4
\$\begingroup\$

I'm designing a Mosfet based switch that uses a low current reed switch to control either 24V or 12V LED strip lighting. Having some issues with the protection circuits. My specifications are as follows:

  • 12-24V DC input
  • 2A Max output current
  • Reverse polarity protection
  • Short circuit protection

My current circuit diagram: Circuit Diagram

Operation as follows:

  • Input & output at the bottom left, MSTBA4.
  • Switch connected top right SWT1
  • Zener D1 ensures the gates are not driven above 15V to protect them from high voltage damage
  • Q1 is connected in anti-series (drain to ground) and serves as reverse input protection, when reverse input the gate is pulled to ground through R1 turning Q1 off. During normal operation Q1 is turned on and bypasses its internal body diode.
  • Q2 is the main switch, when the switch closes then Q2 is turned on and connects the load. (Low side switched)
  • F1 is a Bourns MF-R250 PTC resetable fuse rated 30V 2.5A (trip current 5A) This is a problem as I will explain shortly.
  • D2 is a Bidirectional TVS diode to absorb any transients.

The design works but there are a few issues I would like to improve, mostly with the current protection. The PTC takes about 5A to trip in a reasonable time, the 'fets are ok with this as they can handle 10A (thermal considerations aside). The problem comes with using a supply with current limiting such as a lab supply. For example with a current limit of 3A on the supply, the input voltage drops before the PTC kicks in. Once the input voltage is low enough the gate voltage on Q2 drops and causes it to leave saturation and enter triode region. Now Q2 starts to dissipate all the energy itself and fails! When working with a 24V Meanwell PSU this isn't an issue (NES-50-24), however I'm worried about this switch being used with other supplies...

The goal is to keep the cost and complexity as low as possible. I've considered some options:

  • Add a comparator circuit for undervoltage protection. i.e. when input is <12V, pull Q2 gate to ground. > Problem here is even with hysteresis this causes the circuit to oscillate as the PSU recovers and then trips the protection again.
  • Add over current protection with a current shunt and small op-amp + comparator to turn Q2 off. Again same problem with oscillation, also increases complexity.

Any suggestions?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ So, the protection oscillates with a crap power supply. Why is this a problem. You can use a capacitor to keep this at a low rate (maybe once per second). \$\endgroup\$
    – Andy aka
    Jun 20, 2016 at 16:55
  • \$\begingroup\$ @Andyaka, would it be fair for me to assume (perhaps even stipulate to the customer) that a decent DC PSU will shutdown when the output voltage drops below a certain boundary of normal operation? If so then I am simply over complicating things for myself and the design is good \$\endgroup\$
    – Rambo
    Oct 3, 2016 at 14:11
  • \$\begingroup\$ I believe so but please do try and ensure the rate at which it switches isn't annoyingly higher than once per second. \$\endgroup\$
    – Andy aka
    Oct 3, 2016 at 14:17

1 Answer 1

1
\$\begingroup\$

Just some ideas, but you could try the following...

1 - Replace Q1, D1, C1 and R1 with a single Schottky diode for reverse current protection.

2 - Replace Q2 with a BJT constant current source, for which you could repurpose D1 and R1.

These ideas eliminate two components and any ESD concerns, while providing inherent current limiting.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ This isn't an improvement to OPs circuit: Your suggestion would cause an additional voltage drop of around 1,1V (where ~0,4 V is lost in the schottky and 0,7 V in the bjt). In a short circuit the BJT would also die very quickly from overheating unless a massive heat sink is added for it, since it would try to maintain the nominal output current while having the full input voltage over it. \$\endgroup\$
    – jms
    Aug 2, 2016 at 10:01
  • \$\begingroup\$ Agree with @jms on this, I went with the twin 'FET solution to minimise the losses in a schottky. \$\endgroup\$
    – Rambo
    Oct 3, 2016 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.