Snubber Circuit for DC Power Switch

[LukeZ]

Hi all,

I've built a seaplane where I use differential thrust for steering on the water. There are four motors on the plane (GWS Speed 400s); when I turn the rudder full right I have a limit switch that opens the circuit to the outmost motor on the right wing, and vice versa with the left. Cutting out just one motor of the four really gives me some great turning, you can see a video of it by clicking here: http://www.kansasflyer.org/Images/eZone/Solent/DiffThrust.wmv

So basically my outboard motors are wired to the NC contact on the limit switch. The rudder servo "opens" the switch.

Now, my concern is that arcing across the switch contacts might eventually degrade my switch or cause it fail. This would spell bad news if it failed in flight for some reason.

Each of my motors on this airplane draws roughly 10 amps at full throttle (FYI, I'm using a 3s lithium battery). I don't usually taxi at full throttle but switching 10 amps would be the worst case scenario the switches will have to deal with.

The limit switches I'm going to use are DigiKey part #450-1445-ND (http://rocky.digikey.com/scripts/ProductInfo.dll?Site=US&V=450&M=1478603-3). They're rated for 16A DC, so maybe right there that takes care of what I need.

Otherwise, I think an RC snubber circuit on the switch would be wise, the thing is I can't find how to figure out which values of resistance and capacitance I should use. I see all kinds of options thrown about but no real empirical method for going about determining the optimal or even viable values for the purpose.

If anyone can shed some light, even basic theory that might guide me in picking my components, I'd sure be grateful.


Luke

[Trevor_G]

A snubber will only absorb the inductive spike when you switch off. A better way of achieving this is to wire a diode across the motor. This will be far more efective in removing the spike.

Try a 1N5822 - Cathode (white band) to +ve and anode to -ve. Be careful not to connect the motor the wrong way round at any time or you will blow the diode.

Don't forget although you run at 10A at startup there will be a much bigger surge and a 16A switch may struggle.

Trevor

[LukeZ]

As I understand it the inductive spike comes from the collapsing magnetic fields on the coil of the motor, when power is removed. I suppose it would be possible for this spike to travel back down the wire to the switch and cause arcing across the contacts - they would already be open by then but perhaps not open far enough to prevent arcing.

What I am thinking of specifically is when I turn the motor on. Let's say my ESC is already at full throttle, my motor is disconnected because I'm in a turn, then I straighten out and the contacts within the switch snap shut. Surely there is going to be an arc across the contacts as they approach each other and the current leaps across the small space between them. I don't see how a diode across the motor will have any effect on this phenomenon, but a snubber circuit should take care of it. I'd just like to know how I should go about determining the values to use in my snubber.

Incidentally, the snubber circuit should also take care of back-emf from the motor when the switch is opened, or so it seems to me.

Luke

[john whitehead]

A snubber will only absorb the inductive spike when you switch off. A better way of achieving this is to wire a diode across the motor. This will be far more efective in removing the spike.

Try a 1N5822 - Cathode (white band) to +ve and anode to -ve. Be careful not to connect the motor the wrong way round at any time or you will blow the diode.

Don't forget although you run at 10A at startup there will be a much bigger surge and a 16A switch may struggle.

Trevor

I agree with the above, go with a diode as described..... A diode with ultra fast switching characteristics or try TranZorb transient suppressor.

[jeffs555]

What I am thinking of specifically is when I turn the motor on. Let's say my ESC is already at full throttle, my motor is disconnected because I'm in a turn, then I straighten out and the contacts within the switch snap shut. Surely there is going to be an arc across the contacts as they approach each other and the current leaps across the small space between them. I don't see how a diode across the motor will have any effect on this phenomenon, but a snubber circuit should take care of it. I'd just like to know how I should go about determining the values to use in my snubber.


