Hi, I could really use a bit of tutorial in MOSFETS, I can't seem to predict their behaviour...
Circuit attached & screenshot with a variety of different N type MOSFETS being simulated in LT Spice
I am wanting to switch a battery (V1) over a 3kohm (R1, R2, R4, R5) load.
The MOSFETS are switched on by a 5V signal (V2; from Arduino)
Why is it that some of the MOSFETS will allow the whole battery voltage to be applied to the load and others will only allow a portion?
I'm guessing they are not being fully switched on? Sorry, this is a really basic question, but I am observing it in my circuit (3.7V battery when drained to 3.3V goes down to 2V when switched) and I was trying to simulate to see what happens.
Without looking up the numbers of those fets, my first guess would be you have some that are spec'd as "logic level" and some that are the regular version. The regular versions don't really start to switch on until something like 4v on the gate while logic level fets switch on with a lower gate voltage (e.g. IRF520 vs IRL520 - the IRL520 is a "logic level" fet. Take a look at the gate voltages for the FETS in your example and I think you will see what I mean.
dlloyd:
For all MOSFETs, connect the load (3K) from V1 to the Source, connect Drain to GND.
Not quite. You accidentally got it backwards.
For N-channel MOSFETs, (which these are), connect the source to ground, connect the load between the supply voltage and the drain.
Regarding threshold voltages, all of these MOSFETs should be suitable at the given supply gate voltage.
The highest Vgs(thr) is 2.2V.
I've modified the *.asc file and attached it. Give it a try.
(I saved the plot settings, so you just need to open it and click on "Run".)
dlloyd:
Yep. Can I blame whoever named those pins?
The arrow's a giveaway, but often that's accidentally drawn in backwards as well.
(I cheated and referred to the datasheets. )
I usually draw P-channel MOSFETs vertically flipped as the note in your image says.
Personally, I reckon that the inherent drain->source, (or source->drain), diode should always be drawn in as well. It's direction is yet another indication of whether it's NMOS or PMOS. It's also easier to mentally go over circuit operation when the diode is explicitly drawn.
Wawa:
OP has drawn source followers.
The voltage on the sources is the gate voltage minus the mosfet's threshold voltage.
Leo..
The question indicated that he didn't really want source followers, since he expected the full supply voltage across the load. That's why I swapped the circuit around to common-source.
Personally, I reckon that the inherent drain->source, (or source->drain), diode should always be drawn in as well. It's direction is yet another indication of whether it's NMOS or PMOS. It's also easier to mentally go over circuit operation when the diode is explicitly drawn.
Yes I think the diode should be include as it gives a very big hint if the FET is correctly drawn in the schematic.
Note the P MOS FET in the power supply of the UNO Rev 3 drawing, the diode plays an unique part in the circuit function.
.
Wawa:
OP has drawn source followers.
The voltage on the sources is the gate voltage minus the mosfet's threshold voltage.
Leo..
Hi, thanks everyone for your input - I should have been more specific - I am trying to switch on a DC-DC step-up dual voltage regulator. It accepts a >3V input and outputs +/- 30V.
On the input side, there is a (+) and (-) input
On the output side, there are the step up (+30V) and (-30V) output. There are also two other ground pins that appear to be connected to each other and appear to be also directly connected to the (-) INPUT.Does this sound right? Or is this a 'virtual ground'?
The reason why I arranged the FETS in this way was so that I could keep my "output ground" connected to the ground that I am using across the rest of the circuit.
So connecting it the way suggested does indeed let me switch it on and off, however in the off state, I have 0.86V across the load (the DC-DC converter). I am using this MOSFET:
rj87:
Hi, thanks everyone for your input - I should have been more specific - I am trying to switch on a DC-DC step-up dual voltage regulator. It accepts a >3V input and outputs +/- 30V.
On the input side, there is a (+) and (-) input
On the output side, there are the step up (+30V) and (-30V) output. There are also two other ground pins that appear to be connected to each other and appear to be also directly connected to the (-) INPUT.Does this sound right? Or is this a 'virtual ground'?
The reason why I arranged the FETS in this way was so that I could keep my "output ground" connected to the ground that I am using across the rest of the circuit.
Hope that makes sense
For some reason, I didn't get a notification for this post. Only spotted it when you posted the last reply. I'll address each in turn.
Using MOSFETs, the best/easiest way to keep your output ground connected to your input ground is to use a P-channel MOSFET, (high-side switching), rather than N-channel. Without a separate high-side driver, you can't use an N-channel because to turn it on fully you need about 5V more than the supply voltage.
