PWM for mill power feed, MOSFET question?

[The Artful Bodger]

Hi, this is really an electronic question but there is a machinist component!:)

I have made a power feed for my mill using a DC electric motor, nominal 48V, and I am building a PWM controller for it. The open circuit voltage on the PWM controller is about 70V and the circuit works well. However the MOSFET gate takes continuous current, I thought that was a capacitive interface and current should be very low once the device is turned on?

The current I am talking about is in a pull up resistor between the gate and the 70V supply.

The MOSFET seems to be fully 'on' as it does not heat, it is just the pull up resistor that gets got, OK I can use a higher wattage resistor but I would like to know why there is any current on this leg.

thanks

[Jeremy_BP]

What size is the resistor?
When the PWM controller pulls the output low, it may be sinking some current.

[MaxHeadRoom]

A FET is a voltage driven device, there should be no (measurable) Gate current?
The FET should be switching from 0 to +dc.
Is this your own design or out of the IR app notes etc, or a proven design?
Max.

[Black_Moons]

The current I am talking about is in a pull up resistor between the gate and the 70V supply.



No mosfet on the face of earth is designed to take 70v to the gate, if it survived at ALL, its because theres a 20v~ zener internal to the mosfet from gate to source designed to save it from static/ringing/etc.

Mosfets are 'capactive' input, and have a *VERY SMALL* max Gate-Source voltage tollerance of usally about 20v (Check your mosfet datasheet for the exact voltage)

Its usally recommended to drive mosfet gates with about 12v, unless they are logic level fets, then 3~5v will do fine.

"I once was a mosfet like you, But then I took 70v to the gate"

[J Tiers]

No mosfet on the face of earth is designed to take 70v to the gate, if it survived at ALL, its because theres a 20v~ zener internal to the mosfet from gate to source designed to save it from static/ringing/etc.

Mosfets are 'capactive' input, and have a *VERY SMALL* max Gate-Source voltage tollerance of usally about 20v (Check your mosfet datasheet for the exact voltage)

Its usally recommended to drive mosfet gates with about 12v, unless they are logic level fets, then 3~5v will do fine.


+1

While that CAN work, if the voltage is limited by a zener, either one built-in to the device (not all have that) or external, there is no reason to do it, and EVERY reason NOT to do it.

The gate drive you need depends on what PWM frequency you will be using. Usually you won't need anything more than 1 kHz or so, but some like to have it out of the audible range.

At low frequencies there won't be an issue. At higher switching frequencies and voltages you need to take more care to drive the gate hard enough to switch fast and cut power dissipation in the mosfet.

[The Artful Bodger]

OK, I guess it is the internal zener sinking the current.

Thanks for the comments everyone.

[darryl]

I'd be interested to see what circuit you'e using. To me it doesn't seem right that there should be 70 volts available at the gate pull-up resistor.

JT is right- there should be considerable gate drive current available to overcome the gate capacitance and ensure a fast turn on and off, but that can come from drive circuitry operating from a lower voltage supply, say 12 to 15 volts. Then you don't need to cause the gate zener to conduct.

If there's something in the design that is beneficial to doing it the way you have, I'd like to think about it.

[The Artful Bodger]

It is a very minimal circuit.
I looked at some simple schematics which showed the full supply voltage going to the gate via a resistor. I used a 10K resistor to pull the gate up and I am using an NPN transistor to pull the gate terminal to ground. It all worked and the MOSFET was not showing signs of heating, it is just that the current in the resistor seemed wrong.



I rebuilt having thrown out the burned resistor and put another resistor, 1k, from the gate to ground which as expected steadies the gate voltage at 6.3V.

It all works and the only explanation I can conclude is that the original resistor was much less than 10K.

http://farm8.staticflickr.com/7174/6747102129_5c100c5fee.jpg (http://www.flickr.com/photos/25239206@N06/6747102129/)
schematic (http://www.flickr.com/photos/25239206@N06/6747102129/) by aardvark_akubra (http://www.flickr.com/people/25239206@N06/), on Flickr

[Evan]

6.3 volts on the gate isn't sufficient to ensure full turn on for most power MOSFETS. You need to supply between 10 to 15 volts to ensure full turn on of the gate. Most MOSFETS are limited to around 20 volts or so on the gate. The bias current can be very small for medium to low frequencies. No more than a milliamp should do. You can use a simple voltage divider from V+ to V-. With 70 volts use 47K to V+ and 10K to gnd. That will give about 15 volts in the middle which connects to the gate to hold it on.

