Post by thread:MOSFET paranoia

Hello fellow Picers You need the diode across the motor to remove the back emf from the coil during the "OFF" period. This diode should be able to take the current generated by the coil (Check the data sheet for the coil value and work out the worst case energy stored during "ON" period.). The diode you mentioned across Drain - Source is inherent part of the MOSFET manufacture. This diode is (On N channel MOSFET) forward biased from source to drain. You have two problems using this device with high current application. (1) Vgs of this device is 4.5V for fully ON ie Rds = 007 ohm. If the Vgs is less than 4.5V you might be operating device in the linear region of the device and Rds could be serveral ohms, and power = Ids * Rds ? PIC high level worst case output is (Vcc-0.7)V, with 5V, Output = 4.3V Solution; Use an opto-coupler between PIC and MOSFET. Apart from solving the drive voltage problem, you could sperate the ground planes and reduce noise on the PIC supply. (2) During the OFF period the voltage on the Drain could go to 24V due to Back emf. The breakdown Vds voltage of the device is 30V. This is will stress the device. Solution; Try use a device with low Rds and atleast 40V Vds or use a 20V zener acrss drain and source. Regards, Jay At 01:26 22/04/00 -0700, you wrote: {Quote hidden}>I'm using a PIC 16C74A to drive a large 12VDC motor (an automotive radiator >fan), using one of the PWM units to drive a logic level FET. The FET is on >the low side, and has a diode between the drain and the +12 source. > >So far I have tested this using a small motor, and it appears to be working >OK. But before hooking up a larger motor, should I be installing better >diode protection? I have seen schemes where a zener is wired from source to >drain to keep the FET from seeing more than its rated voltage during >switching. I have also seen schematics with a diode from drain to source, >presumably to eliminate the chance of drain voltage falling below source. > >Are the extra diodes necessary? The specific FET is an IRL2203N. > >thanks for any comments > >

Jay.R.Vijay-Indra wrote: > You need the diode across the motor to remove the back emf from the > coil during the "OFF" period. This diode should be able to take the > current generated by the coil (Check the data sheet for the coil value > and work out the worst case energy stored during "ON" period.). Thinking here sounds woolly. Is it a motor, or a coil? What's all this about energy stored by the coil? The current that a "back" diode across a coil will carry will be - the current the coil draws. No more than that, and that value you should already know. A fast diode is not required because practically all diodes are fast to start conducting, "fast" diodes are all about how quickly they turn off. In this case, they have plenty of time to turn off when the "back EMF" current falls to zero by itself. Unless of course you are using PWM and need to turn the coil on again *before* the current has decayed. Now, motors behave differently. Permanent magnet motors generate a back EMF which opposes the applied voltage, and this is related to the motors rotation, which changes relatively slowly. In fact, the inductive kicks due to switching power on and off relate as much to inductance in wiring as to the (leakage) inductances of the armature, and a major problem is commutation spikes as the commutator contacts. Suppression of a motor thus consists of series inductors at the motor itself, a capacitor across the supply side of these, and diodes across the H-bridge driving it. Note these diodes are actually mounted at the H-bridge, not the motor, and may be part of the FETs. Of course, such diodes internal to the FETs are usually specified at the same current as the FETs, exactly as you require. > PIC high level worst case output is (Vcc-0.7)V, > with 5V, Output = 4.3V Only when loaded. The PIC will pull a MOSFET gate fully to 5V, but it may not do so fast enough - it is for this reason; the ability to drive the gate capacitance, that you may need special drivers. > Solution; Use an opto-coupler between PIC and MOSFET. Apart from > solving the drive voltage problem, you could seperate the ground > planes and reduce noise on the PIC supply. Opto-couplers can help, but only if indeed you are supplying separate ground *and* Vcc to the PIC and motor. It is as well to remember that common opto-couplers are rather *slow*. Otherwise it is simpler to ensure that the PIC is feeding a high- impedance driver input which simply cannot impress impulses back on the PIC and which is referenced to the power stage ground instead of the PIC ground. > (2) During the OFF period the voltage on the Drain could go to 24V due > to Back emf. The breakdown Vds voltage of the device is 30V. This is > will stress the device. If you have a diode across a 12V relay coil, fed from 12V, the back EMF seen by the switching FET will be 12V from the supply plus 1V (or so) across the diode. How do you calculate 24V? If there is no diode, then the back EMF across the coil can go to practically any figure, limited only by the breakdown voltage of the switching device. Now using a back EMF diode across a relay or solenoid coil significantly slows its release (as the current decays slowly). If the release must be faster, then a resistor in series with the diode will develop a proportional voltage and speed the current decay. A resistor equal to the coil resistance will develop back EMF equal to the supply voltage so that the FET sees twice the supply voltage on release. Note again that this applies differently for a motor - the back EMF consists of a spike due to parasitic inductances which rapidly settles to a value proportional to the rotation speed. The abovementioned diodes to clamp the former do not affect performance, whilst clamping the latter (by switching on one opposing low-side device) rapidly brakes the motor. > Solution; Try use a device with low Rds and atleast 40V Vds or use a > 20V zener acrss drain and source. The latter is indeed an alternative to a diode across the coil, and functions similarly to a diode in series with a resistance less than the coil resistance. -- Cheers, Paul B.

