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I am trying to connect the output of a Digitally Programmable Step-Up Boost Converter (TPS61045) with a capacitor through a switch. Basically when the boost converter turns on it turns on at the median value. So I need to turn on the boost converter, send commands to lower the voltage to the minimum voltage (~3.6V). And then turn on the switch to connect the boost converter output to the capacitor to charge it. Then I ramp up the voltage of the boost converter to 19V. The purpose of this is to reduce the in rush current on the battery powering the boost converter by ramping the voltage in 64 steps. I'm also looking for a load switch to then discharge the capacitor onto a resistor after the capacitor is fully ramped to 19V.

Does anyone have any ideas of what switches to use for this application? I was looking at integrated load switches but tried using one (FDC6326L) and it wasnt working well.

I want to control the load switch with an arduino 3.3V logic level.

I was working with Power Relays for a bit but want to move towards FETs.

Thanks,

EDIT: I'm using 4 100uF multilayer ceramic caps that reduce by about 35% in capacitence when charged to 19V

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  • \$\begingroup\$ Welcome to EE! Can you go into more detail on why the FDC6326L didn't work for you? \$\endgroup\$
    – mbedded
    Sep 17, 2020 at 16:01
  • \$\begingroup\$ I think was operating above the power dissipation requirements of it. ~1.2W and its rated for .7W \$\endgroup\$
    – David Lin
    Sep 18, 2020 at 18:59

3 Answers 3

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For something like this, I would typically not use a programmable swich and just use a simple high side FET switch. On high side switches where the the power rail is significantly higher than the switching logic rail, you will need a pullup and an extra transistor to handle the switching function. The reason is because the logic level at the gate won't be high enough to cut off current to the high side FET and it will always be at least slightly on. Here's a typical configuration although you can replace the BJT with a FET as well.

enter image description here

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welcome to Stack Exchange.

You are probably overthinking this problem, most capacitor charging supplies use controlled current to charge the capacitor, if you do it in a series of little steps, then your converting will just be running at the current limit anyway on the beginning of the steps, so just wire the capacitor directly, and set the convertor to 19v and let it do it's thing; and you don't really need a load switch , just disable the convertor and leave your capacitor attached to the convertor, this will still leave a voltage divider 5megohm resistor discharging your capacitor, is that a problem.

The TPS61045 using peak current circuitry, so will quite happily run on "current limit" as it is already doing this anyway. TPS61045 block diagram

, this will limit peak switch current to 375mA, because of the variation of battery current with ON time of the boost switch, the battery current will probably be ~ 100mA initially, then 200mA as cap gets to 6v, then 300mA as cap gets to 12v and above. You could also drive the enable pin with a slow PWM at 500Hz to reduce total current (rather than the complex programming arrangement, that some might call elegant, but look at the code in a years time and ask how does that work?), this also allows interchanging with other boost converters.

When you discharge your capacitor, you will also be discharging C2 if you discharge below 3v Wiring diagram

To discharge your big capacitor, just use a logic level mosfet connected to ground. As in this partial schematic, (with optional 1ohm resistor to measure discharge current). U1 is your Arduino or whatever.

You can also make a constant current discharge arrangement by increasing the 1ohm resistor to say 10ohms, and the 10 ohm resistor to zero, with a gate drive of 3v, and a 1v threshold MOSFET, you will have 2v across 10ohm or 200mA constant current discharge.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Hi BobT, Thanks for answering my question. I am working trying to reduce the current to <200mA to prevent brownout issues. This is why increasing the output of the boost converter slowly, to prevent high current draws is really nice as the boost converter cannot slow the current as much as I would like. I was hoping that a load switch or something along the lines with soft start capabilities could help. I'm also worried that if i dont have a switch from my boost converter to the cap, the battery will flow directly to the capacitor or vice versa. \$\endgroup\$
    – David Lin
    Sep 18, 2020 at 18:55
  • \$\begingroup\$ Could you go into more detail about the constant current discharge, This seems like what I am trying to get towards. \$\endgroup\$
    – David Lin
    Sep 18, 2020 at 19:04
  • \$\begingroup\$ Most single boost converter topologies will have a minimum output voltage about equal to the input voltage, so you do need to disconnect the converter to discharge the capacitor. The TPS61045 has an integral load switch for that purpose. I've posted another answer with a constant current discharge circuit. \$\endgroup\$
    – BobT
    Sep 18, 2020 at 22:44
  • \$\begingroup\$ Read this about emitter resistors too electronics-tutorials.ws/amplifier/emitter-resistance.html and en.wikipedia.org/wiki/Common_emitter \$\endgroup\$
    – BobT
    Sep 18, 2020 at 22:52
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Capacitor discharge:

  1. Charge the capacitor with sw1
  2. Drive gate of MOSFET with 3v
  3. As the Vth is 2.5v, there is 0.5v left across R1, this defines the discharge current of 5mA (100ohm is 10mA per volt), (change resistor value to change discharge rate)
  4. The capacitor discharges at the constant current until about 0.5v, after that it tails off.

The circuit also works with a NPN transistor (but you might want to add a resistive divider to the base, to set base voltage at ~1.6v , this causes the emitter voltage to be 1.0v and a discharge current of 10mA for a 100ohm resistor)

Here are voltage displays from this sim: enter image description here

You should be able to run this sim yourself by clicking on the schematic , press simulate near bottom left then press "run time simulation"

schematic

simulate this circuit – Schematic created using CircuitLab

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