You should also put a 10 ohm resistor in series with the collector lead of the transistor and the gate. Together with the gate capacitance of the MOSFET that will slightly slow down the turn off time and greatly reduced ringing of the motor circuit which could cause damage. That will help out the clamp diode if it isn't a super fast type.

[The Artful Bodger]

6.3 volts on the gate isn't sufficient to ensure full turn on for most power MOSFETS. You need to supply between 10 to 15 volts to ensure full turn on of the gate. Most MOSFETS are limited to around 20 volts or so on the gate. The bias current can be very small for medium to low frequencies. No more than a milliamp should do. You can use a simple voltage divider from V+ to V-. With 70 volts use 47K to V+ and 10K to gnd. That will give about 15 volts in the middle which connects to the gate to hold it on.

You should also put a 10 ohm resistor in series with the collector lead of the transistor and the gate. Together with the gate capacitance of the MOSFET that will slightly slow down the turn off time and greatly reduced ringing of the motor circuit which could cause damage. That will help out the clamp diode if it isn't a super fast type.


Thanks Evan. The only value I can find on the little data sheet for the IRFP450 MOSFET is 'gate threshold' min 2 max 4 volts. I assume this is a threshold and the gate voltage must be above that (i.e. not necessarily between the values)? The absolute gate to source voltage is +/-20, so 15 volts would be hard on, more or less, for this device?

[darryl]

Make the 1k resistor a 2.7 k instead, and the 10k should probably be a 1 watt size. When the gate is pulled down, the 10k resistor is dissipating 1/2 watt, so it's prudent to use the larger size. With a 2.7k gate to ground, the gate is sitting at about 15 volts when not pulled down, so at least you get a full turn-on. However, the higher the gate voltage goes, the less current is available to drive it quickly, because the 2.7k will be robbing the current. With the 1k in there it turns on even slower, and if your motor loads down, the mosfet will heat up because it's not being fully turned on.

If it works, it works, so I won't fault you for not making any changes. But if it were me, and stuck with the 70 volt source to supply gate turn-on, I would change the 10k resistor to a 15 or 18k, and use a 15v zener diode instead of the gate to ground resistor. All of the drive current is then available to give a fast pull-up of the gate. If you can safely rely on the internal zener, then you might make the gate to ground resistor a 3.3 or 3.9k. Just having this resistor in circuit will take some load off the internal zener diode.

I just did a quick look, but I didn't find any specs for the safe zener current. I typed in a few mosfet numbers I have, but didn't find this spec. If you have it for the particular mosfet you are using, it should be about 10 ma minimum in order to work safely that way.

I'm not aware of any circuit where the gate zener is used to keep the voltage to a safe level. Normally the drive circuit has a maximum voltage it can put on the gate, and that's below the zener voltage.

Just reading your last post- the threshhold voltage is that at which the device begins to turn on. In a power switching circuit like that, the gate should be getting a minimum or 11 or 12v, and 15v is a better value since the mosfet will then be acting more like an on-off switch, which is what it's supposed to do in a circuit like that.

Looking at the threshhold values again, 2 volts or less is where the gate needs to be for the device to be off. Better is to drive it right to ground level, as that more quickly draws the gate charge out to turn the thing fully off. In a switching circuit like that, the gate threshhold voltage doesn't really come into the picture. It would be important if it was to be a class A amplifier circuit, but you certainly don't want any operation in class A for this application. Limiting the gate voltage to 6 volts or so- if the load current is high, the device will be in class A and will get hot.

[The Artful Bodger]

Thanks Darryl, the circuit is very much a bread board at the moment and I will be paying attention to your comments when I come to build it again, if all goes well I will have three to do.

The real thrust of this project was to experiment with using a PC serial port to produce the PWM wave form. It works quite well but of course the logic is inverted (in that the motor runs when the serial channel is idle) so I will have to attend to that.

John

P.S. I just read your last comment, yes the MOSFET got hot when I tried to drive it directly with the signal from the serial port.

[Evan]

Feeding the gate 15 vdc will in no way be hard on the device. A MOSFET gate is a capacitor. It only draws current when it is charging and supplies current when it discharges. Once in either state no current flows through the gate since it is insulated from the rest of the device (providing the maximum gate voltage is not exceeded).

