Can smaller trace width limit the current flowing through the PCB?

Question

I am a computer science student and I am a newbie in electronics.

I have designed a PCB for SIM800L, and I forgot to change the trace width of the copper rails. By default, the routing width was set to 0.254mm. Now once my PCB arrived I have soldered all the components to the PCB, and I have found that my design failed.

I suspected on the trace width of the PCB because since SIM800L module doesn't come with an onboard voltage regulator, an external power supply adjusted to the voltage between 3.4V to 4.4V (Ideal 4.1V) is required. The power supply should also be able to source 2A of surge current, otherwise, the module will keep shutting down.

Unfortunately, this has happened to me as well. So as I have no prior experience in PCB designing I just wanted to confirm whether it is because of the routing width or else there is any other reason for that.

Answer 1

This might not be the only issue, but is almost certainly is one of them: the LM317 cannot supply 2A (its output transistor simply cannot pass that much current).

But even if it could, it probably would go into thermal shutdown. It is a linear regulator and linear regulators work by turning excess voltage to heat. That means the higher your input voltage is relative your output voltage, the more heat is produced for the same amount of current. You have your regulator set to 4V and are feeding it 12V. That is an excess of 8V At 2A, that is 16W worth of heat which is a lot. The thermal resistance of the larger packages in the datasheet indicate that it would be somewhere around 600C temperature rise. The LM317 goes into thermal shutdown long before you ever get anywhere near that. Probably somewhere around 125C which is only 300mA.

So if the current surge lasts too long and your capacitor is too small to provide enough reserve charge for the entire duration, the LM317 will try and help out, overheat, and protect itself by going into thermal shutdown.

When linear regulator datasheet says it can supply 2A, it means 2A under the best conditions, not all conditions.

And yes, thin traces cannot carry as much current because they have higher resistance so they heat up more and produce more voltage drop, which can either cause the trace to overheat or the voltage to droop too low.

Answer 2

Supply transient problems in GSM radios are common.

My recommendation is to beef up the traces carrying high current by soldering wires on the surface of the PCB parallel to the traces. Another option is to scratch the solder mask off and add a thick layer of solder on top of the traces.

It's also possible that the problem is in your external power supply. Your input capacitor C1 should help you in this case. If you believe this is the problem try using a larger capacitance.

If you have access to an oscilloscope, look for transients on the supply voltage at the module pin and then upstream.

Answer 3

100uF capacitor consumes lots of inrush current at its start-up and I don't think LM317 can support with those short traces.
Secondly, LM317 is an very good linear regulator with tight voltage regulation at its output and so you can move those huge capacitors to its input side and keep low capacitance (~10-20uF) at its output side.
Irrespective of it, you need minimum 12mil trace thickness for carrying 1A of current. So based on maximum current carrying capacity of GSM module you can setup trace width accordingly. Maybe 20mil would be a ideal choice.

Someone suggested SMPS buck IC instead of LM317 and in case if you are going to replace then you can keep 100uF capacitor as it is. Keeping SMPS IC will help in improving efficiency of the overall system but with additional cost. It is up to you to choose. If you are going to use SMPS BUCK IC, then please go through its datasheet parameters clearly.

Apart from those, I would say some more improvements on the schematic what you have given here.

1. For LED you have connected 300Ohm which corresponds to ~15mA for 5V. We don't need that much for LED. May be you can consider to reduce the LED current to ~5mA and increase the resistance accordingly (1kΩ-2kΩ). This change improves your overall circuit efficiency.
2. For ADJ pin of LM317 you need minimum 100uA current only. But at present you have huge current (4.1V /320 Ω (220+100)). So increase the resistance of 220Ω to 2.2kΩ and 100Ω to 1kΩ, which also helps in improving efficiency of the overall module.

Please comment back for any questions.