Electronics > Projects, Designs, and Technical Stuff
DC Lighting with MOSFETs help
sile:
Circuit Description:
I am trying to control some 12V DC lighting, (resistive). I am using a high side MOSFET controlled by a TLC59108F. http://www.ti.com/lit/gpn/tlc59108f. That device is a LED driver that uses PWM to control brightness. It's controlled by I2C and it's outputs are 17V tolerant. So, my thought was that I could use this to control my MOSFET. I have a 1k pullup resistor on the gate of the mosfet to 12V. I also have 1 of the outputs of my LED driver connected to the gate. The LED driver is a current sink so it just pulls the gate to ground when needed.
Everything works as expected... but only for a short time. Then the LED driver chip stops responding to commands. After looking at it with the scope there are some large negative voltages appearing on the output pin of the LED driver. As much as -10V.
Currently I am just prototyping this on a bread board and I am using an N-Channel MOSFET because that is all I have at the moment. I plan to change this to a P-Channel when I actually build the thing. Not sure if this would change anything or not. The only practical difference I see at a functional level is the Vth voltage drop across the MOSFET.
Tested fix
Adding a current limiting resistor between the LED driver and the gate of the MOSFET 22Ohm brings the negative voltages under 2Volts. And with a shotkey diode from ground to the pin of the LED driver it eliminates the negatives voltages all together.
Questions
1. What causes this negative voltage? The only thing I can think of is possibly the discharge of the gate capacitance into the LED driver creates such a high current that the inductance in the wires drives the voltage negative. If this is the case, would this be significantly less once on a PCB?
2. Is there any other effective way to eliminate this problem? I don't really want to add 2 more parts to each control line. The plan is to have 4x of these LED drivers with 8x channels on each. With 2 more parts per channel that is 64x more parts.
3. Will changing the part from N channel to P channel make any difference?
Thanks for your help in advance.
jahonen:
Yes, parasitic non-freewheeled inductance is probably the cause. Even with a PCB it can be a problem, unless you carefully minimize areas of all critical current loops.
Regards,
Janne
Psi:
might be a good idea to check that your grounds are connected in a star arrangement
am2pgs:
Adding a schematic of your circuit would help. Not sure about the negative voltage yet as I am not sure I undrestand your circuit.
I assume you have 12V connected to Source, the Gate to the LED driver and the Drain to the output.
If this is true, then the circuit should work with a P-Channel Enhancement MOSFET, since the VGS should be negative to turn it on (note there should be a pull up resistor on the gate to turn it off).
If you want to use a n-channel, I think the easiest would be to use it as a low side switch, connecting the Source to Ground and your load between the 12V and Drain. Applying a positive voltage to Gate relative to Ground will then turn on the switch.
to use an n-channel as a high side you will need a voltage on the gate that is higher than your source (i.e. VGS must be positive). Often they use charge pumps to boost the drain voltage and then they use it to turn on the FET.
Another thought: Why not use a PWM output from the processor (or whatever) directly to control the FET through a small PNP tansistor? i.e is the LED driver needed at all?
Zero999:
Yes, a schematic would help and I agree with the parasitic inductance theory.
The resistor should help but if the output of the IC is CMOS it should have enough series resistance and clamping diodes to limit the negative voltage pulses to a safe level.
Navigation
[0] Message Index
[#] Next page
Go to full version