Pull-down resistors for MOSFETs

[VoodooRoller]

Hi everyone,

*** Background ***
I started my first electronics project about 6 months ago and had to delve into the world of transistors to solve a few problems. Now my circuit is filled with MOSFETs acting as custom-designed logic switches and mini-relays for various functions (came up with a lot of cool new ideas).

The project has been on the shelf for a few months but now I am getting back on it.

One query I have is about pull-down resistors for the gates of my MOSFETs. As I understand, through both my own observations during prototyping and extensive Googling, applying a voltage to the gate and then taking the voltage supply away can--in layman's terms--leave some voltage 'stuck' at the gate which tricks the MOSFET in to staying active.
Adding a 'pull-down' resistor between the gate and ground can 'drain away' this 'stuck' voltage and restore proper operation.

*** Question ***
So the actual question is: do I have to use a separate pull-down resistor for each MOSFET or can I use a communal 'resisted ground' rail for these, with diodes on the bases?

To put what I mean in to pictures:-

Individual pull-down resistors:

One communal pull-down resistor on a rail:

The triggers are usually 12-14V and can be continuous sometimes for hours at a time so I want to minimise the current draw from grounding the triggers, hence the huge 10M resistors used which work in practice and only draw ~1mA through the 'short' created.
Can I go higher than 10M or will performance be sacrificed? If I have, say, 50 MOSFETs then that will be ~50mA that is just uselessly being shorted from +ve to gnd either through one communal resistor or through 50 individual resistors, so the lower I can get that figure the better.

Thanks for your time, and sorry if it's an easy question - I'm still learning

-VR

 

[duke37]

In the second diagram, when one input is high, the resistor will not pull down any of the other inputs.

What do the diodes do?

With 10M across 50 fets, the total resistance (all inputs high) will be 10M/50 = 200k
With 12V input, I = 12/200 mA = 60uA. Are my calculations right?

 

[(*steve*)]

And added to that... Are the source connections of the mosfets connected to, or at a very low potential with respect to, the ground you have shown?

As pointed out, the diodes do nothing useful. Remove them and have one resistor per mosfet, probably connected between gate and source.

 

[duke37]

There will be drive current only when the fet is turned on which presumably will be switching a much higher current.

Also, 10M seems a high value. It may well work in Alice Springs but not in Darwin with its higher humidity.

You will need high quality printed circuit board.

 

[VoodooRoller]

In the second diagram, when one input is high, the resistor will not pull down any of the other inputs.

Ah, interesting, I did not know that.

What do the diodes do?

I put them there as insurance against any, uhhh... 'backwards' voltage. Not needed in the first (individual) diagram but on the second (communal) diagram I assume that one trigger signal would activate all of the MOSFETs without the diodes i.e. if trig1 was at +12V and trig2 and trig3 were both open circuit.

With 10M across 50 fets, the total resistance (all inputs high) will be 10M/50 = 200k
With 12V input, I = 12/200 mA = 60uA. Are my calculations right?

Ah so one resistor with multiple 'inputs' shares out its resistance equally between them? Guess I should have known that lol
Also your calculations are correct; checked mine and they are wrong

And added to that... Are the source connections of the mosfets connected to, or at a very low potential with respect to, the ground you have shown?

A vast majority of them hit the ground at some point, although a couple go through resistors first (these are part of the current working prototype and function properly though).

As pointed out, the diodes do nothing useful. Remove them and have one resistor per mosfet, probably connected between gate and source.

Like this?

Does that not effectively turn the MOSFETs in to normal transistors?

Also at a glance I can see that in my circuit I use an NPN and PNP pair as a SPDT relay before the load (both throws connect +ve to the load but with different pre-set resistances). Should I connect the pull-down resistor for those to ground instead of to the source?

There will be drive current only when the fet is turned on which presumably will be switching a much higher current.

Also, 10M seems a high value. It may well work in Alice Springs but not in Darwin with its higher humidity.

You will need high quality printed circuit board.

How do I go about roughly calculating the highest value resistor I can use in the intended environment? It will be in a project box and covered from rain but otherwise will be outside, and here in England the weather varies from really dry to really wet (sometimes in a matter of hours).
As stated I'm looking at minimising the current draw through the pull-down resistor when the MOSFET is on.

I was planning on making my own PCB by toner transfer and chemical etching, but I haven't done much research into that stage of the project yet (want to get a complete stripboard beta prototype made up first).




Thanks for all your help so far

 

[duke37]

You should come to Derbyshire, they tell me it is balmy here, I think they meant the weather.
A pack of silica gell to absorb the moisture may help but the box needs to be sealed for this to work.

A bjt transistor will have a gain of up to 1000. A fet will have a gain of millions even with a resistor across the input. You have not said how the fets are driven, you may not need the resistors at all.

 

[VoodooRoller]

The signal wires for the bases of the MOSFETs do not all use the same type of source.

Some of the signal wires come from a pre-existing circuit that I'll be tapping in to, and will use the +12V supply to certain loads to detect whether they are switched on or off (they are all +ve switched). These signal wires when in the "off" state are, naturally, connected to the load which is connected to ground in the pre-existing circuit.

Without the resistors they will short out when on and either blow the fuse or bypass other load components in the circuit.

 

[(*steve*)]

The signal wires for the bases of the MOSFETs do not all use the same type of source.

OK, you may need to do different things for different signal sources. Also you may need to do different things for different loads, especially if you're turning them on and off rapidly.

Some of the signal wires come from a pre-existing circuit that I'll be tapping in to, and will use the +12V supply to certain loads to detect whether they are switched on or off (they are all +ve switched). These signal wires when in the "off" state are, naturally, connected to the load which is connected to ground in the pre-existing circuit.

A number of potential issues here. Firsly you need to ensure that the Vgs(max) of your mosfet is adequate, OR you need to provide protection for the gate.

You also sound like you're taking this signal from the power applied to a load. Is this load inductive? Is the power supply clean? Can the load ever be disconnected?

It may be simpler to provide protection for the gate. This is achieved by placing another (low value) resistor in series with the connection to the gate and a zener diode across the gate/source resistor. I would advise choosing a mosfet with a 20V (or higher) Vgs(max), using a 15V zener, and a 220 ohm series resistor for your 12V signal. (this assumes you're not powering a heavy load or switching it a lot.)

Without the resistors they will short out when on and either blow the fuse or bypass other load components in the circuit.

No, he meant that you might be able to remove the resistors. For the cost of a couple of resistors, I wouldn't bother removing them.

Perhaps you can tell us more about the loads you're switching, the sources of the signals to turn the mosfets on and off, and whether you've chosen any mosfets as yet.

edit: here is an image that shows gate protection. Note that there is no gate/source resistor shown. If present, it goes across the zener.

(It's the picture in the middle )