I strongly disagree with Jay's response on several points. Here's why: 1. PIC outputs are CMOS. So when the outputs are not driving much current into the load, the output voltage rises very close to the supply voltage. In your case, the output voltage should be within about 0.1 V of the positive supply. You can count on that because you are driving a MOSFET gate (purely capacitive) and (I hope) resistance to ground to make sure the MOSFET is off when the PIC output is tri-state, as during reset. A good resistor value to use would be 10K. You will have no problem getting to 4.5V on the gate, using a 5 volt supply. 2. Rds(on) of these FET's does not change quickly when you change the gate voltage. It's NOT a case where 4.5V means it is "on" and 4.3V means it is "not guaranteed on". While 4.5V gives you 0.007 ohms or less guaranteed, 4.3V will give you a higher Rds(on) - probably around 0.009 ohms. The "linear region" Jay mentions is for much lower gate voltages. Look at the data sheet. The very first graph shows that this device will conduct about 100 Amps with 4.0 volts on the gate. Where the graph is flat is the "on" region, and where it has slope and curvature is the "linear" region. With 4.3V on the gate, you will just have slightly higher power dissipation than if you had 4.5 volts. I assume you are talking about currents like 10 amps, so the power dissipation is less than 1W. Who cares if it's 0.9W or 0.7W? You don't even need a heat sink at that level. If you are using 20 amps, you may need to put a little heat sink on it. 3. This device is 'avalanche rated'. When you exceed the breakdown voltage of the device, it begins to conduct. This is called avalanche breakdown. In some FET's this is very destructive, but this device is designed for it. It even has specs for how much energy it can absorb, how much current it can handle, etc. It acts very much like a zener diode (a 30 volt diode, in this case). Operating it near 30V won't stress the device very much. 4. The back EMF of the motor can't exceed the supply voltage unless the motor is mechanically driven to higher speed than the no-load speed. It's called EMF when you use the thing as a generator, and back EMF when using it as a motor, because it opposes the supply voltage, reducing the input current. When the motor is running free, the back EMF is slightly less than the supply voltage, causing the supply current to be very low. If you increase the supply voltage, the motor speeds up until the back EMF again nearly equals the supply voltage. If you slow down the motor with a mechanical load, the back EMF falls, and the current increases. When the FET is off and the motor is still spinning (from momentum) the EMF makes the voltage on the FET lower than the supply voltage, not higher. This is fine, unless you somehow make the motor spin fast enough to generate more than the supply voltage. This would then want to put current backward through the supply and the FET. As Jay mentions, this FET has a body diode which will conduct this direction, and it is perfectly acceptable to use this. Your supply might not like this, though. You have to evaluate the supply to see. 5. The reason there's often a diode across the motor is not to protect from back EMF (which would reverse-vias the diode anyway). The problem is that the motor winding has quite a bit of inductance. When you suddenly turn off the FET, this inductance will force the current to continue to flow for a short time. If you don't provide a path for this current, one will be provided for you. The voltage can rise to several hundred volts, destroying your electronics. What usually goes is the transistor. This is EXACTLY what the avalanche rating is for. Bipolar and non-avalanche-rated FET's can't handle this, but if you stay within the specs, an avalanche-rated FET can provide the path for this current without suffering any damage. The alternative is to put the diode across the motor. This provides a safe path for the current. It is exactly like the "catch" diode in a relay driver. If you are really pinching pennies (or space-limited) and don't want to use the diode, you have to run the calculations to see if you are within specs of the avalanche rating of the transistor. IMHO, it's a lot easier to just put the diode there. Jay's first paragraph describes how to select a diode. For a 1-off, I'd probably just use a 1N4004. Its surge current rating is sufficiently high to handle most 10-20 amp motor loads. For production, run the numbers and see if this is a good choice. The other thing Jay mentions is noise. Motors with brushes generate LOTS of electrical noise. That's why radio-controlled toys usually have capacitors across the motor leads, and sometimes from each lead to the case. This plus good layout should handle the noise, but if not, opto-isolation might be a good solution. One last point - If you are using this in an automotive environment, choose a higher voltage FET. The electrical system in a car is very harsh. At times, the alternator can surge to 40 or 45V. That's why all automotive electronics are specially rated - they have to handle