0
\$\begingroup\$

I've the following schematic for a buck converter charger circuit.

enter image description here

This is getting power from a 10Watt Solar Panel and charges a 7AH 12V Sealed Lead Acid battery.

I'm controlling the BC817 with PWM from an ATMega2560 at 62.5 Khz and usually the current is around 500 mAmp.

My issue is that the IRF4905 gets extremely hot during charging. With a small aluminum heat sink attached, it does reach temperatures near to 90 degrees Celsius, which is concerning.

The IRF4905 mosfet is designed to handle current up to 74A, so I really don't understand why this is happening here.

Any ideas are welcome.

Thank you!

|improve this question
\$\endgroup\$
  • 3
    \$\begingroup\$ The IRF4905 mosfet is designed to handle current up to 74A That 74A is actually irrelevant, if that PMOS has a Vds of 20 V and a Id of 1 W it will dissipate 20 W. If that PMOS has a Vds of 0.01 V and an Ids of 70 A it will dissipate 0.7 W and get barely warm. In your case probably the PMOS isn't switched on properly (so that is will have a low on resistance) and/or there is a more fundamental issue with the whole circuit. \$\endgroup\$ – Bimpelrekkie May 31 '18 at 17:48
  • 1
    \$\begingroup\$ 90 deg C isn't that hot - what ambient are you running it in and what is the thermal resistance of your heatsink. \$\endgroup\$ – Andy aka May 31 '18 at 18:13
  • \$\begingroup\$ @Andyaka Ambient Temperature is around 30 degrees C. The heatsink is 11X11X6mm No info on thermal resistance of the heatsink \$\endgroup\$ – Lefteris May 31 '18 at 18:19
1
\$\begingroup\$

Your driver is not the best for the job, that 470 Ohm resistance is too much and the gate voltage will charge smoothly, thus keeping the MOS too long in the linear region. Here's a quick'n'dirty change that can rectify that:

gate

I have made certain assumptions in there, about the supply voltage and the output load, but it shows what I mean. Can you spot the differences? If not, here's a plot of the instantaneous power dissipations of both MOSes (yours is the black trace):

pwr

This is for the first `ma of the simulation, so it's right in the beginning of the transient, but relevant, nonetheless. The average readings are 12.515W (black) and 2.512 (red). Nothing calculated, just a quick example.

Note, however, that what I did tries to maximize both the rise and fall times, so it's up to you to decide which and how by adding appropiate series resistors.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ The supply Voltage from the Solar panel is typically around 20V. Output load is like I mentioned around 0.5A \$\endgroup\$ – Lefteris May 31 '18 at 18:22
  • \$\begingroup\$ @Lefteris Then I am not far: the 20V is what I assumed, the load is 12V/7A (as you mention it). Still, those are details that I just tried to add them to be closer to your case, but, ultimately, it's the driver that's the problem (unless there's something else). \$\endgroup\$ – a concerned citizen May 31 '18 at 18:25
  • \$\begingroup\$ @Lefteris Bah, mistake, I read ampers but I did the load, so the currents are lower and the dissipation drops, but the conclusion is the same. In fact, the new values are ~800mW and 70mW, same large difference \$\endgroup\$ – a concerned citizen May 31 '18 at 18:27
  • \$\begingroup\$ Question is, can 800mW generate that much heat? The numbers don't add up, right? \$\endgroup\$ – Lefteris May 31 '18 at 18:48
  • \$\begingroup\$ @Lefteris Don't forget that the heating increases Rdson and can get twice as much, and there are also stray capacitances to consider, diode also heats up, also affects negatively the dV/dt, then the local ambient rises, the convection of air counts, the thermal resistance of the package, itself, there are plenty of factors that will add up, so take my numbers with a fairly large grain of salt. \$\endgroup\$ – a concerned citizen May 31 '18 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.