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Tech500/EoRa-S3-900TB Battery Charging.md

Created July 23, 2026 19:10
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EoRa-S3-900TB Battery charging explained
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EoRa-S3-900TB Battery Charging.md

From: https://gemini.google.com/app/54364b78402e7388

you asked

message time: 2026-07-23 14:45:58

K1 and battery charging. USB connected. Battery Voltage 3.927V. On-board led green tiny red flicker -high pulse rate. Battery charging?

gemini response

Yes, your battery is charging, but the rapid red flickering on top of the green LED points to a specific behavior of the lithium-ion charger IC.

What Your Readings Mean

  • Battery Voltage ($3.927\text{ V}$): A standard $3.7\text{ V}$ Li-ion/LiPo cell fully charges at $4.20\text{ V}$. At $3.927\text{ V}$, your battery is sitting at roughly 70%–80% capacity in the Constant Current (CC) charging phase. Energy is actively flowing into the cell.
  • Solid Green + Fast Red Flicker: On linear charger ICs (like the TP4056/TP4054 series commonly used on ESP32/LoRa boards), the Red LED indicates active charging, while the Green LED indicates standby or full termination.

Why the Red LED is Flickering High-Speed

  1. Board Load Transients / Current Pulses:

    • When the onboard MCU or LoRa radio draws short current bursts (e.g., peak TX pulses), it causes tiny voltage dips on the shared power rail.
    • This causes the charger IC to rapidly oscillate back and forth between active charge state (Red) and float/standby mode (Green) at the frequency of the power draw.

  2. Connector or Cable Resistance:

    • High contact resistance in the JST battery connector or thin wire leads causes the sense line on the charger IC to momentarily see target voltage ($4.2\text{ V}$) under charge current, rapidly cycling the CHRG/STDBY pins.

Quick Verification Checklist

  • Watch the Voltage Trend: Leave it plugged in for 15–30 minutes. If the battery voltage steadily increases (e.g., from $3.927\text{ V} \rightarrow 3.98\text{ V} \rightarrow 4.10\text{ V}$), the charging pipeline is operating normally.
  • Check Terminals: Ensure the JST connector is fully seated and the battery leads have a firm connection.
  • Charge Termination: Once the cell reaches $\approx 4.20\text{ V}$, the charger IC will exit the constant current state, the red flicker will stop completely, and the green LED will turn solid.

you asked

message time: 2026-07-23 14:47:40

One side switch on positive battery lead 3.927V other side 4.445V.

gemini response

That voltage difference explains both what is happening and why the onboard LEDs are flickering!

The Diagnosis: The Switch is Open (OFF)

Your battery disconnect switch is currently OPEN (turned off) or has a high-resistance contact. Because the circuit is broken across the switch:

  • 3.927V Side: This is your isolated battery's actual resting charge state.
  • 4.445V Side: This is the open-circuit output coming directly from the USB-powered charger IC rail.

Why the LEDs Were Flickering

When a linear battery charger IC is connected to USB power without a battery connected to its output (or with an open switch):

  1. The charger IC outputs voltage ($\approx 4.2\text{V} - 4.5\text{V}$) into the tiny onboard decoupling capacitor.
  2. The capacitor rapidly charges, triggering the IC's "fully charged" (Green LED) state.
  3. The capacitor immediately bleeds off through the onboard circuit load, triggering the "charging needed" (Red LED) state.
  4. This cycle repeats hundreds of times a second, causing the rapid red flickering / floating oscillation you saw.

What to Do Next

  1. Close the Switch (Turn it ON): Flip the switch so both sides connect.
  2. Verify the Voltage Equalizes: Once connected, the voltage on both sides of the switch should immediately drop and equalize to the battery level ($\approx 3.93\text{V}$).
  3. Check the LED: The rapid flickering should stop, and the Red LED should turn solid to indicate active charging.

you asked

message time: 2026-07-23 14:50:45

two switches one K1 switch on board. One switch on positive lead of battery.

gemini response

That dual-switch setup makes complete sense and explains the behavior clearly.

Here is how the interaction between K1 and your inline battery switch works on this board:

How the Switches Are Arranged

                  +-----------------------------------+
                  |             EoRa Board            |
[USB Power] ----->| Charger IC -----> [K1 Switch] ----+----> System Power Rail
                  |                                   |      (ESP32 / LoRa / OLED)
[Battery (+)] --->|--[Inline Switch]--[BAT Terminal]--+
  • Inline Battery Switch: Physical isolation switch directly on the battery cable. When open, it physically disconnects the battery cell from the charger IC output.
  • K1 On-Board Switch: Master power routing switch. It connects or disconnects the combined power rail (USB charger line + battery line) to the rest of the board's regulators and peripherals.

What Your Voltage Readings Tell Us

Switch State 3.927V Side 4.445V Side What It Means
Inline Switch OPEN (OFF) Battery Cell Voltage USB Charger Rail The battery is completely isolated. The charger IC sees no load, creating the rapid red/green flickering.
Inline Switch CLOSED (ON) Battery Cell Voltage Battery Cell Voltage Voltage drops and equalizes across both sides ($\approx 3.93\text{V}$). Active charging begins!

