- Sat Feb 07, 2015 2:48 pm
#9204
Hi,
I will do a new revision and make the power traces wider and place the module at the edge so that the antenna part will protrude the PCB
There is already a flash button (connected to GPIO0) to put the module into flash mode. The power LED is both a load for the voltage regulator (minimum output current required to maintain regulation) and also an power indicator for the module (the ESP-led only flashes some of the time)
With other projects I use a P-CH Mosfet for reverse polarity protection. In this case the 3V3 voltage regulator will prevent any damage and also the MOSFET-approach will add cost, effort and parts.
Using a Schottky-diode would only drop the voltage about 0.4V, the AMS1117 @ 250mA has a drop of about 0.8V so 1.2V together. When 5V is applied even a normal diode would be ok. There are still the usual disadvantages (power dissipation, leakage current) when diodes are used.
markbee
lethe wrote:The traces look a bit thin to me, it might be a good idea to make them wider (esp. the power traces).
I would also rotate the ESP-12 by 180° and put the button on the other side, where the LDO is. That way you can mount the ESP, so that there's no PCB underneath (eliminating any risk that your board might affect antenna performance).
I will do a new revision and make the power traces wider and place the module at the edge so that the antenna part will protrude the PCB
lethe wrote:You also might want to add a reset button, for easier flashing and I don't see much need for that power LED you added, since the ESP-12 already has one (or doesn't it? there is a LED mounted at least).
There is already a flash button (connected to GPIO0) to put the module into flash mode. The power LED is both a load for the voltage regulator (minimum output current required to maintain regulation) and also an power indicator for the module (the ESP-led only flashes some of the time)
lethe wrote:As for your protection diode: the diode will typically drop about 0.6V (depending on the actual diode used and the current draw). With a 5V input, that will drop your input down to 4.4V, but the LDO has dropout voltage of about 1.2V. While I like the idea of the reverse polarity protection, a simple diode is not ideal in this situation. Using a MosFET would be better: http://hackaday.com/2011/12/06/reverse- ... h-a-p-fet/ (you can also use a cheaper n-FET, when you put in in the ground line).
With other projects I use a P-CH Mosfet for reverse polarity protection. In this case the 3V3 voltage regulator will prevent any damage and also the MOSFET-approach will add cost, effort and parts.
Using a Schottky-diode would only drop the voltage about 0.4V, the AMS1117 @ 250mA has a drop of about 0.8V so 1.2V together. When 5V is applied even a normal diode would be ok. There are still the usual disadvantages (power dissipation, leakage current) when diodes are used.
markbee
Electronics blog http://lookmanowire.blogspot.com
