I want to push back a bit on multiple LDOs in parallel. The "18650 Battery Shield V3" has three XC6206 3.3V regulators in parallel, with no diodes. The schematic is here:
I have a couple of these, but I haven't tested the 3.3V output under significant load. But many thousands of these modules have been sold, and I've never heard of a problem with the 3.3V output.
Of course the outputs will all be a bit different, so the regulator with the highest output will supply the load until its output starts to sag, and then the next one will kick in. At least I assume that's what happens.
Anyway, I would suggest that you test the original setup to see if it works. You just want to be sure they supply enough power, and don't get hot. I suspect it will work just fine.
You will not need diodes unless your regulators both source and sink current. Most LDOs just source current, so when the voltage is higher than they are set for, they just shut down. But you can test that with your meter. The only backflow should be a tiny current through the feedback voltage divider.
Edit: Instead of diodes, inserting a resistor in series with each output might help balance the current loads among the LDOs. I'm thinking very low value resistors, like an ohm or two. Or maybe a resistor only in the low-current LDO's output. If you're lucky, the output of the big LDO will be the higher one, and then it will supply all the current when it's enabled, with no resistor needed.
I will try the parallel configuration with different resistor configurations, because I figured out that my power mux consumes 40uA and I cannot find the GLF4001.
[quote="Wawa, post:21, topic:1131434, full:true"]
Are you sure you need that second LDO.
A buffer capacitor across 3.3volt VCC can be used to bridge those short high-current peaks during WiFi. 1000uF (I use 2 x 470uF/6.3volt tantalum) will quiet down an ESP8266 to <100mA (no peaks).
I doubt an ESP32 would be much different.
Leo..
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```you mean capacitors between 3v3 and and vcc (vcc is battery/usb)
my first test pcb arrived. I ordered it bevore I started that converatsion. The schematic is attached (the diodes will probably be omitted, or replaced by resistors if neccessary, now they are just shorted). It behaved interestingly. I designed it in a way that allowed me to to connect/disconnect the ldo's to 3V3. Everything work's as expected when I only connect the 250mA LDO (mcp1700). I get a current draw with nothing connected to the output of 1-5uA. When I additionally enable the 500mA LDO (without it's output being connected to 3V3) I get 40-50uA what is also expected. It also works fine when I connect the 500mA LDO's output to 3V3 and have it enabled. What is strange is when I have both LDO's output connected to 3V3 while the 500mA LDO is not enabled. That is basically the default configuration that I would use it in. But then the circuit draws ~26mA. My guess would be that the 500mA somehow draws that current through it's output. But I have no Idea. Maybe someone has a idea that explains that and more importantly a idea for a soultion. I could use a mosfet to connect the 500mA LDO only to 3V3 when it is actually used. But maybe you have a better solution. Thanks for your help
The block diagram in the datasheet for the ME6211 shows mosfets to the output from both Vin and Vss. So it seems this is one of those regulators that can both source and sink current. I guess the lower mosfet is turning on partially and sinking the 26mA. But that's just a guess.
If you use the diodes, the output will no longer be 3.3V, but more like 3.0V if they are schottky diodes. A mosfet should work, but you would need to think about its orientation. I suspect the drain of a P-channel mosfet would need to be connected to the ME6211 output, and the source to the 3.3V rail. Otherwise, current would still flow through the body diode when the mosfet is off.
Edit: But I think the simplest solution is to just have two MCP1700 LDOs in parallel, always on. You're still under 3uA quiescent current.
I got ~26mA when I connected the outputs of bove LDO's ouput pins to 3V3. The 250mA LDO has no enable pin and the 500mA LDO's enable pin is not connected what turns the LDO off. When I connect the enable pin to 3V3 i get 40-50uA what is expected.
also when I use a diode on the 500mA LDO I would have the problem that the voltage off the 500mA LDO is lower, so the 250mA LDO will supply all the current.
I will look at the parallel approach. I was just affraid, that only one LDO would supply the current. But I will watch the video. and 3uA quiescent current sound fantastic
the video conlcudes, that the mcp1700 works fine in paralled. The load gets not completly equally shared, but he got more than 500mA output. And for my application that should work. I will try it tomorrow and report back. Thanks for the vid, quite informativ.
so, I soldered another pcb. This time I didn't solder the 500mA LDO, instead I soldered another MCP1700t on to the first one (which makes them operate in parallel). Without everything connected, I got a current draw of ~6uA which is fine. When I connected the esp32 and it woke from deepsleep I saw a spike to over 300mA. That resulted in a crash with only one MCP1700t, but with the prallel MCP1700t it works. The video above explains a bit why you can paralled the MCP1700 better than normal ldo's. So, I guess that's the final solution, thanks you for your help.
It might also be good to have a hefty capacitor on the ESP32's input, just to minimize any dip that might otherwise take place. Can you look at that point with a scope to see if it stays constant?
