Im currently trying to design a 12v to 5v step down PCB which revolves around the LM22673 chip (Buck Switching Regulator) to provide 5v 3A power to power two USB ports for a Raspberry Pi 4 and Screen in a car.

I sent my first circuit board request to a vendor in China before realising, oh whoops, I've used 6mil trace widths on the power, the circuits going to melt quite quickly. So I'm redesigning. I can certainly up the trace widths using 1oz copper to 35.3mil.

However I've now realised something I cannot see in the specs. I've got the Output current, but not the Input current. So how am I supposed to know what the trace widths should be for this? The betting man in me is looking at the (relatively) huge inductor and thinking that this is where the bulk of the work occurs and everything before the inductor will probably still be minimal amps.

I also have some TPS2001C chips too for additional USB ports

Any thoughts to help a beginner?

Simon

 

Hi,

A quick estimate for a buck is the input current is the output current times the reciprocal of the step down ratio. In your case the step down is 12:5 so the input current is 5/12 times the output current.
Then there is efficiency.
If you want to take that into consideration also, then multiply that figure by 1/Eff where Eff is the efficiency.
So if you have 80 percent efficiency then Eff=0.8 so the input current is:
inputcurrent=5/12*(1/0.8)*outputcurrent
In your case that would mean about 1.6 amps input current for normal operation. It could be higher during startup but should not last long.

That's a quick estimate really.

 

Hi Simon, I tend to use this as a reference:

But generally I would ALWAYS use much thicker tracks than calculated. If you have the room, use as big as you can.
For power calculations.. your output power is 15W... the buck regulator should be about 90% efficient.. so a worst case assume say 80%.. so your input power should be 15W x1.2 = 18W...@12V=1.5A..

Make sure you try and follow the PCB layout guidelines for the regulator, usually this means locating the regulator diode, inductor, and input and output capacitors as close to the IC pins as possible, and using copper pour or very thick tracks.

 

Hi Simon, I tend to use this as a reference:
View attachment 195888
But generally I would ALWAYS use much thicker tracks than calculated. If you have the room, use as big as you can.
For power calculations.. your output power is 15W... the buck regulator should be about 90% efficient.. so a worst case assume say 80%.. so your input power should be 15W x1.2 = 18W...@12V=1.5A..

Make sure you try and follow the PCB layout guidelines for the regulator, usually this means locating the regulator diode, inductor, and input and output capacitors as close to the IC pins as possible, and using copper pour or very thick tracks.

Hi,

I assume that shows temperature rise so the ambient temperature would have to be added to any result obtained from the chart.

 

There are loads of track width calculators, the problem was really trying to work out the input current - and I don't particularly want to splash out on PSpice.

Thank you for the advice though.

Looking at the board design again however, it is interesting seeing how other manufacturers do things, and quite often it seems they use copper pours for voltage and ground. So Im using some polygons in AutoCad Eagle which allows me to visualise the layout better.

 

Another small question, given that I know I've made a mistake on the design, do you think it's worth proceeding with the assembly of the circuit board or should I try and get them to cancel? Im in two minds. On one hand, yes, the board won't work. On the other hand, it'll give me a reference point to see how things work before I commit to the second version. I should at least be able to test many of the other features and even I just test some LOW POWER USB's, it'll be good for PoC?

 

In a buck regulator all of the current for the output comes from the source during the on time of the switch. The average input current is lower than the output by the ratio of the duty cycle, but the peak current is the same as the output current.

Bob

 

In a buck regulator all of the current for the output comes from the source during the on time of the switch. The average input current is lower than the output by the ratio of the duty cycle, but the peak current is the same as the output current.

Which means that the current to the input filter capacitor will be the average current but the current from that cap to the output will be equal to the inductor/output current.