BQ2403x and DPPM

I should also mention that right now I'm only using the battery input to the bq2403x.  The USB and AC adapter inputs are at 0V.  The battery voltage is at 4V but like I said above the output voltage gets stuck around 1.8V.

Do you have a schematic that you can post?  Are you powering the IC through the AC or USB input?  What is at 1.8V, the OUT pin or the BAT pin?  Is a battery present?

Is your DPPM voltage set to 1.8V?

8664.Doc1.docxI'm only powering the IC through the battery right now.  The DPPM voltage is set at 3.8V and the battery input is at 4V.

If only the battery is connected, then what you are seeing is not DPPM.  DPPM reduces the battery charging current (from the input source) to allow the output voltage to remain above the DPPM threshold.

If the battery is at 4V and the OUT pin is at 1.8V, then something else is going on.  How much current is your buck/boost drawing?  Is the battery capable of delivering this much current?  Try testing with just a power supply connected to BAT and see if that affects the operation.

I do have a power supply hooked up to the battery input and it's not hitting its current limit; the input voltage to the bq24032A battery pins stays at 4V.  Once the IC starts limiting its output voltage to 1.8V it doesn't matter how small of a load I put on output, it stays around 1.8V.  I had assumed this problem was DPPM related because the datasheet says "with most any capacitive load onthe output, the part can get stuck in short-circuit mode" and recommends putting a capacitor from the DPPM pin to ground to delay short circuit protection (I guess short circuit protection and DPPM are unrelated then).  The bq24032A feeds right into the input capactors of the buck/boost converter so I thought this might be the problem.  I tried tying the DPPM pin to ground directly and that does keep the IC from limiting the output voltage which seems to confirm I'm in some sort of short circuit protection mode.  Is it a bad idea to just change the circuit such that R6 is a 0 Ohm resistor which would keep the DPPM pin at 0V all the time?  This would eliminate any short circuit protection but are there other issues I'm not considering like shorting the 100uA DPPM current source or eliminating other DPPM features.

Just to add a little clarification to exactly what the output voltage looks like:  It is a sawtooth shape that rises to 1.85V and then drops straight down to 1.42V and repeats at a 500Hz frequency.  I've tried varying the DPPM capacitor from 1nF to 10nF to 100nF.  I see the same output signal no matter the capacitor value.  If I reduce the load enough such that the 10mA constant current source (enabled during short circuit mode to limit current from the battery to output) is sufficient to power the buck/boost the output signal changes shape but still oscillates and doesn't allow the output voltage to rise above 1.85V.  Hope this may help somehow. 

A scope capture of Vbat, Vout, Vdppm, and Iout when the output voltage drops to 1.8V would help discover what is going on.

It sounds like you are in short circuit protection.  About how much current are you drawing from the battery when this occurs?  To enter short circuit protection it would take typically 5A on the OUT pin (40 mOhm FET trips at 200mV).  This is far above what the part can handle and thus your power topology would need to be redesigned to avoid this.

Could you send the schematic of your buck/boost?  I would guess that it has a UVLO circuit on it that trips at about 1.4V falling and 1.85V rising--this is what is causing the sawtooth on the OUT pin.

I get locked up in the short circuit protection when the output voltage of the buck/boost is around 1 - 1.5A.  The output voltage is 3.6V and with the input being about 4V and some reasonable efficiency I would expect the input current to the buck/boost is pretty close to the output current which is pretty far from 5A.  The buck/boost (LTC3533) also has an input current limit min of 3.5A and typical of 4.5A so its short circuit protection should kick in before the battery charger.  The UVLO for the buck/boost kicks in at 1.6V but has some hysterisis.  The datasheet for it doesn't cover this though, its only depicted in the block diagram.  I'm not sure I understand how the UVLO for the buck/boost could cause the battery charger to get stuck in short circuit protection but the voltage levels are in agreement with UVLO voltage levels.  The buck/boost schematic is below and I'll see what I can do about getting scope captures so you can see what I am.  Thanks

That is a rather low current, but looking at it with a scope will definitely help.  In addition, there could be a lot of ripple on top of the average current.

The UVLO of the switcher explains why you get the sawtooth on the output.  Please measure the current during this case too--I would expect it to be around 10mA.  Short circuit is entered and the BAT FET is turned off.  The switcher pulls down its input until its UVLO kicks in (around 1.4V).  The 10mA source is turned on, charging the OUT pin back up.  Once the UVLO is reached, the switcher turns back on and pulls the bus down immediately.

You can save yourself some money by using the TPS63010 family instead of the LT part.