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My brother (the SWM) wants me to help him with an electronics project on his car. We will be driving a high-current, non-inductive load that will pull up to 13 amps. We will control the load using PWM directly from the CCP module on a PIC processor. I'm using this PWM schematic (linked to from here) as my foundation. The project is über secret... cannot disclose the details of the load. <p>I'll begin by saying that I read the FET article in the recent Nuts & Volts mag, although it was a little advanced (at least the terminology) for me to fully grasp. I have two questions. My first question revolves around the schematic. What is the purpose of the two diodes and two resisors (100 and 10K) near the FET?<p>My second question is, how do I choose a FET? I'd like to control it directly from the CCP port on the PIC. I've done a modest amount of searching on Google and found little information on this particular subject. Randy at Glitchbuster has listed some FETs and recomends an IRL530N which is described as "N-ch MOSFET logic-level gate voltage, 17A, 100V TO-220". From this description, I'd guess that the following aspects of a FET are important to consider:
- logic-level gate voltage
- current rating
- voltage
- package type<p>Randy caters to people who know what they want, so he may have left off some specs that might be obvious to his typical clients. Is there something else I should look out for? The only other spec I've seen on MOSFETs is a resistence rating. Do I need to consider the resistence? I want to ensure that a 5v logic signal will allow maximum voltage/current to flow to the load.<p>Thanks for any insight,

The 10K and 100K resistors form a voltage divider from the 555 output to the FET, putting apprx. 1 volt of drive on the FET gate. I am not familiar with the particular specs of your FET, but would assume it is to eliminate drawing gate current at high levels here. The two diodes are to eliminate destructive transient spikes upon FET turnoff. These would generally only be used when you have an inductive load (relays, motors,etc.) Only one diode needed for this purpose when load is grounded.

Save yourself a lot of effort. Use the IRL540N from glitchbusters. It is good for 36A at 100V. But the part you care about is the fact that is is logic level triggered. You can hook it up just the way you said and it will work fine. You may need a heatsink. <p>Connect one side of your load to +12V. Connect the other side to the drain lead of the MOSFET. Connect the source lead of the MOSFET to ground and connect the gate to the logic output of the PIC.<p>If your load becomes inductive you will need to deal with a kickback diode.<p>See http://www.irf.com/product-info/datashe ... rl540n.pdf<p>If you want to do it exactly right, in an automotive sense, see http://www.irf.com/product-info/datashe ... ps0151.pdf These intelligent power switches were made for the auto industry.

Hi Matt,
The 100 ohms is just put there to let the 100k turn the FET OFF completly, not as an active voltage divider. Normally the FET works best with the highest drive voltage. The diodes is, as said, there to protect the transistor. They do two different things, one short the inductive (positive)peaks from inductive loads, the other protects the transistor against negative voltage peaks, if any.<p>TOK

Hi Matt,
Forgot to comment on your other questions. The two key parameters is Rds(on) related to Vgs. For the two FETs named here the IRL540N is 2.5 times better in this regards than the IRL530N. If you use a PIC at 5 volts to drive this transistor you will have an output voltage at about 4v, which is the lowest Vgs in the data sheet.<p>At 13A current you will heat the 530 with 25W, and the 540 with only 10.5W. This is due to the Rds(on) value 0.15 ohm (530) and 0.063 ohm (540). The 530 will also drop your output voltage with 2V to .8V for the 540. This is the max. values, they may be lower for the unit in use. If you raise the Vgs to 10V the numbers is much better but power loss is still more than double for the 530.<p>TOK

Sorry about the error Matt
After reading Gorgons reply, I rechecked the print and saw that I mistakenly read the 100 ohm as 100K. It did appear very strange to me as usually FET's want more drive than that, but as I said I was not familiar with this particular FET, so I let it slide.. Its a good thing someone is always looking over your shoulder on these replys to catch stupid mistakes such as this.

You can also add a 470 ohm pull-up resistor on the gate of the IRL540N. This will help guarantee a rapid transition to full turn-on.<p>Connect the other end of the resistor to +5V.<p>This will add 10 mA of sinking current to the PIC but that is well within the pin capability.

If you're going to drive a 12V@13A load with PWM, I'd recommend an external driver. Sure, the PIC can drive a logic level FET, but it (probably) can't source enough current to turn the FET on fast enough to avoid heating as it moves through the linear range.<p>When you apply voltage to the gate of a HEXFET, it behaves sort of like a voltage-controlled resistor. That means that for the amount of time the FET takes to go from cutoff to full saturation, it is dissipating power and heats up. If you were just turning it on or off occasionally, the PIC would be sufficient. But since you're going to PWM it at (I assume) many kHz, the FET will be turning on and off thousands of times per second, heating up each time it goes on/off. A FET driver will force enough current into the gate to slam the FET on and off so fast that the heating will be negligible.<p>I recommend you first try the PIC with the FET on a big heatsink. If it gets too hot at max ambient temp (remember if it's in a car, the air temperature may be way up there), then investigate FET drivers.

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by rstofer:
You can also add a 470 ohm pull-up resistor on the gate of the IRL540N. This will help guarantee a rapid transition to full turn-on.
<hr></blockquote><p>This will also raise the PICs off output voltage to 0.25V - .3V depending on the temperature. This is getting very close to the gate threshold voltage that is min 1.0V and might increase the leakage/ temperature problem as mentioned.<p>If the PIC only solution doesn't work, use an external driver. <p>TOK