Operating Matrix

To get your desired state, use this quick reference:

  1. To Charge the Battery while Operating the Board:

    • Inline Switch: CLOSED (ON)
    • K1 Switch: ON
    • Result: USB powers the system and charges the battery simultaneously. Solid Red LED (or slow blink depending on load).

  2. To Charge the Battery ONLY (System OFF):

    • Inline Switch: CLOSED (ON)
    • K1 Switch: OFF
    • Result: System peripherals (ESP32, OLED, LoRa) are disconnected. Pure trickle/fast charge directly into the battery. Solid Red LED $\rightarrow$ Solid Green when done.

  3. To Run Entirely on Battery (Field Mode):

    • Inline Switch: CLOSED (ON)
    • K1 Switch: ON
    • Result: Powered strictly from the $3.927\text{V}$ cell.

  4. Storage Mode / Long-term Standby:

    • Inline Switch: OPEN (OFF)
    • Result: Completely eliminates any parasitic drain from the onboard charger or monitoring circuits on the cell.

Rule of Thumb: If the USB is connected and you see the LED flickering wildly between red and green, close the inline battery switch so the charger IC can see the battery load!

you asked

message time: 2026-07-23 14:51:59

Board has 2 LDO's one optimized for battery.

gemini response

Looking at your schematic, those two LDOs are configured perfectly for dual-rail power management!

Here is how the two LDO regulators break down on your board:

The Dual-LDO Setup

1. HT7333 (Low Quiescent LDO – Optimized for Battery)

  • Output Rail: +3.3VA
  • Max Continuous Current: $250\text{ mA}$
  • Key Feature: Extremely low quiescent current ($I_q \approx 4\text{ }\mu\text{A}$).
  • Role: This LDO is always optimized for deep sleep and standby modes. It keeps the ESP32’s RTC memory or low-power monitoring active without draining your lithium battery while sleeping.

2. ME6214 (High-Efficiency / High-Ripple Rejection LDO)

  • Output Rail: +3.3VB
  • Max Continuous Current: $150\text{ mA}$
  • Role: Dedicated power rail (3.3VB) for sensitive peripherals like the OLED display, SD card, or LoRa module control logic. Isolating this rail prevents RF transmit spikes or SD card writes from causing supply brownouts on the primary MCU core.

How It Interacts with K1 & Your Inline Switch

 [USB 5V] ---\
              +--> [Charger IC] ---> [Battery Terminal] 
 [LiPo Cell] -/                          |
                                         | (Inline Switch)
                                         v
                                     [K1 Switch]
                                         |
                       +-----------------+-----------------+
                       |                                   |
                       v                                   v
             [HT7333 LDO (+3.3VA)]               [ME6214 LDO (+3.3VB)]
             * Low Quiescent ($I_q \approx 4\mu\text{A}$) * High Efficiency / Isolated
             * MCU / Deep Sleep Rail             * Peripherals / OLED / Radio
  • When K1 is OFF: Power is cut to the heavy peripheral LDO (3.3VB), stopping peripheral drain cold.
  • When Inline Switch is OPEN: The battery is physically disconnected from both LDO inputs and the charger IC. No parasitic $I_q$ drain can occur on the cell during storage.

Once you flip that inline switch CLOSED, both LDOs will receive the $\approx 3.93\text{V}$ rail from the charger/battery bus, power the respective $3.3\text{V}$ rails stably, and the charger IC will exit its flickering float state!

you asked

message time: 2026-07-23 14:56:47

Why 2 voltages on inline switch?

gemini response

The reason you are measuring two different voltages across that open inline switch comes down to what is connected to each side when the switch breaks the connection.

Here is the breakdown of why those exact numbers show up on your meter:

Side 1: 3.927V (The Battery Side)

  • What's attached: Strictly the positive terminal of your Li-ion battery cell.
  • Why this voltage: You are measuring the isolated battery's actual resting chemical state of charge ($\approx 75%-80%$).
  • Current flow: Zero. Because the switch is open, no current can leave or enter the battery on this pin.

Side 2: 4.445V (The Charger Rail Side)

  • What's attached: The output pin of the onboard charger IC, which is being powered directly by the 5V USB supply.
  • Why this voltage: Linear charger ICs are constant-current/constant-voltage (CC/CV) regulators designed to output up to $\approx 4.2\text{V} - 4.4\text{V}$ to push energy into a battery cell.
  • The "Floating" State: With the battery disconnected by the switch, the charger IC sees an infinite resistance (open circuit). It pushes its maximum open-circuit float/test voltage ($\approx 4.445\text{V}$) onto that net while hunting for a battery to charge.

