistance properties and the graph i was reading showed that at 5v on the gate it would give something like 1OHM (I may be wrong on the figure).

Does this mean that the mosfet will cause a lot of current to be drawn, increasing power consumption?

The MCU will provide 0v=Low and 5v=High like you assumed correctly.

From my understanding if the mosfet is open fully it will have very little resistance, am i correct here? If so does a logic level mosfet open enough at 5v to not cause much current impact on the circuit being switched? Do i have to look at graphs to see this property or is there a figure to find on the tables of the datasheet?
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
When an N channel MOSFET is operated as a switch, it's source is usually grounded. When you apply sufficient Vgs, the MOSFET is fully on and is operating at it's minimum Rds. So the correct way to view this is the drain is near ground.

If the MOSFET was operated as you're implying (load on the source), voltage developed across the load would decrease Vgs and cause it to start turning off.
When i mentioned putting the load on the source that was for a P Channel Mosfet, which i'm finding much simpler to understand due to the fact all i would have to do is give the gate 5v to stop the circuit flowing and then drop the mcu pin, grounding gate, to start it flowing. I think. But as i keep reading, the N Channel mosfet is just generally a better option. I'm finding more P Channels available at a lower price point and much easier to get my hands on.
 

Thread Starter

Sam Matthews

Joined Jan 16, 2016
178
Just a quick follow on example of Alec_t's. Lets say that Vds = 5v, Vgs = 5v and source is connected to gnd so Vs = 0v Vd = 5v and Vgate = 5v. From the datasheet of http://uk.farnell.com/fairchild-semiconductor/bss123/mosfet-n-100v-sot-23/dp/9845321RL it shows that with Vgs = 5v the Rds = 1.25(ish). So i would do drain current = 5v/1.25OHM = 4A. Is this right? Now this may be a stupid and dumbass moment but does the 4A drain current mean i can pass up to 4A through the drain or what?
 
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