It is the electric field from the charge on the gate that either allows current to flow or "pinches" it off, called "pinch off". When the electric field is between pinch off and full on the device exhibits a resistance from the source to drain that is proportional to the gate voltage. Once full on the resistance is whatever the characteristic minimum resistance is for each particular part, often in the milliohms region for power devices. When full off the resistance is in the megohms.

It works very much the same as a vacuum tube. It is the solid state equivalent. The gate acts as a control grid. Once the voltage is either high enough or low enough the device either conducts fully or not at all. It is a voltage controlled device, not current. The only current that flows in and out of the gate is the charging voltage of the effective capacitor.

[J Tiers]

Any of the above will be OK for low frequencies.....

Yes, you SHOULD have at least 10V on the gate for a solid turn-on. That prevents problems.

Specs for your part (an old part number, but still current) are at

http://www.vishay.com/docs/91233/91233.pdf

A few items to consider:

1) the "gate threshold" or "turn-on voltage" known as "VDS(on)" is given for a very small current, around 1 mA.

You MUST provide substantially more than that to get the device conducting solidly. A mosfet with 2 to 4 volts threshold will take somewhere aroung 5 to 6 volts minimum to pull substantial current.

You want the mosfet turned ON, and no foolin around. For illustration of the issue , there will be a curve given in the spec sheet that shows the "typical" relation of gate voltage to maximum possible current. But don't count on it, just give the thing 10 to 15 V and KNOW it is "on" solidly. Figure 2 and fig 6 on teh data sheet show this, although typically the 6V would be just enough for reasonable current (10A or so).

Failing to do that risks having the mosfet "partly on", with substantial voltage across it, which will lead to heating and failure.


2) You want to turn the mosfet "on" fast, and "off" fast. The more time it takes to turn on and off, the more power is dissipated in the mosfet.

Your circuit is good for the "off",because the transistor pulls it right down. For the "on" transition, it is slower, but still should be OK for low frequencies.


3) I do not like the zener only approach, I prefer a resistor....the zener is a high impedance for any sort of transient that can reduce teh gate voltage and possibly partly turn off the mosfet.... and with hand wired circuits that sort of thing is quite possible to have.

a 1K resistor is a good source impedance for low frequencies, better, I think, than 2.7K or higher. The divider with 10K and 1K will have a source impedance of about 910 ohms.


4) with 1K though, you will have a slowish turn-on time, 10 to 20 microseconds depending on the voltage. Ok if you don't do it very often (low frequencies).

if you wanted to switch faster, you would want to change the drive setup.

All I have time for right now.....

[dp]

Substitute an incandescent lamp for the motor and measure the voltage at D to ground. With the MOSFET turned on it should be the SAT voltage of the device.

What do you do with the back EMF when the motor is coasting?

[dian]

hm, to which terminal do i connect the incandescent pull in resistor?


http://i973.photobucket.com/albums/ae218/romandian/komode013.jpg

where does the zehner pinch off come out?

[Black_Moons]

http://www.micropik.com/PDF/irfp450.pdf
Note the charts page 4.
Output charactistics and transfer charactistics.
Both show that it basicly hardly does *anything* at 6v, and really needs at least 8v before it starts showing any decent amount of proformance, and 10v would be better.

Also note the gate is speced 20v +-, So anything over 10v and up to 20v is fine.

That said, if you wanna PWM at much speed, your driver circuit really sucks. You really should be using a 12v supply for the mosfet and a more direct driver.
Note the 'input capacitance' of that fet is 2.6nF, thats pertty high for such a weak driver circuit so don't expect fast switching speeds or low mosfet disipation (Use a heatsink!)

Also, Any fet that does not scream 'LOGIC LEVEL MOSFET' on page 1 of its datasheet, is NOT to be used at under 10v gate voltage, Even if it says the threshhold voltage is 2~4v or whatever, threshhold voltage is just when the fet starts doing *anything*, Not when its actualy usable for switching.

[Evan]

Putting a 10 ohm resistor in the gate lead will stop hard turnoff without increasing dissipation. Together with the transistor turn on time it will still give a time constant far faster than needed for the intended application. It prevents the sudden turnoff that MOSFETs exhibit from causing high order harmonics in the ringing of the motor coils. It also prevents high frequency oscillations. The resistor should be placed as close as is practical to the gate lead of the MOSFET so that the inductance of the gate circuit is minimized.