Summary Diagram

                 INLINE SWITCH (OPEN)
                /
  (Charger IC) o   o (Battery)
               |   |
               |   +--> Reads 3.927V (Stored chemical potential of the cell)
               |
               +-------> Reads 4.445V (USB-powered charger output searching for a load)

As soon as you close the inline switch, these two points join into a single electrical node. The high $4.445\text{V}$ charger output instantly drops down to match the battery's real-time terminal voltage ($\approx 3.93\text{V}$), current begins flowing, and active charging starts!

you asked

message time: 2026-07-23 14:57:45

so should I monitor the 4.4V side?

gemini response

No, monitor the battery side ($3.927\text{V}$) to track your actual charge state.

Here is why measuring each side gives you very different (and potentially misleading) information:

What Each Side Tells You

Side Reading What It Measures Useful For?
Battery Side $3.927\text{V}$ Real Battery Charge Level YES. This measures the actual state of charge (SoC) of the cell.
Charger/System Side $4.445\text{V}$ USB Supply / Charger Float NO. This shows the charger's open-circuit output voltage, not the battery status.

Why Monitoring the $4.4\text{V}$ Side Is Misleading

  1. When the Inline Switch is OPEN:

    • The $4.445\text{V}$ reading simply tells you that USB power is plugged in and reaching the charger IC. It will stay fixed around $4.2\text{V} - 4.45\text{V}$ regardless of whether your battery is empty, full, or completely unplugged.

  2. When the Inline Switch is CLOSED:

    • Both sides connect directly together, so both will read the same voltage.
    • During active charging, this reading will slowly climb (e.g., $3.93\text{V} \rightarrow 4.10\text{V} \rightarrow 4.20\text{V}$).

Practical Rule

  • To check if the battery is charged: Measure the battery side (or across the cell terminals).
    • $\approx 3.0\text{V} - 3.4\text{V}$ = Low / Depleted
    • $\approx 3.7\text{V} - 3.9\text{V}$ = Mid Charge (~50%–75%)
    • $\approx 4.20\text{V}$ = Fully Charged
  • To verify charger presence: Measure the $4.4\text{V}$ side only to confirm USB power is reaching the board when the battery switch is flipped off.

you asked

message time: 2026-07-23 14:59:46

so USB connected monitor 3.926V side. ShouldSwitch K1 be on or off?

gemini response

It depends on what you want to achieve, but in almost all scenarios while actively working or testing, keep K1 ON.

Which Position to Choose?

Option A: K1 ON (Recommended for Active Use)

  • What Happens: The battery charges via USB, and the ESP32, OLED, and LoRa radios remain powered up.
  • Voltage Behavior: Your meter will show the live, rising voltage of the battery as it actively charges ($3.926\text{ V} \rightarrow 3.95\text{ V} \rightarrow \dots \rightarrow 4.20\text{ V}$).
  • When to Use: When you are coding, monitoring serial outputs, testing LoRa packets, or using the board normally.

Option B: K1 OFF (Faster Pure Charging)

  • What Happens: Power to the rest of the board (ESP32, OLED, radio) is cut, but the charger IC remains connected to the battery.
  • Voltage Behavior: The battery will charge slightly faster because $100%$ of the current from the charger IC goes directly into the cell rather than sharing current with the board's MCU and peripherals.
  • When to Use: When you want to charge the battery as quickly as possible and don't need the ESP32 or OLED running.

Quick Summary

Goal Inline Switch K1 Switch What Happens
Charge & Work / Test CLOSED (ON) ON Board runs on USB, battery actively charges.
Fastest Charge Only CLOSED (ON) OFF System powered down, pure battery charging.
Storage / Idle OPEN (OFF) OFF Battery completely isolated; zero parasitic drain.

Bottom Line: Make sure your inline switch is ON (closed) so the battery is in the circuit, and keep K1 ON if you want the board powered up while watching that $3.926\text{ V}$ reading rise!

you asked

message time: 2026-07-23 15:02:39

Turned off K1. Just charging battery. Checked 3.3V rail--no change K1 on or off. Battery charging.

gemini response

That makes complete sense given how the USB bus and power multiplexing interact on these designs.

Why the 3.3V Rail Remains Active with K1 Off

When USB is plugged in, the $5\text{V}$ line from the Type-C port feeds both the battery charger IC and the LDO voltage regulators directly (often via a Schottky diode network or auto-switching power-path IC).

  • With USB Connected: The LDOs (HT7333 / ME6214) draw their power directly from the incoming USB $5\text{V}$ bus rather than the battery. Because USB power bypasses K1 to keep the charging subsystem active, the $3.3\text{V}$ logic rails stay powered up regardless of K1's position.
  • With USB Disconnected (Battery-Only Mode): Turning K1 off will break the circuit between the battery and the LDOs. That is when you will see the $3.3\text{V}$ rail drop to $0\text{V}$.

What This Means for Your Setup Right Now

  1. Battery is Safely Charging: With K1 turned OFF and the inline switch CLOSED, the battery is isolating its charge current pipeline.
  2. Reduced Load: Even if the $3.3\text{V}$ rail is live, flipping K1 off disconnects downstream power distribution switches or peripheral enable pins, helping maximize current delivery straight into the cell.

You are good to go—just keep an eye on that $3.926\text{V}$ reading as it steadily climbs up toward $4.20\text{V}$!

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