Logic Level Mosfet? - Page 1
Forum section: electronics / beginners

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 07:31:45 am
content: Hello,I need suggestions on which type of mosFET to use.I will be driving 8 LED assemblies, each of which draw 246mA each at 12.8 volts. Since they are arrays built for automotive use, they have resistors built in. The base of the mosFET's will be connected to Logic Gates, so the input to the base of the mosFET's is only 5volts.So, I need a mosFET with a continuous draw of 2A. It will be wired like the picture. I need them to have a diode to help with any noise issues if they arise. A small surface mount device as well, NO through hole.There is a reason for the diodes at the bottom of the LED's. I actually have more LED arrays, and they all have a common ground connection, so the diodes there keep the from turning on/off for the rest of the circuit (which has SCR's in it) Basically, the LED's assemblies have two colors, 3 wires out. One is ground, and one is for power for each color. So the ground is common.Anyway, I have drawn the diodes to illustrate the voltage drop in the circuit, as well as the transistor at the bottom. There is a reason that transistor needs to be there, so that the other LED arrays in the circuit (different color) can be turned off via another transistor, via another logic gate.In simulation, I can't get more then 54mA or so through each LED. I don't know why. Maybe because the simulation program I am using are not Logic Level mosFET's? Is it just that they are not turning on all the way? I can't edit the perimeters for them in the simulation program.I even tried driving the mosFET's in the program using a transistor to switch on the mosFET, and I get 212mA through each LED, UNTIL I put in the diodes and transistor at the bottom of the schematic, then it drops to like 23mA. The diodes will be a schottkey diode with a voltage drop of around 0.42volts.The DIP switch is so that each LED can be turned off in the array for situations where all 8 LED's won't be used. I need it to connect to the bases of the mosfet, because the DIP switch can only have a little current through it.Would it be better to replace the NPN Transistor with an N-Channel mosFET? I am not sure on the values needed.Thanks for any help given. I appreciate all your guys help over the last couple months. I've been racking my brain with this circuit, and though I've learned alot, there's still more to learn.

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 07:35:54 am
content: Oh, one other thing.I noticed that as I switch the LED's off, the other LED's connect gain more mA through them, thus the LED gets brighter. I can't have that, they need to stay the same brightness, regardless of how many are switched on (or off). I don't understand why the circuit acts this way, to me, it makes no sense, unless the transistor at the base of the schematic is limiting the amount of current flow somehow.

author: Psi
author_rank: Super Contributor
timestamp: August 02, 2014, 08:01:58 am
content: Quote from: Falcon69 on August 02, 2014, 07:35:54 amAll your current for all leds goes through the Q9 NPN transistor.You don't have enough base current to saturate it so the more LEDs you have ON the more of a voltage drop your losing across the transistor and the less voltage left to run the LEDs (so more LEDS on = more current = more voltage lost in transistor = dimmer leds).You can lower R10 to get more base current as long as U1 can supply it.Add up the total LED current needed (all on) and check the DC gain of that Q9 transistor.That will let you work out the absolute min base current needed to fully turn on that transistor. Multiply it by two and that's what you need on the base.Or you could simply swap Q9 for a logic level N channel mosfet and change R10 to ~100R or just remove it entirely, but 100R is safer.(A mosfet only needs gate current to charge up the gate capacitance. Once the gate is charged the fet is fully on and it doesn't need more current to keep it on so you don't get the problem)

author: mij59
author_rank: Frequent Contributor
timestamp: August 02, 2014, 08:22:51 am
content: Hi,Q1..Q8 are not use the proper way, drain and source have to be swapped.Then you also need to change the gate circuit.

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 08:28:39 am
content: thank you PSI.I changed the circuit to have a NPN transistor at the bottom of the schematic for each LED assembly. That fixed the problem with the amperage changing in the remaining connected LED's when one (or more) are switched on/off. I am getting 126mA through each LED now, however, that's half of what it needs to be for those LED assemblies.

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 08:56:40 am
content: thank you mij59. I have made that change. Did not notice I had done that with those mosFET's.What do you mean change the gate circuit?

author: mij59
author_rank: Frequent Contributor
timestamp: August 02, 2014, 09:31:05 am
content: Quote from: Falcon69 on August 02, 2014, 08:56:40 amYou can't drive the mosfet with 5V, you'll need a pull down transistor.

author: Rick60
author_rank: Contributor
timestamp: August 02, 2014, 10:03:52 am
content: Q1 in this circuit is unecessary , shorting Drain to source will have exactly the same effectbasically when pin 4 on U1 is low , q9 is off so the fet is not used , when high q1 will be somewhere between hard on and the middle of it linear region depending on Vgs so in reality it goes beween hard on and maybe hard on so its a wireOn a P fet l the source is connected to +12,8v so to turn the fet hard on you need to set the gate to as near to 0v as you can get it , you have 12.8-5 so 7.8v a logic level fet should be hard on a normal one may not, as the circuit is drawn there is nothing limiting the currents so the fets can't be hard onso in reality you need to drive the gates via an open collector system ie add a transistor to pull the gates to 0v this also will stop the 100K/7( if all switches are on ) resistors trying to pull u1 pin 4 to 12v which u1 may not like

