P-channel MOSFET as high-side switch gets hot

Question

So I'm trying to control my front lights of my car using an Arduino Nano. In total there are 14 lights (including foglights and blinkers etc.). I'm using two 75HC595 shift registers to control these lights. In my car I also have 14 LEDs so I can see which lights are on. To switch the lights I wanted to use MOSFETs, but because all the lights have a common ground I must use them as a high side switch -> P-channel. I'm using the IRF4905. The lights I'm switching are 55W. At 12V the current thats flowing is around 5A, which should not be a problem, right? The problem is that the MOSFET gets really hot after a short amount of time.

This is the circuit I used to switch one light (all lights would be a huge picture). Transistor is BC547C. Relay is the standard one in my car. Couldn't find lightbulb in Fritzing so I used a Resistor (R6 is lightbulb). The value of R6 is not correct

When the light is on there is about 11.52V across the (+) terminal of the battery and negative side of the light, but there is 10.85V across the light itself. So am I correct if I say there is 0.67V across the MOSFET? I measured the current and it was about 2.58A.

\$0.67V \cdot 2.58A = 1.7W\$ at the MOSFET

\${{0.67V}\over{2.58A}} = 0.26\Omega\$ while \$R_{dson} = 0.02\Omega\$

If I'm doing something wrong in my calculations please tell me. Is there something wrong with my circuit or do I have a faulty batch of MOSFETs (tried 3 different)?

Edit: I've already tried to remove the 74HC595 in the circuit but it's still running hot.

Answer 1

The problem

The /OE (output enable) pin of the 74CH595 toggles the output drivers between two states: driving the outputs (high or low) and high impedance (letting the output lines float). It is active low, so driving /OE low causes the 74CH595 to drive the outputs high or low depending on the data you have shifted in, while driving /OE high causes the outputs to float undriven.

You have tied the /OE pin to the 5 V supply trough the 10 kΩ resistor R29, so the output drivers are always in the high impedance state.

Ideally no current would flow in this state, and consequently the unnamed NPN bjt would never conduct any current, but in practice there is always some leakage current present:

There is no such thing as a "BC447C", so I assume that the transistor is either a BC447, a BC547C or some other similar part. If say +1 μA was to leak from the 74CH595 output to the collector of the BJT, the transistors DC current gain (likely somewhere between 100 and 600) would cause it to conduct from 100 μA to 600 μA of current from the collector to the emitter. For example a collector current of 300 μA would lead to a voltage drop of 3 V over the 10 kΩ resistor R7, leading to a Vgs (gate to source voltage) of -3 V.

This hypothesis is consistent with what you saw (Ids of 2.58 A and Vds of 0.67V), the gate voltage must be very close to the threshold voltage (Vgsth):

The fix

• Add a pull-down resistor between the base of the BJT and ground to stop stray currents from turning it on.

• Remove the unnecessary 10 kΩ resistor R29, and connect the /OE pin directly to ground.

• The LED would light up extremely dimly in its current configuration. Put it in parallel with the transistor base, with a separate current limiting resistor.

• Add a bypass capacitor for the 74HC595, as close as possible to the 74HC595.

• If this is going to be installed in a car, I would add transient protection for the MOSFET gate in order to protect it from any voltage spikes (the gate oxide is easily destroyed by overvoltage). A 15 V TVS diode would be ideal, but a zener will work as well.

The components you have chosen are just fine for what you are doing.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

We can now say that the p chan has low enough on resistance and the gate source drive volts are adequate .There is no protection for your gate so you could be damaging your fets from power supply spikes .Place a say 18V zener between gate source and try again.Another possibility is parasitic oscillations which a scope would prove or disprove .I think you should place a gate resistor in series with the gate of say 1K ohm to eliminate the possibility of parasitics. Your 12V rail should have some decoupling caps also.

Answer 3

I think, that the correct answer would be that the MOSFET is getting hot because it should be.

Let's make some calculations: According to datasheet of IRF4905, it has junction to ambient thermal resistance \$R_{ja}\$=62°C. Let's assume ambient temperature \$T_a\$ of 25°C and dissipated power of 1,7W. Now, according to this site: http://www.rohm.com/web/eu/tr_what7 to calculate junction temeprature you have following equation: $$T_j = T_a + R_{ja}*P$$ When you do the calculations for your case, you get junction temeprature 130°C - pretty hot, but still not hot enough to cause any damage to transistor. Still if you want to put it in some compartment, I would suggest adding a heatsink. Following circuit design advice given by jms is also a good idea.

Answer 4

My idea is that the ground point of MOSFET source pin is inductive as much as kicking the source voltage up and drop the gate voltage under theresold after the rising edge, causing a new inductive kick after rise of Vgs, preparing the conditions nearly same with the first kick, enough to make a stable oscillation. So, the recipe is to reduce the rising speed of turn-on voltage OR improving the ground connections.

If you first turn on the lights by the relay, then turn on the MOSFET and then turn off the relay, and if MOSFET is doing good, then this makes a nice hint supporting my idea.