Electronic – Voltage drop over MOSFET

mosfetswitches

I have a question about high-side switching a load using an p-channel MOSFET.

I'm using the following configuration to control the backlight brightness of a LCD panel:

enter image description here

Q1 = N-channel logic level MOSFET so it can control the p-channel MOSFET using a microcontroller.

Q2 = SI2319DS p-channel MOSFET

Now because the backlight has a very specific voltage range (between 29,5V to 30,2V) I want to calculate the voltage drop over the source-drain terminals of Q2.

The datasheet of the SI2319DS states that for a Vgs of -10V Rds(on) is about 65 mΩ. This would mean that my voltage drop would be:
Vds(drop) = 65 mΩ * 120 mA = 8 mV.

Is this really the case? it seems so low.

Best Answer

Welcome to the miraculous world of low Rdson MOSFETs.
Yes, the results you have calculated will be ABOUT correct.
Data sheets give typical values in graphs and large print claims unless otherwise noted. At high power levels where the device is heated significantly by the switched current, actual Rdson can be up to about twice the 25C value for SOME devices. But at low currents compared to Idsmax, the Rdson will not vary much during operation due to heating.

Look at the graph at top left on page 3 of the data sheet. This shows the device current / voltage drop characteristics for various gate voltages. The Vgs = -10V "curve" is close to a straight line and intersects Vds = 1V at about Ids = 16A, or Rdson = 1V/16A = = 62.5 milliOhm =~ 65 mOhm as claimed.

Your chosen MOSFET is able to easily carry 3A or about 25 x your design current - so is not at all stressed and is liable to run at near ambient temperature. Thermal resistance on FR$ PCB is 166 C/W worst case.
Device dissipation at 120 mA and say 80 milliOhm Rdson is
P = I^2R
= 0.120^2 x 0.08 = 1.12 mW
for a temperature rise of aboyt 0.0012 x 166 C/W =~ 0.2 C :-)