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 04:22:33 pm
content: The switches are there to turn off the individual LED's, if they are not wanting to be used. I'm trying to design the circuit to accommodate the use of 1-8 LED assemblies.And yes, I understand that I would have 7.8 volts going into the gate. I don't want that, that is why I need a Logic Level mosFET, but I am not sure which one. Mouser shows 19 available, but all have different values. Do I just pick the one with the lowest RDSon? If I use this schematic, that means I would need to select one that has a continuous current capability of at LEAST 246mA, preferably twice that, as that is what the LED's draw.http://www.mouser.com/Semiconductors/Discrete-Semiconductors/Transistors/MOSFET/_/N-ax1sf?P=1z0y3dt&Keyword=logic+level&FS=TrueThe transistor at the bottom of schematic IS needed for the rest of the circuit. The ground of these LED's will be connected to another transistor and controlled by another logic gate, (and more mosFET's above it feeding power to another set of LED's) to turn on a different set of LED's, which in the assemblies have the same ground. I can't break open the LED assemblies to rewire them to make them NOT common ground, so I need to find a way around this.Q1 IS needed, as it drives the first LED which will never be selected as OFF. (actually, the first 3 LED's will probably never be selected as OFF). In simulation, if the 100k resistors are not there, then when the switch is selected as OFF, it will not turn off the mosFET, thus turning off the LED. FYI, changing the resistor to 10M still turns off the mosFET, but changing to 10k does not.

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 06:17:59 pm
content: Would this one work?http://www.mouser.com/ProductDetail/Vishay/SiA817EDJ-T1-GE3/?qs=%2fha2pyFadujrHSVKQWQhPfgtIVUOTZLVOO6ycSFadcg%3dI like it, comes with a separate built in schottkey diode with a forward voltage drop of about 0.31v when less then 500mA applied to it. And it is cheap at just .23 each when ordered in multiples of 100.The RDSon isn't as great as others, but I think it will work at 80m ohm at -4.5voltsIt is 30 volt, so should be fine for an automotive application.The Gate-Source Threshold is low at about -0.95volts. I noticed in simulation when I lowered this value, I got more mA to the LED's.Can you maybe suggest a better one perhaps? I am lost when trying to understand all this data stuff, but from my understanding, I need it to be able to handle at least 500mA (twice the rating of the draw of the LED assembly array), be about 24volts or higher (twice the vehicle voltage), and a low RDSon, is this correct thinking?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 02, 2014, 10:03:06 pm
content: As shown, your circuit does nothing:When the logic gate output is high, all MOSFETs are off and all BJTs are on. No current flows.When low, all MOSFETs are on but all BJTs are off.Tim

author: Falcon69
author_rank: Super Contributor
timestamp: August 02, 2014, 11:22:22 pm
content: Okay, I redrew it.In simulation, with the mosFETs set at a threshold of -0.95v, The LED's Forward Voltage at 12volts (via a 10m ohm resistor, as the simulation shows error with no resistor......the assemblies have them built in), all the Schottkey Diode's Forward Voltages set at 0.31volts, hFE set at 100 on all NPN transistors, I am getting 11.45volts and 227.53mA through each LED.The problem I have, which probably isn't a big deal, as the existing incandescent lights draw WAY much more, is, I have 1.83A at the source voltage (the input of 12.8volts) and 1.99A at the ground. That's wasting 160mA through this circuit.I tried changing resistor values here and there, but it's all inverse of each other, change resistor value to a lower amount, too much current at ground vs current at source. Higher amount for resistors makes the LED's less mA passing through them.I am using Falstad's Simulation program found atwww.falstad.com.One other thing, if I get logic level mosFET's, I can do away with transistor/resistor Q9 and R9, correct?I have included the file for the Simulation I was using. Just open the applet from Falstad, import the file, and run it to see.Maybe I am doing something wrong with it. I just can't seem to get 246mA through the LED's and have the forward voltage of them set at 12.8v instead of the 12. That drops the mA down to 119.66mA and the voltage through them at 11.99v. The data sheet on the LED assembly says they draw 246mA at 12.8 volts, no other information is given on the data sheet for the lights.Thanks again Guys.Jason

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 03, 2014, 04:49:13 am
content: Each 100 ohm base current limiting resistor is drawing some 35mA...I don't get why you picked PMOS for the high side switches (which are more expensive anyway) and NPNs for the low side switches (which require gobs of base current when on). Why not NMOS instead of NPN? Or just one switch at all (leave the loads common to +V or GND, switch the other side).Tim

author: Falcon69
author_rank: Super Contributor
timestamp: August 03, 2014, 05:03:50 am
content: I tried NMOS for the low side switch, but could not get it to work properly.I need to switch each LED off at the high side, but I can make it so one switches off all of them for the low side, but when I tried that, it messed with a number of different things, like consistency in amperage or NOT enough amperage through LED's.