this input voltage. I think you've made a good choice for the FET. A cap or three on the motor (AT THE MOTOR) and a single catch diode should work great. >Jay.R.Vijay-Indra wrote: >Hello fellow Picers > >You need the diode across the motor to remove the back emf from the coil >during the "OFF" period. This diode should be able to take the current >generated by the coil (Check the data sheet for the coil value and work out >the worst case energy stored during "ON" period.). > >The diode you mentioned across Drain - Source is inherent part of the >MOSFET manufacture. This diode is (On N channel MOSFET) forward biased from >source to drain. > >You have two problems using this device with high current application. > >(1) Vgs of this device is 4.5V for fully ON ie Rds = 007 ohm. If the Vgs is >less than 4.5V you might be operating device in the linear region of the >device and Rds could be serveral ohms, and power = Ids * Rds ? > >PIC high level worst case output is (Vcc-0.7)V, > with 5V, Output = 4.3V > >Solution; Use an opto-coupler between PIC and MOSFET. Apart from solving >the drive voltage problem, you could sperate the ground planes and reduce >noise on the PIC supply. > >(2) During the OFF period the voltage on the Drain could go to 24V due to {Quote hidden}>Back emf. The breakdown Vds voltage of the device is 30V. This is will >stress the device. > >Solution; Try use a device with low Rds and atleast 40V Vds or use a 20V >zener acrss drain and source. > >Regards, > >Jay > > > >At 01:26 22/04/00 -0700, you wrote: >>I'm using a PIC 16C74A to drive a large 12VDC motor (an automotive radiator >>fan), using one of the PWM units to drive a logic level FET. The FET is on >>the low side, and has a diode between the drain and the +12 source. >> >>So far I have tested this using a small motor, and it appears to be working >>OK. But before hooking up a larger motor, should I be installing better >>diode protection? I have seen schemes where a zener is wired from source to >>drain to keep the FET from seeing more than its rated voltage during >>switching. I have also seen schematics with a diode from drain to source, >>presumably to eliminate the chance of drain voltage falling below source. >> >>Are the extra diodes necessary? The specific FET is an IRL2203N. >> >>thanks for any comments >> >>

I see so many posts around this subject, and I think it deserves a little discussion about. Lets start with my thoughts and doubts. I need to reckon I never measured such thing, but I believe that using a simple diode in parallel to the coil, will generate a back emf current that will be different than the current loading the coil. One could say that what we have "stored" into the magnetic field it is much more a relation of Power-vs-Time than current. A low value resistor in series with the diode, allows the use of lower current diodes, so lower cost. If the impedances are similar, you could consider the loading current as to be around the same of the back induced current. Now this is the tricky thing; One can think that when connecting the coil to a low impedance power supply, the generated magnetic field will be a direct relation of the final (steady DC) current through the coil. The (stored) magnetic field will be the same if you keep the DC current flowing for 5 seconds or 5 minutes. It means that the retrieved back-emf will be the same once the loading current is steady. So Power-vs-Time is difficult to formulate. When the magnetic field collapses, it just collapses and disappear. Any metal in the middle of this field will generate an induced voltage, if this metal physical shape goes according to some rules, defining mostly voltage polarity (electrons at one side, lack of electrons at another). If a electrical load is applied to that polarity nodes, an electrical current will be flowing from one node to another via the load. The magnitude of this current is just a Ohms Law, Current = Voltage / Resistance. If this metal shape is a ring, there will be no polarity side, but a complementary voltage loop, the same as if the load above is just a short circuit. With a *very low* resistance load, the current tend to be *very high*. Now, energy can not be created, so, to understand better how the magnetic field interacts with a metal coil, lets think of two coils exactly the same, electrically isolated (as a transformer). If you apply a steady (DC) electric current in coil "A", a magnetic field would be developed. Removing that current, that field will collapse *INTO BOTH COILS*. If you don't have any load at any coil during the field collapse, both coils would present a high voltage, lets imagine 5000 Volts. This generated voltage level can be seen as the maximum quantity of electrons the collapsing magnetic field can offset at the metal, like a saturation thing. Now, if you apply a load to any coil, the other coil (the open one) will not present 5000 Volts any more during the field collapse. This is because the loaded coil "unloaded" its voltage (electrons at one side, lack of electrons at another) through its load. This electrons flow open space for more "induced voltage effect" from the collapsing field. Think as if there is an "impedance" to the collapsing field to generate volts, if you steal some of the generated volts you reduce this "impedance", the field feels easier to generate more volts at that coil. You can think as if the loaded coil sucks more collapsing magnetic field than the unloaded (open) coil. What happens here is that if you have a load at *any* coil one of this multi-coil system, the overall back emf will be reduced. At the end, the stored energy should flow to the easier path. Ok, What about to save this (magnet) back emf energy, instead to just waste it away over a diode? +12Vdc----o--------o--------------------. | | | | | A B | | '------. .