Ordinary diodes are pretty slow and the more current they are rated to handle the slower they are. For this sort of application the gate resistor will slow the turn off time just enough to allow the diode time to conduct before the ringing rises too high. In this application it is best to use a Schottky diode as they are far faster than the usual power diodes. The main limitation there is the max voltage spec since Shottky diodes are usually pretty low voltage devices. A good source is from a computer power supply. Those have Schottky diode pairs thare are usually rated at a minimum of 100 vdc or higher.

This is the output circuit that I have been using in the PWM driver for the crossfeed/leadscrew drive on my lathe for years. It can easily handle much higher voltage such as 70 volts. It should work fine with the MOSFET you have too. The RC circuit across the MOSFET is a snubber that slows down the rise and fall of the motor ringing. The clamp diode is from a computer power supply.

http://ixian.ca/pics9/mosfetdriver.gif

[J Tiers]

Ordinary diodes are pretty slow and the more current they are rated to handle the slower they are. For this sort of application the gate resistor will slow the turn off time just enough to allow the diode time to conduct before the ringing rises too high. In this application it is best to use a Schottky diode as they are far faster than the usual power diodes. The main limitation there is the max voltage spec since Shottky diodes are usually pretty low voltage devices. A good source is from a computer power supply. Those have Schottky diode pairs thare are usually rated at a minimum of 100 vdc or higher.



To be fair, diodes with tens of amps rating and sub 100 nS recovery times are readily available for a couple bucks in qty one, parts such as RURP3060, etc.

Not that they are needed here.

The rule is generally that if you don't do something very often, i.e. low frequency, you can afford more dissipation when you do it, because the average power is still low. Naturally the "event" must be reasonable, not beyond ratings, etc.

The faster you switch, the more important speedy switching is. For very low dissipation, quasi-resonant zero voltage switching is very effective, if you can do it.

Again, not needed here., assuming, as seems to be the case here, that switching frequency is quite low. You might get away with nearly any diode and not have overheating.

At faster speeds, the problem is real, and must be taken into account.

DP:
Coasting isn't an issue unless the voltage exceeds the power supply, in which case it will charge the supply through the mosfet intrinsic diode. That would be an "overhauling load", which should not be the case with a feed drive, they normally cannot back-drive (ballscrews maybe).

Back EMF will be poled positive to the upper terminal.

[dp]

DP:
Coasting isn't an issue unless the voltage exceeds the power supply, in which case it will charge the supply through the mosfet intrinsic diode. That would be an "overhauling load", which should not be the case with a feed drive, they normally cannot back-drive (ballscrews maybe).

Back EMF will be poled positive to the upper terminal.

The system is a PWM so the coasting time is that time that the motor is not powered. That can actually be the majority of the time. The armature (DC motor) acts as a generator most of the time in that situation, and the load impedance approaches zero, to the ouput voltage is going to be maximum all during the coasting period. That can be a lot of voltage.

Evan's circuit suggestion adds a snubber diode which helps for current, but the total voltage is still the additive output of the motor and the power source. This is a consequence of a single stage driver vs a circuit with an active pull-down/pull up.

I think too the driver circuit for the MOSFET device is probably drawing too much current at Vce Sat than a FET requires hence the high wattage required for the collector resistor. This is another place where active pull up/pull down can remove some high dissipation components. Zener diodes or three-pin voltage regulators are helpful here. No need to drop 70 volts across the collector resistor when driving the MOSFET.

[darryl]

Since we're talking about driving mosfets, a few here might be interested in the little chips I'm finding on some of the flat screen tv circuit boards. There's a little 8 pin chip called S21850, made by IR, that is called a high side driver. I don't pretend to fully or even significantly understand it, but it apparently does all the chores of driving the mosfet properly, including allowing the gate and source terminals to go up to 600 volts high. In other words, you can put the load from source to ground rather than from + to the drain terminal. This is equivalent to the emitter follower configuration with bipolar transistors.

The chip is driven by logic level voltages. Could be quite useful, except for me it's too small. It's a PIA for me to deal with this close lead spacing now. The chip itself looks to be a tad less than 1/4 inch square, and there's eight leads coming out of it. Somewhere in the specs is says it has a 4A output current capability- how's that for charging and discharging the gate capacitance!

[Clark]

I used an IR high side driver on an off line switcher I designed for Physio Control's defibrillators in ~1996.

They work, but it is hard to watch them work. Where are you going to put the scope ground?

[J Tiers]

The system is a PWM so the coasting time is that time that the motor is not powered. That can actually be the majority of the time. The armature (DC motor) acts as a generator most of the time in that situation, and the load impedance approaches zero, to the ouput voltage is going to be maximum all during the coasting period. That can be a lot of voltage.

