Is this an effective circuit for replacing a 48V 5A mechanical on-off switch?

Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Hello, I noticed that there are not many 48V 5A mechanical on-off switches (such as toggle or rocker). From what I read this is because of the challenges associated with DC arcing when a circuit is opened under load. The ones I did find were extremely bulky and expensive.

So, I would like to consider MOSFET options. I don't have much experience with this sort of power switch, so I am seeking feedback or criticism before I buy parts to experiment with.

It's probably obvious to most but MYSW is imitating a mechanical on-off switch in the circuit below.
1673398902464.png

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A little pricey but I suppose another option would be an BTS50085-1TMB (attached) with a mechanical switch between input pin and ground. This offers a lot of nice protection features but costs 8 bucks and is out of stock everywhere I looked.
Update (2 minutes later) 2021-01-10 @9:09 GMT-5: I found BTS500851TMBAKSA1 in stock at Mouser. I like this option for the protection features, but would still like to know if my circuit works well in theory.
 

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sghioto

Joined Dec 31, 2017
5,368
I like this option for the protection features, but would still like to know if my circuit works well in theory.
Generally a N channel mosfet such as the IRLH5020 is wired as a low side switch. You have it connected as a high side switch.
The simulation shows appx 6.3 amps through R1 which would equate to 12.6 volts leaving 35.4 volts across M1. At 6.3 amps M1 will be dissipating a whopping 223 watts. Not a practical design but does work in theory.
What exactly is that mechanical switch used for, meaning what is the load and voltage?
 
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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
... M1 will be dissipating a whopping 223 watts. Not a practical design but does work in theory.
What exactly is that mechanical switch used for, meaning what is the load and voltage?
Ah!, right; I need to remember to check power dissipation. So re-situating to reduce dissipation, the load would probably draw only 1.5 amps during normal use, so I changed the resistor value to 8 ohms.

Screenshot 2023-01-10 231106.png

The resistive load in my circuit represents 6 LM317HVs. 5 are being used in a current limiting capacity to charge capacitors to drive solenoids and 1 more LM317HV is being used to render 5V for a micro. Each will draw approximately .25 amps at 48VDC (not the one for the micro, but for simplicity sake ill say all draw .25A @48VDC). This still gives M1 a heavy 56 watts.

I want to use this switch as the ON/OFF (main) power switch for a solenoid controller board that is powered by a 48V DC SMPS. The board draw will draw up to 1.5 amps.
 

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sghioto

Joined Dec 31, 2017
5,368
Suggest using the BTS50085 as it is a high side switch. This will eliminate the dissipation on the mosfet.
Post a schematic of the controller board.
 

Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Suggest using the BTS50085 as it is a high side switch. This will eliminate the dissipation on the mosfet.
Post a schematic of the controller board.
OK sounds like a good tack.
The LTspice model attached does a good job of demonstrating the current draw of the whole board. Looks like about 2 amps. The LM317 used to regulate 5V for the micro wont draw much I don't think.
I provided the Eagle schematic, lots of stuff not relevant to this much, but attached nonetheless. Its based on the Teensy LC and uses the Teensy bootloader chip. Also U2 can be ignored I just had it on there for the stencil when i ordered the pcbs.
I would be switching power to the whole board where it comes in at J13.
Im bringing 5v in at jp4 and jp5 externally but for future iterations there will be an LM317HV converting 48 to 5V for the micro and encoder, button, octocoupler, etc.
 

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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
I would stay with the BTS50085. At 2 amps the dissipation would be less then 50 milliwatts.

Depending on the total current draw the 5 volt regulator may require a heatsink or a series resistor to drop the input voltage to a more manageable level.
Thanks!
 

Kirkster

Joined Aug 20, 2019
1
Ah!, right; I need to remember to check power dissipation. So re-situating to reduce dissipation, the load would probably draw only 1.5 amps during normal use, so I changed the resistor value to 8 ohms.

View attachment 284970

The resistive load in my circuit represents 6 LM317HVs. 5 are being used in a current limiting capacity to charge capacitors to drive solenoids and 1 more LM317HV is being used to render 5V for a micro. Each will draw approximately .25 amps at 48VDC (not the one for the micro, but for simplicity sake ill say all draw .25A @48VDC). This still gives M1 a heavy 56 watts.

I want to use this switch as the ON/OFF (main) power switch for a solenoid controller board that is powered by a 48V DC SMPS. The board draw will draw up to 1.5 amps.
Hello Mike, Rather than using a semiconductor circuit why not keep it simple and purchase a small circuit breaker rated for your load. The control voltage in a diesel electric locomotive is 75 volts DC and the breakers are rated 80 VDC if I remember correctly. The cab in these locos are loaded with these breakers with various amp ratings. They are quite small, reliable and provide circuit protection. Airpax makes them.
Just sayin…
 

Irving

Joined Jan 30, 2016
3,793
Your circuit is flawed because your MOSFET is not turned fully on due to your load being on the source side. For a grounded load you need a P-channel MOSFET with the load on the drain side. The circuit below meets your needs. Any reasonable 60v, 10A P-channel device will do. Power dissipation in the MOSFET when turned on is under 0.5W so no heatsink needed for a TO-220/TO-247 or similar cased device. There is a short 50W pulse at switch on/off as is common with MOSFETs but unless you're switching at high speed this is of no consequence. Total dissipation at your example 20Hz is under 300mW average. D1 is a 10v zener to protect the gate from excessive voltages, while R2 ensures M1 turns off quickly. R3 should be a 1W device as it dissipates around 0.5W when the switch is closed.

1673645401229.png
 

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Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Hello Mike, Rather than using a semiconductor circuit why not keep it simple and purchase a small circuit breaker rated for your load. The control voltage in a diesel electric locomotive is 75 volts DC and the breakers are rated 80 VDC if I remember correctly. The cab in these locos are loaded with these breakers with various amp ratings. They are quite small, reliable and provide circuit protection. Airpax makes them.
Just sayin…
Hi Kirster, great idea. I appreciate you input. I'm not sure how many on-off operations circuit breakers are rated for (maybe typical is 2000) but I suspect its less than most on-off switches. Also I would like the product to have a sort of polished end-product look when I put it in a case, where the circuit breaker has more of an industrial aesthetic. For this reason the BTS50085 seems the best idea so far, but in terms of rugged, simple functionality, you're on the money.
 

Thread Starter

mike__MecE

Joined Feb 20, 2022
69
Your circuit is flawed because your MOSFET is not turned fully on due to your load being on the source side. For a grounded load you need a P-channel MOSFET with the load on the drain side. The circuit below meets your needs. Any reasonable 60v, 10A P-channel device will do. Power dissipation in the MOSFET when turned on is under 0.5W so no heatsink needed for a TO-220/TO-247 or similar cased device. There is a short 50W pulse at switch on/off as is common with MOSFETs but unless you're switching at high speed this is of no consequence. Total dissipation at your example 20Hz is under 300mW average. D1 is a 10v zener to protect the gate from excessive voltages, while R2 ensures M1 turns off quickly. R3 should be a 1W device as it dissipates around 0.5W when the switch is closed.

View attachment 285188
Thank you very much Irving this looks great. I will pick up a P-channel mosfet and try this on perfboard.
 
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