----. --- === S||S | A Diode --- S||S | | | S||S | | | .------' '----------' Gnd | | | Gnd |----' //---------|<---. |----| | Gnd The proposition here is that the coil B should have less wire coils than coil A. During the field generation, the voltage generated at the coil B will not be enough to bypass the diode+12V so it will not interfere with the magnetic field and/or with the coil A impedance, but yes at the collapsing field time, and so it will return current to the Voltage source, battery, whatever. The biggest generated voltage at the A coil during the collapse will be 12.7V, so the transistor should stand 24.7V VCE rupture. Is it enough for a Sunday? :) Wagner. "Paul B. Webster VK2BZC" wrote: > > The current that a "back" diode across a coil will carry will be - the > current the coil draws. No more than that, and that value you should > already know. A fast diode is not required because practically all > diodes are fast to start conducting, "fast" diodes are all about how > quickly they turn off. In this case, they have plenty of time to turn > off when the "back EMF" current falls to zero by itself.

On Sun, 23 Apr 2000, Wagner Lipnharski wrote: <snip> {Quote hidden}> Ok, What about to save this (magnet) back emf energy, instead to just > waste it away over a diode? > > > +12Vdc----o--------o--------------------. > | | | > | | A B | > | '------. .----. --- > === S||S | A Diode > --- S||S | | > | S||S | | > | .------' '----------' > Gnd | | > | Gnd > |----' > //---------|<---. > |----| > | > Gnd > > > The proposition here is that the coil B should have less wire coils than > coil A. During the field generation, the voltage generated at the coil B > will not be enough to bypass the diode+12V so it will not interfere with > the magnetic field and/or with the coil A impedance, but yes at the > collapsing field time, and so it will return current to the Voltage > source, battery, whatever. The biggest generated voltage at the A coil > during the collapse will be 12.7V, so the transistor should stand 24.7V > VCE rupture. This works well for *Ideal* transformers. But there's always a little "leakage" inductance that won't couple to the secondary. So some protection will still be required. If you're using the transitor to which Donald referred, I see no problem with this circuit. Otherwise you'll need a snubber like those designed for flyback converters. (An antiparallel diode by itself will nullify the energy recover benefits of the secondary.) I guess another problem is adding that winding to the motor :). > Is it enough for a Sunday? :) Nah, we've come to expect it from you Wagner! Scott

Wagner Lipnharski wrote: > I need to reckon I never measured such thing, but I believe that using > a simple diode in parallel to the coil, will generate a back emf > current that will be different than the current loading the coil. You need to go back to theory. The coil itself is a resistor in series with a theoretical inductor. Whileever the current is constant, the EMF across the theoretical inductor is zero. The EMF across this component is always proportional to the rate of change of current, in fact, E=L.dI/dt (where the dot means "times"). Damped only by the parasitic capacitances, when you turn off your switch, that voltage rises to *whatever* voltage is necessary to limit the change in current. The *whole concept* of back EMF hinges on the inductor's behaviour in keeping current constant. And at the very moment of the switch turning off, be assured, the current *will* be the same as it was instantly before the switch turned off. From that point, the decay of current (i.e., it smoothly reduces) depends on the back EMF developed. With a diode directly across the coil, the back EMF is applied to the coil's intrinsic resistance and since this resistance is the same as that which limited the steady-state current before and the current is indeed the same, then that internal (unseen) back EMF is the same as the voltage previously across the coil and the current decays over the same curve as it built up when the switch first turned on (making pull-in and drop-out times similar). If you place resistance in series with the diode, or return it to a higher voltage the current decays quicker accordingly, but the decay only ever starts from the previous coil current. OTOH the *average* diode current is quite low unless as I stated, you are using PWM. Diodes have a (recurrent) surge rating well above their continuous rating and these ratings have a lot to do with power dissipation and an integrating interval for same. > One could say that what we have "stored" into the magnetic field it > is much more a relation of Power-vs-Time than current. A low value > resistor in series with the diode, allows the use of lower current > diodes, so lower cost. That is an intuitive observation which holds out in practice not because the *initial* current is lower; it certainly is not, but because the diode may be rated higher for infrequent surges. > If the impedances are similar, you could consider the loading current > as to be around the same of the back induced current. That is woolly. There is no "back induced current", just one that decays. {Quote hidden}> Ok, What about to save this (magnet) back emf energy, instead to just > waste it away over a diode? > > +12Vdc----o--------o--------------------. > | | | > | | A B | > | '------. .----. --- > === S||S | A Diode > --- S||S | | > | .S||S. | | > | .------' '----------' > Gnd | | > | Gnd > |----' > //---------|<---. > |----| > | > Gnd An interesting idea, but you actually gain very little due to the resistance of coil B. Scott mentions that leakage inductance will give you grief and please note that if you have fewer turns on B than A, the transformer action makes the voltage on the FET *more* than twice the supply voltage e.g., if B has half the turns and half the resistance, then the induced EMF in A will be twice the supply voltage, *added* to the supply voltage giving three times the supply voltage across the FET. -- Cheers, Paul B.

> If you place resistance in series with the diode, or return it to a > higher voltage the current decays quicker accordingly, but the decay > only ever starts from the previous coil current. Paul, There is something wrong here because your conclusion doesn't make sense. Adding R to reduce a time constant doesn't compute. At the moment before you turn off the inductor, the current flowing is determined by the coil R and the applied supply voltage. When you turn off the switch, the field starts to collapse and *attempts* to generate current flow in the same direction it was flowing when it was on. Current can only flow if there is an EMF to drive it. Obviously, if it is an open circuit, current can't flow so the assumption that "it will be the same" doesn't hold. Rather, the collapsing field generates an increasing potential difference across the coil. If you have a diode across the coil then the voltage will only rise to 0.7V before current starts to flow and will be held at that level. The maximum current that can flow in that situation is Vdiode / Rcoil and the instantaneous current (and therefore the discharge time) can be determined from E(1-e^(L/R)). If you increase the R in that circuit it can only increase the time to discharge the field. Since you can only increase the time constant, the only way you can reduce the time to discharge is by allowing the voltage across the coil (circuit) to go higher either by a zener in series with the catch diode or by using the avalanche characteristics of a switching fet. Steve. ====================================================== Steve Baldwin Electronic Product Design TLA Microsystems Ltd Microcontroller Specialists PO Box 15-680, New Lynn http://www.tla.co.nz Auckland, New Zealand ph +64 9 820-2221 email: spam_OUTstevebTakeThisOuTtla.co.nz fax +64 9 820-1929 ======================================================

Ah! my teritory again, OK so you nedd to drive a FET to control a blower motor, good! But alas! Oh yeh, there are some real gotchas in this one! Firstly, You are using the FET to drive on the low side, bad boy! There are some problems with this that may not seem apparent, a) The drain of the FET will always be at 12V and you can not connect this to chasis (I am assuming that this thing will draw 15Amps plus) b) A fuse will be required unless you have protection in the FET c)FETS (The cheap ones) fail mode is ON, thus the FAN will run always (Not a nice condtion) d) As you are low side switching you will also need to supply two heavy wires, one on the drain and one on the source Second a) As you are low side switching you are suseptable to the common failure mode of short to ground (Around 95% of all failures) b)This item is not acceptable to the eurporean market (Low side switch, only high side is protectec from shorts to ground, VBATT and short on load) Third Unless this FET is located in a shield and sufficent sheilding of the drain, source and load wiring is performed this device will not pass EMI requirements. In this case it is better to mount the device in the motor. You will fins that the drain wire can be no longer than about 150mm before you will fail this test Fourth I have seen some people suggest a 40V process on the FET, ignore this, FORD is looking at bringing back 80V load dumps, GM has a 60V dump, also check to sensure that the reverse charaterisitcs of the FET Averlanche diode are at least that, that the motor current is expected to be Dennis > {Original Message removed}