LTC3862-2 Step Up problem, output voltage drop

Question

I'm designing a solenoid powering and control device which is centered around the LTC3862-2 Step Up controller (see datasheet). I have taken as a design start point the typical application shown in page 38 of the datasheet. I have only modified the resistor divider values in order to get a 40 V output (316 kOhm and 10kOhm) instead of the 80V shown in this typical application.

My requirements are:

• Vin= from 6 to 8.4 Volts
• Vout= 40 Volts
• Iout= 2.5 Amps

Plus, the basic circuit schematic is shown below:

^{simulate this circuit – Schematic created using CircuitLab}

The problem comes when I activate the solenoid. Instead of having the 40V/2.5A from the step up, there is a sudden decline from the moment M1 closes from 40V to 24V, plus the current is only about 1.5A. As a result, the solenoid is far from moving the mechanism it is meant to. When the load is not connected, the output is stable at 40V.

Additionally these are other measures from the experiment:

• Battery voltage under load= 8.3V @ 5.5A (LiPo battery can deliver more than 30 A continuously).
• M1 transistor source(GND) to gate voltage 4.8V (transistor datasheet)

Do you know what is happening or the reason of this behavior? Can you think of a solution to avoid this output voltage drop?

Any help is very much appreciated. Thank you in advance!

----Edit: Schematics----

Hi guys, sorry for not posting thee aactual schematics. Here I attaach the typpical application from the IC datasheet as well as the schematic from the board I have manufactured.

From this schematic, I have only made this minor modifications:

• Changed output caps to 100uF 63V since output voltage for this application is 40V.
• Input capacitors changed from 50 to 16V since input voltage range is 6-8.4V.
• Resistor divider to FB pin in order to have 40V output (12.4 kOhm to 10 kOhm and 796 kOhm to 316 kOhm).

This results in this schematic:

Hope this sheds more light. Thank you very much!

Answer 1

Pay particular attention to pages 23 and 28 of the datasheet, where it describes selection and placement of current sense resistors RU7 and RU9.

The value of these resistors is critical. If they have even milliohms too much resistance, this will limit maximum inductor current, and output current capability will suffer. Their physical placement is equally important. Be sure that the SENSE- and SENSE+ signals are taken close to (at, in fact) the actual sense resistor terminals.

I would also strongly recommend testing those sense resistances with the regulator and MOSFETs out of circuit. Use a current source to pass a known (large) current directly through the sense resistor, and measure the potential difference at the corresponding SENSE+ and SENSE- pads where the regulator would be. Use Ohm's law to verify that the resistance is indeed the value you expect.

Answer 2

Your layout may have a bit to do with it (you can share it if you want more info). As well as needing bulk capacitance on the solenoid.

The boost converter has a finite source impedance, as does the layout. By adding a large capacitance right next to the solenoid, you effectively get an additional very low impedance source (for a limited time). I would suggest adding several caps in parallel to create around 1000uF as a start. That will keep the total ESR lower. To get better values, you'll need to measure the current waveform through the inductor at turn on. If you don't have a current probe, you can place a non-inductive 0.1Ω (or smaller) resistor in series from the M1 source to ground connection, then use ohms law to convert to amps.

^{simulate this circuit – Schematic created using CircuitLab}