Please draw a schematic of your intended setup, with as much info as possible. I can foresee at least one other problem if your input voltage might be as low as 3V.
rj87:
So connecting it the way suggested does indeed let me switch it on and off, however in the off state, I have 0.86V across the load (the DC-DC converter). I am using this MOSFET:
Using the 'low-side' switching shown in my modified version of your LTSpice circuit, if the gate of the N-channel MOSFET is grounded by the drive signal, there will be 0V across the load. (The MOSFET will not conduct at all.)
As I mentioned in my last reply, draw diagrams, so we can see what's going on without having to guess.
The last we knew was:-
I am wanting to switch a battery (V1) over a 3kohm (R1, R2, R4, R5) load.
I imagine the load is really going to be lower resistance. What current will the DC-DC converter need to draw, and provide at +30V/-30V?
OldSteve:
The arrow's a giveaway, but often that's accidentally drawn in backwards as well.
(I cheated and referred to the datasheets. )
I usually draw P-channel MOSFETs vertically flipped as the note in your image says.
Personally, I reckon that the inherent drain->source, (or source->drain), diode should always be drawn in as well. It's direction is yet another indication of whether it's NMOS or PMOS. It's also easier to mentally go over circuit operation when the diode is explicitly drawn.
Agreed.
Just to be complete the body diodes are only found on power MOSFETs, where its an integral
part of the vertical current flow design. Small MOSFETs (such as in VLSI/logic chips) are completely
symmetrical and there is no distinction between source and drain at all. Higher voltage and higher
power devices are more complex in design to get better performance and are thus asymmetrical,
in particular allowing the drain-gate voltage to be much greater than the gate-source voltage
maximum.
Drawing the gate lead near the source is the indication that the device isn't symmetrical BTW, the
symbol for VLSI MOSFETs is different.
OldSteve:
Using the 'low-side' switching shown in my modified version of your LTSpice circuit, if the gate of the N-channel MOSFET is grounded by the drive signal, there will be 0V across the load. (The MOSFET will not conduct at all.)
As I mentioned in my last reply, draw diagrams, so we can see what's going on without having to guess.
The last we knew was:-
I imagine the load is really going to be lower resistance. What current will the DC-DC converter need to draw, and provide at +30V/-30V?
Thanks very much for your help. The DC-DC Converter draws ~100mA at 4V so I have put it as a 40 ohm resistance on the attached diagram. I also have a fast diode connected up (don't know if it's necessary) since there seems to be a very large inrush current.
So now I have both the IN- and the OUT-GND connected together at the DRAIN of the MOSFET.
It switches, however there is always a certain voltage across the IN+ and IN-, even when OFF.
rj87:
Thanks very much for your help. The DC-DC Converter draws ~100mA at 4V so I have put it as a 40 ohm resistance on the attached diagram. I also have a fast diode connected up (don't know if it's necessary) since there seems to be a very large inrush current.
What do you expect the diode to do?
So now I have both the IN- and the OUT-GND connected together at the DRAIN of the MOSFET.
It switches, however there is always a certain voltage across the IN+ and IN-, even when OFF.
I don't know where to start. So you're saying that the connections aren't as shown in your schematic diagram and that instead the battery negative connects to the MOSFET drain, (And the battery positive connects directly to the +Vin terminal of the DC-DC converter?) So in effect the battery is connected straight to the DC-DC converter? (The 40 ohm resistor in your diagram.)
This is how I interpret what you're saying:-
Please draw a full accurate schematic, exactly as you now have things connected. Forget LTSpice, just use a pen and paper.
Edit: Just one other thing - how much current do you need to draw from the DC-DC converter's +30V and -30V outputs?
Do you have a link to it's product page/datasheet?
can someone help me how to make a switch using PMOS or NMOS?
I have an adapter which can out 12V 1.25A that would connected to my 5800mAh 42.9Wh battery pack.
how can i use arduin mega so that the switch which im going to make will automatically turns off when the battery pack is full at 7V. then will turns on when the battery pack reaches zero V.
the idea is that, the switch will turn on the connection between the adapter to battery pack, then on the battery pack there`s a PWM ckt which will read the LED of the battery pack, and as soon as all the LED is on, the pwm will send signal to arduino to turn off the switch. then vice versa, as soon as the battery pack reaches zero volts, the pwm again will send signal to arduino to turn on the switch.
please please someone with a brillian mind could help.
)
