designpcb
I've had a very simple PCB designed to supply the ESP32 WROOM module with power using a voltage regulator that supplies it with 3.3V.
However, this does not seem to be enough to run the firmware application I loaded to it before hand.
Reading from the docs It tells me GPIO 0 should be high by default when not connected (which it is not.) However when I measure it is at 1.4V (floating)
What am I missing?
A few things about your basic circuit before I address your specific questions.
i. Every IC needs a decoupling capacitor. For this circuit, that applies to the linear regulator and the microcontroller. For the microcontroller, as with most ICs, a 0.1uF capacitor between VCC and GND is typical. The capacitor should be as close as possible to the physical pins of the chip. The linear regulator takes a little more thought. Most regulator manufacturers will specify what kind of capacitors and what value to use on the input and output. This particular one does not, but they do provide a test circuit that should be good enough for your purpose:
ii. Every kind of linear regulator has a "dropout voltage" (VDO). The dropout voltage is the minimum difference in voltage between the input and the output. Your L78L05 has a dropout voltage of 1.7V. That means the minimum input must be:
$$V_{INmin}=V_{OUT}+VDO=5V+1.7V=6.7V$$
Your 6V battery pack is less than 6.7V, which will not allow the regulator to regulate to 5V. Instead, you'll get something closer to 4V. That may not be a problem for this particular circuit, but you should be aware of that. There are plenty of linear regulators available with much lower than 1.7V VDOs.
Now on to your specific questions.
One possible complication here is if your circuit contains high-current components that might activate during programming. It is likely the programmer will not be able to supply enough current to keep the voltage stable if there is a lot of current draw from elsewhere. In a situation like that, it would be required to use your own power supply during programming.
You don't need to isolate the programmer when using the battery as long as the programmer's voltage supply is off. The only stipulation here is that the ground references of your circuit and the programmer are tied together. Otherwise, no special isolation is necessary.
If you do use the programmer's power, putting a diode on VOUT isn't a bad idea, but in may not be necessary. Unfortunately, the datasheet for the L78L05 does not give a absolute max voltage on the VOUT pin relative to the VIN pin. However, I have routinely applied external voltages to the output pins of unpowered linear regulators without a diode in between without problems. Without a specification in the datasheet, it's hard to know for sure. But I don't suspect you'll have a problem.
Yes, those are all valid concerns about adding a diode.
If you use the battery pack for power instead of the programmer, there are no more concerns.
Using resistors to isolate the other components from the programmer's pins is a common approach. However, you still need to be aware of how much current might be drawn from the programmer's pins. When a pin on the programmer goes high, the LED will light up. With a 100 Ohm resistor, a 2V LED may sink 30mA (\$(5V-2V)/100Ohm = 30mA\$). That's probably way more than the programmer's data pins can source. That's undoubtedly why the programmer is hanging up.
Since you can't make the resistors any larger without making the LEDs dimmer, you can use digital buffers to remove the load on the programmer pins completely. Other types of components may need different types of isolation circuitry. But resistors and buffers cover most situations.
simulate this circuit – Schematic created using CircuitLab
I don't think this is supported. Here's a workaround (hack):
From the Schematic,
Tools -->
Annotate Schematics... -->
Reset All -->
Update Changes List -->
Accept Changes (Create ECO) -->
Execute
When Altium reports the number changed, it should equal the number of parts in your schematic. If it won't do it for all parts, find the parts that keep re-annotating to the same ID/designator and manually reset those (double-click the part and click "Reset" next to the "Unique ID" field).
Sync the changes over to the layout via the ECO system.
Altium will warn you that things don't match and allow you to match manually.
If you have changed any net names, you will need to manually fix the affected pours and tracks. Altium will reset anything that no longer matches a netlist name to "No Net" so be careful to check for this or you will have a very fragile design that will cause you trouble later.
Best Answer
I don't know about IO0's behavior(breaking chip is possible), but I think you forgot to pull up EN pin.
Pull up it like this.
You can see the entire image in official datasheet. It is captioned as "Figure 4: ESP32-WROOM-32 Peripheral Schematics".