A "lot" of voltage?

Not really....

The voltage CANNOT get beyond the voltage rails. And the total voltage is limited in any case.

There are two sources for excess voltage....

1) spike voltages due to the "release" of the motor terminal by the mosfet. The motor and circuit inductance will cause the drain of the mosfet to "fly up" to the plus rail, at which point that REQUIRED diode across the motor will conduct and clamp it to the rail.

2) back EMF in the case of a PM motor, for instance. back EMF is limited.

Any motor acting as a generator has a "volts per rpm" constant.... at any speed the volts are some specific number.

AND, the back EMF is always LESS THAN the driving voltage, unless the LOAD drives the motor to a higher speed, as is possible with a hoist, for instance. if the back EMF were higher, the motor would not draw power, and would slow down.

When 'coasting" the motor is slowing, and the back EMF will be even less than when running full speed.
.
.
.
Now, the limits on voltage due to the circuit..........

if the voltage goes positive for some reason, the "catch" diode (which is REQUIRED) will conduct and clamp the voltage to the potive rail, ans mentioned.

if the voltage goes NEGATIVE for any reason, the intrinsic diode which is present in all mosfet structures will conduct and "clamp" the voltage to the circuit common.



Evan's circuit suggestion adds a snubber diode which helps for current, but the total voltage is still the additive output of the motor and the power source. This is a consequence of a single stage driver vs a circuit with an active pull-down/pull up.

Not sure what you mean here....... the snubber is the R-C circuit across the mosfet.

The diode across the motor is not an optional nicety, it is required, and is nothing new whatever, it has been standard for all single sided inductive loads.... relays even....



I think too the driver circuit for the MOSFET device is probably drawing too much current at Vce Sat than a FET requires hence the high wattage required for the collector resistor. This is another place where active pull up/pull down can remove some high dissipation components. Zener diodes or three-pin voltage regulators are helpful here. No need to drop 70 volts across the collector resistor when driving the MOSFET.

It would make sense to use a lower voltage drive.... we presume there is some sort of other control circuitry which must use a lower voltage, and if that is sufficient to drive the mosfet, it would make sense to use it.

That said, the direct drive is effective , even if you and I don't particularly like it as a design practice, and it only "wastes" about a half watt. Many other forms of power supply would waste that much power, and most would be much more complex.

Like most things, it is a tradeoff.


I used an IR high side driver on an off line switcher I designed for Physio Control's defibrillators in ~1996.

They work, but it is hard to watch them work. Where are you going to put the scope ground?

The old solution was to "float" the scope, carefully disconnecting it from safety ground..... Modern scopes, like the tek 720, and many of their 4 channel scopes, have inputs which are isolated to 1200V. I use the 720s at work, and can hook a channel directly to the output as a reference, to see the gate signal directly.

You do need to watch out for the common mode interaction between channels.... sometimes extraneous signals get on the signal, and you have to check that by seeing if they are still there when the probe tip is touched to the ground lead.

Also,when using the High side drive chips, or the high and low side pair (half bridge) chips, you need to put in a 'clamp" to prevent the high side "common" from swinging below the main chip "common". IR has an application note (I can get the number if anyone cares) covering that issue...

If you get in trouble with that, all sorts of bad things happen.... one client could not keep their circuit working more than a few seconds before the IGBTs and drive chip failed.... we found several issues including the lack of that "clamp", and had the thing running all day with no glitches.

[MaxHeadRoom]

There's a little 8 pin chip called S21850, made by IR, that is called a high side driver. I don't pretend to fully or even significantly understand it, but it apparently does all the chores of driving the mosfet properly, including allowing the gate and source terminals to go up to 600 volts high.

It is probably the same function that is now commonly done by the IR2110 or IR2101 in many current designs.
Max.

[The Artful Bodger]

Thanks for all the advice on this topic.

[dian]

o.k., but now, what do you do?

[Evan]

Use the circuit I published. It works and has been working for years.

[dp]

A "lot" of voltage?

Not really....

The voltage CANNOT get beyond the voltage rails. And the total voltage is limited in any case.

Yep - you're right. I reversed voltage output of the coasting motor. It is anti-phase to the power source when coasting.

[darryl]

Going back to post #16, the zehner pinch off voltage could easily come from any of the terminals on the right of that monster. In fact I don't think a zener would have much of a chance hooked up to that.:)