What this NPN transistor circuit do?

Question

I need some help to figure out what this NPN transistor circuit do, how it works and why it's needed in order to make the CSR64215 bluetooth chip work. Here's the schematic:

And here's what MFB stands for in CSR64215 chip (took it from the datasheet):

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I'm asking all these questions because I'm about to use this bluetooth chip in my car and the problem I'm facing right now is that when I'm cranking the engine it turns off and back on quickly, but after that it unables to connect to my phone automatically and it's very frustrating as you have to do it manually after all. So basically I'm looking for a fix for this issue.

Answer 1

The circuit, even with the diode is a piece of junk. The 11ms time constant without the diode was pretty short.

Either regulate the supply so it doesn't drop during cranking or use a proper reset chip that provides a timed reset pulse of appropriate duration for a complete reset for any brownout of the supply.

Answer 2

The only reference to "MFB" in the CSR64215 datasheet is the VREGENABLE signal on pin 40. In the datasheet there's a pushbutton switch to connecting VREGENABLE to Vbat.

The circuit you show is similar to the VREGENABLE delay circuit on page 69 of the datasheet:

How does this circuit work?

When the supply voltage (1V8_INPUT, VBAT in your circuit) becomes active the capacitor will charge. This causes a current to flow through the resistor in series with the capacitor (R7 in your schematic). This then activates the NMOS or NPN which will then pull down the output VREGENABLE (MFB in your circuit).

After some time (about RC = 10 kohm * 1 uF = 10 ms, this is a ballpark figure, not very accurate!), a couple of milli seconds, the capacitor is charged and the charging current becomes zero. Then there is no voltage to activate the NMOS / NPN anymore so it opens and the VREGENABLE output (MFB in your circuit) becomes equal to the supply voltage (1V8_INPUT, VBAT in your circuit).

So your circuit makes MFB high after some delay, when VBAT is switched on.

Answer 3

To expand on my comment and answer the additional question:

Note the bug in this (and many similar) circuits : if you switch off and back on quickly enough, the capacitor remains charged and the requisite delay is absent on the second power cycle, possibly causing mis-operation. The usual fix is a diode to discharge the capacitor, by conducting if (in this example) C (lower terminal) goes below 0V.

Schematic with bug fix.

^{simulate this circuit – Schematic created using CircuitLab}

In normal operation D1 does nothing. But when VCC falls to 0, the charge stored on the capacitor pulls D1 cathode negative (below 0V). D1 then conducts, discharging the capacitor much faster than would otherwise happen, to ensure the Delay occurs as expected when power is restored.