Electrical Engineering Asked by J Liggot on February 6, 2021
I would like feedback on the following circuit design please. I am powering a solenoid using a 3S LiPo battery but want to increase the voltage to 35v. I am using an XL6009 module to boost the voltage from roughly 12v to 35v. The voltage booster specs say it can safely output 4A. Do you see any problems with the following approach?
The LiPo battery is connected to the voltage booster input, of course.
The voltage booster output is connected to a 10,000u capacitor. (The solenoid, which has about 3Ω resistance, is firing for only a fraction of a second.)
I am using a resistor to limit the current running from the voltage booster to the capacitor. I need to keep it under 4A to prevent damaging the voltage booster, especially if something goes wrong like the solenoid is engaged for longer than a pulse. I plan to use a 100W 13Ω (or greater) resistor at 35V, which will allow 2.8A going through the line.
I plan to use a relay between the capacitor and the solenoid to control the firing. I also use a mosfet to engage the relay’s coil from a microcontroller and have tested this circuit. I am using a relay that can handle 10A of current. I gather I might need a relay that can handle about 12A to be safe but figure if I am only firing the solenoid intermittently and for only 50ms, I should be ok. I will use a flyback diode to try to protect the relay from the reverse voltage spike cause by the solenoid’s inductive load.
Does this approach makes sense?
Thanks for the help!
XL6009 has a 4A switch current limit. This means boost inductor current is maximum 4A, let's say 3.5A average.
Of course we're talking about the boost inductor current here, which is the input current, not the output current. The latter will be reduced in proportion to Vin/Vout. So, at 30V output, you will get 1.5A output current maximum. Always take ebay descriptions with a grain of salt...
The chip seems to have onboard current limiter. This only operates when Vout>Vin ; when Vout < Vin the chip is not operating and current flows through the diode from input to output without any limiting.
Now, about the 13R 100W resistor. Since XL6009 has current limiting, the resistor does not protect the XL6009. Its real role is to protect the boost converter diode, which will conduct a large inrush current when the batteries are plugged in while the 10,000µF output cap is discharged.
I would put a 2R2 10-20W cemented wirewound resistor in series with the output instead. It will dissipate pulse power when the battery is connected, then when the capacitor is charged. Depending how frequently the solenoid is actuated, you may be able to use only a 10W resistor, maybe less.
If you don't like thermal management, another option is A buck-boost or SEPIC converter which would eliminate the resistor.
I plan to use a relay between the capacitor and the solenoid to control the firing.
A low RdsON MOSFET will work better, smaller, cheaper, more reliable. Also if you want 50ms, this is too fast for a relay, but trivial for a FET. Make sure you don't forget the freewheel diode, and don't skimp on it! It should be rated for the full current.
Warning: Cheap DC-DC modules on ebay usually come with capacitors which are not rated for the actual ripple current. They tend to fry or pop. Check the caps' specs...
Answered by bobflux on February 6, 2021
This approach will work fine and is widely used for latching solenoids. Remember that the voltage on the capacitor will start to drop when the solenoid is energized, and you must reach the solenoid rated current (not voltage) to get full force. This means you must consider the inductance of the solenoid coil. The catch diode must be capable of handling the full solenoid current. Make sure the diode is in close proximity to the coil.
Answered by John Birckhead on February 6, 2021
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