It takes voltage to cause an arc. When the switch has been open for a while, the only voltage will be the voltage of the battery which is too low to arc across much of a gap. Also, the motor is an inductor, and the main property of an inductor is that current through it cannot change instantaneously. When the switch has been open for a while, the current in the inductor will be zero. When you close the switch, the inductor will try to keep the current at zero, and the current will only rise gradually, so there will be very little current when you initially close the switch.

However, when you open the switch it is entirely different. If you open the switch when the motor is drawing the full 10 amps, you will generate a huge arc. The reason for this is that the property of inductance tries to keep 10 amps flowing through the motor. When the switch opens, the resistance in series with the motor becomes extremely high. From ohms law voltage is equal to current times resistance, so if you try to flow 10 amps through an extremely high resistance, you get an extremely high voltage. The voltage can be tens of thousands of volts which will arc across a long distance.

Like the others said, a diode is the best solution.

[LukeZ]

Jeff, thank you for your lucid explanation. Thinking of the motor as an inductor helps, I should have done that.

If you don't mind though, let me just be clear on what purpose the diode serves. The switch opens, the inductor wants to maintain the same current flowing through it - but with the diode, instead of attempting to draw that current over the widening gap of the switch's contacts, it somehow quenches it across the diode? It seems to me the diode would only come into play when the polarity of the voltage across the motor changes, such as when the electromagnetic fields within the motor collapse.

Sorry if I'm dense - aside from having a functional model, I guess I also like to understand what I'm doing.

The 1N5822 mentioned earlier is a 40v, 3A diode. I suppose those specifications speak to the current and voltage it can "block" in opposition. Should I look for one with a higher current rating, or will that be sufficient?


Luke

[vintage1]

Yup. The diode is more there to prevent huge inductive arcs..the inductance of the motor is like a car ignition coil: Break the circuit and you get a massive spike of several hundred volts as the field collapses.

This is usually oscillatory in nature, as the parasitic capacitance of the motor and its suppressors will cause it to 'ring'..the diode takes care of the energy within one full cycle.

The best diode is a schottky barrier diode of around 3-5A and 30V or so.

[pmackenzie]

J
If you don't mind though, let me just be clear on what purpose the diode serves. The switch opens, the inductor wants to maintain the same current flowing through it - but with the diode, instead of attempting to draw that current over the widening gap of the switch's contacts, it somehow quenches it across the diode? It seems to me the diode would only come into play when the polarity of the voltage across the motor changes, such as when the electromagnetic fields within the motor collapse.

The diode provides an alternate path for the current.
When the switch is closed current flows in the positive terminal and out the negative, out to the battery and back though the switch.
When you open the circuit it can no longer go through the battery and switch, but it can go from the negative terminal through the diode and back to the positive.
The observed polarity at the motor terminal will switch from the battery voltage to the negative value of the forward voltage drop of the diode.

It is normally called a free wheeling diode and is part of all chopper type speed controls including brushless ones.


Pat MacKenzie

[jeffs555]

Pat is correct, and here are drawings to help visualize what is happening.

In the left picture with the switch closed, current flows from battery through the inductor(ie the motor) and back into the battery. The diode is reverse biased so draws no current.

In the right picture, the current flows from the inductor through the diode which is now forward biased. Note that the voltage across the inductor changes polarity, but the direction of current flow through the inductor does not change.

[LukeZ]

Jeff, thanks for the great drawing. I think I understand now. Previously I couldn't see how the negative terminal on the motor became positive - but I suppose relative to the open circuit on the positive terminal, the negative terminal now actually has a positive potential, is that it? And therefore the polarity switches, allowing the current to flow through the diode.

Guys, thanks for taking the time to explain this. I was certain you were all correct at the beginning, but I much prefer to understand the why.

Here's the diode I think I'll get, it's a schottky, 5A, 60v - SB560/4 (http://www.digikey.com/scripts/DkSearch/dksus.dll?Criteria?Ref=91311&Site=US&Cat=33882806).


Luke