Configuring regulator shut-off

Question

I'm building a regulator based on the TPS63020 buck/boost converter, and I'd like to program a low-voltage shut-off. If I'm understanding it correctly, when V_INA sees a falling voltage and it crosses 1.5 V, it should shut off the regulator until V_INA recovers to above 1.7 V.

I want to shut the regulator off when the input voltage (from a battery) drops below 3.1 V to protect the battery from overdischarging. So I put a voltage divider between V_IN and V_INA and configured the resistors so that once V_INA gets an output of 1.5 V and lower, it should shut off. The battery has a max. voltage of 4.2 V.

My math is shown below:

$$V_{\mathrm{INA}}=V_{\mathrm{IN(low)}}\cdot\frac{R_{\mathrm{upper}}}{R_{\mathrm{lower}}}\\ 1.5=3.1\cdot\frac{R_{\mathrm{upper}}}{560\ \mathrm k\Omega}\\ R_{\mathrm{upper}}\approx 270\ \mathrm k\Omega$$

The circuit schematic looks as follows:

Did I do this right? I know the TPS63070 would be better suited to my needs, but it's too expensive for my application as I need to build many of these converters.

Answer 1

The TPS63020 has an EN enable input pin, active high as it's naming suggests.

This can be driven by just a small and cheap voltage monitor IC in a 3-pin or 5-pin package.

There are a great many different parts on the market that you can select from a supplier.

An example is the Rohm BD4931G, a voltage monitor with a 3.1 V ±1% threshold costing something like 40p. When the supply rail is below 3.1 V ±1% its output will drive low, otherwise it is driven high. This part comes from two families, one with open-drain outputs and the other with push-pull, as shown below (modified datasheet diagram).

The BD4931G consumes 0.9 uA typ. and operates correctly down to a 0.95 V supply. The TPS63020 regulator can operate from a 1.5 V supply, so the voltage monitor will reliably cover the full circuit operating range.

Unless you are happy with a delayed switch-on, ensure that you select a voltage monitor rather than a power supply supervisor that delays its output switch-off to generate a reset pulse.

The addition to the TPS63020 circuit is then as shown below. R1 allows an open-drain part to be fitted as well as a push-pull, allowing a wider range of usable parts.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Did I do this right?

Nope.

You're not supposed to apply external voltage to VINA pin. If you look at the block diagram the VINA voltage (supply voltage for internal blocks) is generated by input voltage, VIN, through an internal resistor:

It appears that the chip's UVP level is too low for your application. So what you can do to implement a custom UVP (or UVLO) is to play with EN pin functionality. According to datasheet the chip runs when the EN pin voltage is at input voltage level, and stops when the EN pin voltage is zero. So you can build a comparator that generates 0V when the input voltage is below some threshold:

^{simulate this circuit – Schematic created using CircuitLab}

A comparator is basically an open loop amplifier with very high gain. Therefore a slight difference between the inputs will show itself at the output as one of the rails: If the voltage at inverting input is higher than that at non-inverting input then the output will be zero, and vice versa.

The TLV431 is a 1.24V stable reference when its ref and cathode are shorted. This reference voltage is applied to the comparator's non-inverting input. The input voltage is sampled by R5-R6 divider and applied to the inverting input.

When the sampled input voltage is higher than the reference voltage (presumably 2V) then the output will be zero. So the MOSFET will be off, EN will be high (through 100k) so the IC will be active and running.

When the sampled input voltage is lower than the reference voltage then the output will be high, this will turn the MOSFET on so the EN input of the chip will be pulled low and therefore the chip will stop operating.

NOTES:

• The bottom resistor might be replaced with a potentiometer for easy adjustment and calibration.
• I intentionally did not indicate any part number for the op amp. The op amp should be able to run at supply voltages as low as 1.8~2V for proper operation.

Answer 3

Did I do this right?

According to me your approach is viable since Vina present an inner UVLO function. See the figure below that i took from the IC datasheet.

However you have to keep particular attention to the quiscent current needed by the control section and with 200kohm you are close to the one needed by it. I would find aresistor divider smaller.

Lastly, if you don't use it, you have to pull-up the enable pin. To avoid the device working in power saving mode you have to pull-up Sync pin too.

Another good approach is the one proposed by Rohat in the previous comment.

Best regards.