Trying to eliminate and expensive diode relay

[iso9001]

Hi... I have automotive circuit using a pic that when certain input are given drives a transitor high that then in turn connects the coil of my relay, triggering it.

I know that in order to protect the transistor from emf backfeed I need to and have been using a relay with a diode across the coil. I'm using some green ones from delcity but would MUCH rather use a normal diode that I can find at any autoparts store incase I have a problem somewhere down the line. A reliability issue. Plus these things are expensive!

So... I was wondering since I'm using a the same voltage souce (12 battery) for everything, is it possible to move my diode onto my board and still have it provide the same protection ?

My current circuit is A. and the one I would rather be using is B... The sim says that this will work just fine. But they are not always right so....

Any Suggestions ?

Thanks!

 

[Oznog]

Looks fine. Protection diodes are usually put on the board rather than included on the relay itself.

 

[iso9001]

Ah... a handy little tidbit of info that is

Espessially since a TON of people told me that it HAD to be in the relay across the coil. :?

 

[Sebi]

Just be careful when design the PCB: this line (track from collector to relay) radiated more noise when the relay switch off.

 

[DirtyLude]

Method B is how I do it.

I use standard Bosch relays, and I also use some cheap PCB mounted automotive relays I got from Digikey of all places. Cheap and Digikey, don't usually mix.

 

[JimB]

From the point of view of protecting the transistor from the back-EMF from the relay coil, solution B is probably as good as solution A.

But, if there are other circuits nearby which are sensitive to magnetic fields, use solution A.
The problem with B is that you have a large loop of wire, from the relay to the diode, from the diode to the supply, from the supply to the relay. When you de-energise the relay, a large current pulse will flow in this loop and create a varying magnetic field which will induce voltage spikes into other nearby wiring.

If you find the cost of relays with diodes too expensive, would it not be possible for you to modify a standard relay and solder a diode across the coil?

JimB

 

[Styx]

From the point of view of protecting the transistor from the back-EMF from the relay coil, solution B is probably as good as solution A.

Not entirely correct. Solution B has the advantage that the free-wheel diode is as close to the switching device as possible. This reduces and stray inductance between the diode anode and the switches positive terminal.

If solution A is used and teh layout is very bad, this stay will cause turn-off voltage overshoot's higher than would have occured in Solution B. This will increase turn-off switching losses as well as potentially causing reverse voltage breakdown IF the deive is chosen to have a low peak voltage (say 20V), you can seasily see alot more that that

 

[Roff]

From the point of view of protecting the transistor from the back-EMF from the relay coil, solution B is probably as good as solution A.

But, if there are other circuits nearby which are sensitive to magnetic fields, use solution A.
The problem with B is that you have a large loop of wire, from the relay to the diode, from the diode to the supply, from the supply to the relay. When you de-energise the relay, a large current pulse will flow in this loop and create a varying magnetic field which will induce voltage spikes into other nearby wiring.

If you find the cost of relays with diodes too expensive, would it not be possible for you to modify a standard relay and solder a diode across the coil?

JimB

De-energizing the relay does not cause a large current spike. The current that was flowing through the relay when it was energized simply looks for another path, because it can't stop flowing instantaneously.
Actually, in B the current in the wire to the relay will decay relatively slowly when the transistor is turned off, while in A the di/dt in that wire will be much higher, causing a voltage spike v=L(wire)*di/dt, which is what Styx was alluding to. The EMI from A will be worse than that from B. When the transistor turns on, di/dt will be about the same in both cases.

 

[Styx]

From the point of view of protecting the transistor from the back-EMF from the relay coil, solution B is probably as good as solution A.

But, if there are other circuits nearby which are sensitive to magnetic fields, use solution A.
The problem with B is that you have a large loop of wire, from the relay to the diode, from the diode to the supply, from the supply to the relay. When you de-energise the relay, a large current pulse will flow in this loop and create a varying magnetic field which will induce voltage spikes into other nearby wiring.

If you find the cost of relays with diodes too expensive, would it not be possible for you to modify a standard relay and solder a diode across the coil?

JimB

De-energizing the relay does not cause a large current spike. The current that was flowing through the relay when it was energized simply looks for another path, because it can't stop flowing instantaneously.
Actually, in B the current in the wire to the relay will decay relatively slowly when the transistor is turned off, while in A the di/dt in that wire will be much higher, causing a voltage spike v=L(wire)*di/dt, which is what Styx was alluding to. The EMI from A will be worse than that from B. When the transistor turns on, di/dt will be about the same in both cases.

Yup, just re-read my post and it is quite broken.
Basically with power-electronics you need to keep stray inductance to a minimum

With B there will be some stray inductance between the inductive load and the diode-FET but this just adds to the load inductance and is find.

 

[Roff]

From the point of view of protecting the transistor from the back-EMF from the relay coil, solution B is probably as good as solution A.

But, if there are other circuits nearby which are sensitive to magnetic fields, use solution A.
The problem with B is that you have a large loop of wire, from the relay to the diode, from the diode to the supply, from the supply to the relay. When you de-energise the relay, a large current pulse will flow in this loop and create a varying magnetic field which will induce voltage spikes into other nearby wiring.

If you find the cost of relays with diodes too expensive, would it not be possible for you to modify a standard relay and solder a diode across the coil?

JimB

De-energizing the relay does not cause a large current spike. The current that was flowing through the relay when it was energized simply looks for another path, because it can't stop flowing instantaneously.
Actually, in B the current in the wire to the relay will decay relatively slowly when the transistor is turned off, while in A the di/dt in that wire will be much higher, causing a voltage spike v=L(wire)*di/dt, which is what Styx was alluding to. The EMI from A will be worse than that from B. When the transistor turns on, di/dt will be about the same in both cases.

Yup, just re-read my post and it is quite broken.
Basically with power-electronics you need to keep stray inductance to a minimum

With B there will be some stray inductance between the inductive load and the diode-FET but this just adds to the load inductance and is find.

Styx, I thought your post was fine. I was just commenting on JimB's post.

Ron

 

[iso9001]

Either way. Its all good for me


Oh, I mentioned this is an auto circuit but I was wondering. Even thought the 12V is coming from one source. The 12V I have coming into my pcb is an ignition feed and not direct to the battery... Does this matter ?? (Also, I think I know the answer to this, but would this work if there were 2 batteries (two different 12Vs?))