Electronic Speed Controller

[JdOwNj]

I am building an electronic speed controller for a robotics application. It needs to be able to handle ~300 amps running, ~400 stall(Ford 1.6L starter motor). I realize that im not going to be able to get a single MOSFET that will take this kind of current. Ill have to run 3-4 in parallel, what MOSFETs would you suggest?

[russlk]

A mosfet is better than an IGBT for this low voltage application. You want low RDS, so something like the IRFP044 would be good, but you need about 10 in parallel. Check out this site, which has a 250 amp module: http://www.sensitron.com/partbrowse8.as ... roupid=HLF

[bobsRAC]

The number of devices required is mainly determined by the device junction temperature (Tj) limitations. For a given device current, a maximum Tj is specified (usually between 120 and 180dC). The Tj is above the ambient temperature by a temperature<p>Tja(max) = Rja * Pdiss, where Pdiss = I(max)*I(max)*Rdson(max)/n, Rja = junction-to-ambient thermal resistance in dC/W, n = number of devices<p>where Rja is the sum of the thermal resistances from junction-to-case, case-to-sink, and sink-to-ambient. The sink-to-ambient is given by the heat-sink.<p>By definition,<p>Tj = Ta + Tja<p>Therefore, the number of devices required for a given current I(max) can be determined by<p>n(min) = ( I(max)*I(max)*Rdson(max)*Rja )/( Tj(max) - Ta(max) )<p>As a design example...<p>Given IRFP3703 devices:
I(max) = 500A
Rdson(max) = .0056 Ohm
Rjs = 0.9dC/W
Rsa = 1dC/W (heat-sink)
Tj(max) = 170dC
Ta(max) = 30dC<p>n(min) = ( 500*500*.0056*(0.9+1) )/( 170-30 ) = 19<p>Do note that Rdson(max) is usually specified at Tj = 25dC. As the junction temperature increases, so does the Rdson. The IRFP3703 device is specified Rdson(max) = 0.0028, but that value doubles at Tj = 170dC.<p>------------------------------------------------<p>(Digikey: IRFP3703 $5.58@1, $3.86@10, $3.25@100)
I've used the International Rectifier IRFP3703 (Nch 30V 210A Rdson(max)=0.0028) for high current applications. It comes in a TO-247AC package with typical 0.9w/dC junction-to-sink (greased).

With 20 FETs ($78), you'd get<p>500A(max)/20 = 25A/fet
Tj(max) = 170dC @ 25A
Rds_on(max) <= .00028 @ 170dC, Vg = 10V
Pdiss(max) = 70W @ 170dC, 500A
Pdiss = 56W @ 170dC, 400A<p>For operation longer than 0.1s, the conditions are the same as continuous operation. This requires the dissipation of 70W @ 500A. The junction to heat-sink thermal resistance of .9w/dC dictates that the junction will rise 63dC above the temperature of the heat sink. The IRFP3703 is limited to 170dC for 25A each, so the heat sink must be kept well below 110dC (absolute maximum). Depending on your application, This may be do-able. For operation up to 120dF, a 1dC/W heat-sink is required.<p>Each FET represents a 21nF load, and 20 represent a 0.420uF load. This can be a formidable opponent if you intend to switch this at high speeds. The current required to drive the FETs is proportional to the switching frequency. In this case, you require 4.2uC each time you charge or dis-charge the gate. At 10kHz, Ig(avg) = 84mA. To keep your switching-time dissipation low, switching times (rising plus falling) should be kept to at most 10% of the switching period. Peak gate currents at 10kHz, therefore need to be 1.6A or more.<p>Special-purpose mosfet drivers are manufactured just for these conditions. MICREL, for example, manufactures special-purpose mosfet drivers capable of supplying 12Apk.<p>[ July 09, 2002: Message edited by: bobsRAC ]</p>

[JdOwNj]

Thank you for all the help.... Im gonna start getting the parts together...

