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| BQ24780S: What is the minimum headroom required for this part? Is it a requirement to have 2V minimum voltage difference between the input and output? | |
| Part Number: BQ24780S Our host has a 17V DC input for the built-in charger, which uses bq24780S. Currently, wehave a battery with 15V charging voltage. But what is we use a battery that has a higher charging voltage? We want to know the minimum headroom required for this part. The charging current is around 1.5A. | |
| Hi Lauren, Theoretically, as the battery voltage approaches the adapter voltage, the duty cycle approaches 100%. However, two things get in the way of this. Firstly, the conductive loss creates a voltage drop, so that whatever is your conductive loss voltage drop, that is the voltage differential where you would actually hit 100% duty cycle. Secondly,BQ24780s drives NFET switches using charge pump built with the BTST capacitor. Even if you try to drive at 100% duty cycle you won't quite get there because there is a forced refresh pulse on the BTST circuit. I can't really give you an exact cutoff for the device, because it is very dependent on external components. The BQ24780s does not have an enforced duty cycle cutoff, so you will see it switch right up to the point that it reaches that almost-100% duty cycle point. On the EVM at 800 kHz, this is generally around 0.8-1V for 2A. But keep in mind that as you approach this 0.8-1 V point, you will start to see increased ripple at the input. Around 1.3-1.5V the input ripple will start to increase and it will get worse and worse until you reach that 0.8 - 1.0V point. If you need a really tight dropout, I would recommend that you move to the BQ24800, which has options for 300kHz and 400 kHz switching. The lower switching frequencies give you a little more headroom (maybe 200 mV.) The BQ24800 also has some improvements to reduce the ripple at very high duty cycle. Since there isn't really a hard cut-off, I really think the best way to answer the question is to experiment with the EVM, keeping in mind that changing the external components may change the results slightly in the final system. Regards, Steve | |