Flyback diodes

Other Hardware General Electronics
1

Let's say I have a motor running off 3v, and across its leads I have a flyback diode. Let's also say that across those same leads I have a 3v laser that lights when the motor is on.

When the motor shuts off, will the laser see a potentially damaging voltage spike, or will the flyback diode safely route it back into the motor?

I'm kind of on the fence over what would happen. I know the diode is supposed to protect the rest of the circuit but I don't see why the voltage spike should choose to go through the diode and back through the motor instead of the laser module or my microcontroller. It seems like the motor would have a higher resistance than other paths the electricity could follow.

2

because the forward conduction voltage of the diode is usually less than 0.7 volts so that is the maximum reverse voltage your circuit will experience

3

Even a negitave 0.7V is enough to kill a laser diode.

4

If a negative voltage of 0.7V will damaged the laser, you can't connect the laser to the same leads as the motor. You should use a seperate output line to control the laser.

There is even more to this. Suppose the motor is switched with a very fast mosfet, and your diode is a slow rectifier diode (like 1N4007). In that case the spike would be a lot bigger, since the diode is too slow.

5

Read this on driving a laser
http://www.repairfaq.org/sam/laserdps.htm

6

This isn't a laser diode though, it's a laser module. They have other chips and things on there to drive them. Unfortunately the ones that are cheap and easy to find on ebay (ie, the ones most people will use) don't come with datasheets so I couldn't tell you what is between the diode and the circuit. But I have to assume they've designed them so that if you flip the battery around they don't blow up, since many come with a spring just for the purpose of putting these in laser pointers. They also have what looks like it might be a tiny voltage regulator on there.

But that's besides the point. I do want to know if a laser will blow up, but I also just want a general idea of what will happen. And it sounds like you're saying the rest of the circuit might see a -0.7v spike.

As for the speed of the diode, this is the diode I'll be using:

If the voltage rate of change is what I should be looking at, it looks like it's pretty fast.

As for the motor it will be switched with this Vreg:

I'm not sure how fast I can switch that, but I don't have to be able to switch it fast. It would be nice if I could though. Fast enough to adjust the speed of the motor smoothly anyway. 480hz should be sufficient.

Hm, a chart on page 10 of that datasheet tells me how fast it responds to the enable pin.

Looks like on enable it takes 60us before it reacts, it reaches 3.3v after another 40us, and after another 60us it stabilizes at 3.3v.
And on disable, it reacts immediately but takes over 180us to reach 0v.

This is for a 6.6 ohm load though, it may react much faster with a higher load. Also it may still be stable if switched on and off faster than the time it takes to reach 3.3v or 0v. But if I want my PWM to be linear in relation to the voltage it needs to switch on and off fully.

So if I assume 200us for it to go to 3.3v or 0, that's 0.2 milliseconds, which means theoretically I could switch it on and off 5000 times a second.

As for that diode, ar 1000v/us it seems like it is much faster than this voltage regulator.

7

Thanks for that laser link Mike. Even though I don't need to drive a diode directly it has some useful information on what kinds of things I might find in a laser module circuit.

8

That regulator for the motor is very slow compared to a transistor or a mosfet. So any diode will do. The maximum current through the diode is just a peak, and the peak current is the same amout as the current going through the motor when it was on.

9

You could put another diode in series with the laser if the voltage drop wouldn't affect it that much, couldn't you?

10

Don't have a lot of space on the board, and I don't really expect people to connect multiple things to one board but I put three sets of pins on there so they could connect say three lasers. I just wanted to know if they did attach a motor if it would blow up their laser diodes.

Also, perhaps I should go with a beefier diode if the current back through it will be the same as the current that was going through the motor, (because it seems silly to have a 1A regulator you can never run at 1A) but I'm curious how the current can be the same if the voltage is thousands of times higher? It seems like you ought to have to trade current for voltage.

11

If you put the flyback diode in and had a large capacitor across the laser module's power leads, I would guess the capacitor would probably be able to suppress the voltage spike(or drop).

12

(because it seems silly to have a 1A regulator you can never run at 1A)

No it is not silly to run a component at less that the maximum rating. In fact it is silly to run a component at the maximum rating. For reliable operation you must derate components, that is run them at less. Typically 80% derating is used.

but I'm curious how the current can be the same if the voltage is thousands of times higher?

Thousands of times higher is not going to happen.
However you will extract more energy out of a motor when you are trying to stop it than it takes to keep it going. This is because the momentum it has built up during the time the motor has been on needs to be reduced to zero. A motor does not have this momentum instantly on starting, it builds up and the current while it is doing this is greater than the current to keep it running.

13

scswift:
but I'm curious how the current can be the same if the voltage is thousands of times higher? It seems like you ought to have to trade current for voltage.

Actually it is related to the inductance & resistance of the circuit, on an unloaded circuit it can be tens of KV. but the current will be very very low.(you CANNOT reliably measure it with a multitester)
One thing to note is that there is ALWAYS back emf from a motor even when it is running under power, it occurs whenever a magnetic field collapses or is cut by a metallic object.
If you have a very fast scope you can see this as the motor rotates under power. try scattering some 10nf or 100nf ceramic capacitors about the circuit, as well as the back emf diodes, but note that rectifier diodes really are too slow for this.

14

No it is not silly to run a component at less that the maximum rating. In fact it is silly to run a component at the maximum rating. For reliable operation you must derate components, that is run them at less. Typically 80% derating is used.

Well, yes, I understand derating well, but what I mean is if the diode's rating is half that of the regulator, then I have to run at 80% of the diode's rating, which is a mere 40% of the regulator's rating. So the diode is the limiting factor, and it's a cheap component so I should upgrade it to match what the more expensive components can do.

15

Thousands of times higher is not going to happen.
However you will extract more energy out of a motor when you are trying to stop it than it takes to keep it going. This is because the momentum it has built up during the time the motor has been on needs to be reduced to zero. A motor does not have this momentum instantly on starting, it builds up and the current while it is doing this is greater than the current to keep it running.

Probably not really related to what you're talking about, but I did read a post on flyback diodes which indicated that when you're doing PWM on a motor you can get a lot of heating in the diode. I guess cause if you keep energizing the motor and powering down you have a lot of current flowing in reverse through the diode from the magnetic field collapsing over and over again. That post indicated that even though with something like a relay you could get away with a current rating 1/5th of what you're putting into the relay because you don't switch it often, with something like a PWM motor, even at relatively low frequencies like 5000hz, the diode doesn't have time to cool down and you need at least a rating high enough to handle the current you're putting through the motor.

Though my intuition tells me that half should be sufficient because the motor can't be turning off 100% of the time, nor should the current spike well above the average if you're switching it fast enough that the field never collapses completely.

But just to be on the safe side I decided to go with a diode rated for 1A average and with a lower thermal resistance. And I'll be tying the pads to both copper planes on the board without thermals to provide adequate heat sinking.

The diode:

16

One thing to note is that there is ALWAYS back emf from a motor even when it is running under power, it occurs whenever a magnetic field collapses or is cut by a metallic object.
If you have a very fast scope you can see this as the motor rotates under power. try scattering some 10nf or 100nf ceramic capacitors about the circuit, as well as the back emf diodes, but note that rectifier diodes really are too slow for this.

I have 10uf and .1uf caps scattered about the board. And there's two 10uf caps on the motor controller board itself. I assume these will be adequate. We're only talking about tiny 3v vibration motors.