0
\$\begingroup\$

I bought a rather high-power electric bike recently, and would like to build a small microcontroller IC with GPS receiver and some other tricks inside it.

The MCU part I know how to do. The thing needs to be powered with 5 V, and it would be really nice to pull this from the main battery bank, which is a 20S lithium system, voltage ranging between 68 V and 84 V.

There are off-the-shelf components that convert from a very wide range between 20 V to 100 V down to 5 V. However, I would need some (preferably low-power) mechanism to disengage this device if the battery voltage drops below some set overdischarge limit. Using a mechanical powered relay seems excessive, and those things draw a lot of power just to stay on.

So what I would like to have is:

  • An analog circuit that works an on/off power switch.

  • May need to be initially turned on manually

  • Automatically switches itself off (goes to non-conducting state) as a function of voltage.

  • This threshold voltage would need to be around 68 V.

  • Consumes next to no power when switched off, and as little as possible when on.

Is there some easy(ish) analog circuit design to do this?

\$\endgroup\$
2
  • \$\begingroup\$ Certainly exists, but have you considered a super low Iq DC-DC always running? If left discharged for years, the battery is toast anyway and the battery has self-discharge. Iq << self-discharge = not an issue. \$\endgroup\$
    – winny
    Commented Jul 12, 2022 at 19:51
  • \$\begingroup\$ You'll need hysteresis. You'll need to figure that out numerically what you need. \$\endgroup\$
    – Andy aka
    Commented Jul 12, 2022 at 20:05

1 Answer 1

2
\$\begingroup\$

M1/IC1 form the latching 'thyristor type' power switch. It draws no power when off, and little when on.

The TLV431 (not the right symbol, this was close) is designed as a 3 terminal shunt regulator, that maintains the sense voltage at 1.24v. You could think of it as a high gain NPN transistor, with a very precise 1.24v base voltage. Q1 is used as a cascode current source to get the signal up to the high voltage present at the FET, as the TLV431 is only suitable for 6 or 16 V, depending on manufacturer.

schematic

simulate this circuit – Schematic created using CircuitLab

In order to inhibit switch-on if the battery is low, the ON switch connects Vbatt to the voltage sense divider chain via D3, and conveniently the ON LED to show that something is happening. If Vbatt is high enough, IC1 turns on, and sends a current through Q1 to turn on M1, connecting the load, and maintaining the sense voltage via D2. The ON button can now be released. D1 limits the gate voltage for high voltage battery inputs. D2 prevents the ON switch from powering the load directly.

I'm using a blue or white ON LED for double duty as a regulator, the slope resistance is an order of magnitude better than a Zener at this voltage. Choose R7 for a few mA if it's needed as an ON indicator, or for 100 uA if it's not.

When coming to turn the circuit off manually, large capacitors in the load may hold the load voltage above the battery minimum threshold for long enough to be tedious, pressing the off button and waiting for it to stop. R11 allows you to actively load the output when turning off, choose a smaller value of R11 to discharge faster. The maximum current you can draw from the load is limited by R11 pulse load handling, D2 and the off switch current handling. Don't omit R11, unless you also omit the off switch, as it protects the off switch from a high current from the load's charged capacitors.

Use of D2 and D3 reduces the threshold accuracy slightly, but not significantly. R9 and R10 values are shown for approximately 11 V threshold, about the minimum for nominally 12 V lead acid batteries (this was originally a design for a 12 V switch, but with suitable ratings/values for Q1, M1, R7 and R10 can work for any high voltage). Increase R10 appropriately to drop out at your battery minimum voltage, allowing for the D2/3 and R11 voltage drops.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.