[Juanu777]

Well, I'm here because contrary to my better judgement, I decided not to build the speed control circuit described in the magazine (June Issue) I decided to test my motion control system with a commercially available DC motor speed controller. The first few rounds went OK but then my 8051 controller board went up in flames. The manufacturer of the board said that my signal isolation was the culprit and I do not know enough about the subject to argue back. I thought I was isolated since the voltage I was applying came from a different circuit powered independently. So, would anyone please tell me what is signal isolation and whether I could build the circuit to drive a 90volt 5.5A motor with stop and reverse?<p>Thank you in advance

[bobsRAC]

The motor puts out HUGE voltage spikes when the comutator slides from one contact to the other. The most obvious cause of problems associated with these spikes is the tendency to damage other circuits attached to the same supply. As you noted, connecting this to a seperate power supply is absolutely necessary. Depending on your power supply, if these spikes are large enough, they can exhibit themselves as noise on the power lines and find their way down through other equipment. Therefore, It might be wise to either:<p> 1) Ensure that you've got good bypassing on the 8051 board. This means low-ESR capacitors in close proximity to the chip's power leads. Their effectiveness can be measured with an o-scope. Attache the ground lead to a ground point as close to the chip as possible and probe it's power leads. These should be free from voltage spikes associated with switching noise from the motor.
2) It is also worthwhile to investigate transients comming through the driver circuitury. Again, find a ground lead very close to the 8051 and probe the pins that are driving the motor. Look for any transients that appear when the motor is connected.

These spikes can transfer packets of charge to the base. If the base is driven by CMOS, the packets are likely cause the driver to go into latch-up. Also, these spikes show themselves by generating currents in the ground leads between the controller ground plane and the FETs. These currents can cause the microcontroller problems as well. <p>To prevent issues, use seperate power supplies, use latch-up resistant mosfet drivers, and use extra bypassing on the FET driver. Also most motors in RC applications include a snubber network to reduce transients. The FETs can also be bypasses with a series R and C to snub transients. You'll have to experiment with values until you find an acceptable time-constant. Motor snubbers are simply capacitors connected across the two motor leads, and from each lead to the metal casing of the motor. FET snubbers (as used when driving igntion coils -- this might be a good area to investigate for info on values) are connected in series from the FET drain (low-side switch) or source (high-side switch) to ground to provide a path for transients to be dissipated.<p>If you need further isolation, the extreme in this situation is opto-isolation. A simple digital level opto-isolater will allow the complete electrical discontinuity between the two circuits. The only connection need be to the LED on the controller side and to the photo-transistor on the FET side. No signals can be fed back through the opto.

[bobsRAC]

I took the liberty of investigating the cost of building such a motor controller, and found that a circuit meeting your requirements is possible, and relatively simple to construct. It's cost greatly depends on it's intended use.<p>I designed a bridge rated at 150Vpk, 7.5A continuous at Tamb(max) = 45dC(113dF). For this situation, the circuit turns out to cost around $50 + pcb costs. The size is around 5" x 3" x 3". Smaller size or higher Tamb requires more money .<p>The circuit consists of three stages per switch.<p>The first is a level-converter stage. A current source consisting of a pnp (Q1) whose base is two diode drops below V+, and whose emitter is tied to V+ through a resistor. The resistor value sets the current. The collector of the current-source is tied to the collector of a npn (Q2) in common emmiter configuration. The result of this circuit is a high-gain inverted signal swinging between the Q2's saturation voltage above ground and Q1's saturation below V+. The output (collectors of Q1 and Q2) is clamped at 12v + a diode drop above ground (for the low-side switch) or below V+ (for the high-side switch). This is acheived with the use of two 12v reguators, one supplying 10vdc above ground, the other 10vdc below V+. A diode is connected between the output from the collectors of Q1 and Q2 to the respective 10v rail.<p>The next stage is a bi-polar FET buffer. A dual n-ch/p-ch FET chip is connected in an inverter/buffer configuration (gates tied together, drains together as output, and sources to respective rails) between V+ and (V+ - 10v) for the high-side switch, and between gnd and 10v for the low-side switch.<p>The final stage is the power FET. The output from the bi-polar FET buffer is connected to the gate of the power stage. The drains of the two power FETs are tied together and the sources tied to the respective rails (V+ and GND). These FETs are mounted on a heat-sink as they dissipate 25.25W worst-case (at Tj = 150dC, 7.5Acont).<p>The parts, including the heat-sink, total to just over $40 on digi-key. A schematic can be supplied on request if anyone's interested.