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#1
08-29-2010, 02:48 AM
 EAI New Member Join Date: Aug 2010 Posts: 3
Back EMF

Hello

When connecting a power transistor to a ignition coil how is it possible to stop back emf.

Placing a diode across the emitter and collector or use a zener diode across the emitter and collector.

This circuit is very popular but usually there is so many problems with it as due to back emf destroying the lm555 or the transistor or both.

Thanks EAI
#2
08-29-2010, 02:52 AM
 marshallf3 Senior Member Join Date: Jul 2010 Location: Oklahoma City, OK. Large enough to be modern but plenty of country left around it. Posts: 2,358

The rectifier would go across the coil, cathode to the +12V, anode to the transistor's collector. I'd also use a rectifier with a decently fast switching time, the 3055 is a workhorse but it's old and probably a bit more vulnerable than most.

A schottky diode with decent ratings would be a good choice.

[Post has been edited to describe proper placement, I made a mistake in the original]
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The very first course in engineering school should be how to use Google and Wikipedia
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Last edited by marshallf3; 08-29-2010 at 03:43 PM.
#3
08-29-2010, 05:02 AM
 EngIntoHW Senior Member Join Date: Apr 2010 Posts: 128

How come the rectifier should be connected from Collector (Anode) to Emitter (Cathode)?

Didnt you mean a zenner diode?
#4
08-29-2010, 05:25 AM
 Ghar Senior Member Join Date: Mar 2010 Location: Toronto Canada Posts: 655

Quote:
 Originally Posted by marshallf3 The rectifier would go across the transistor CE junction, cathode to the collector. I'd also use a rectifier with a decently fast switching time, the 3055 is a workhorse but it's old and probably a bit more vulnerable than most. A schottky diode with decent ratings would be a good choice.
Can you justify this?

The entire point of a diode is give the inductor current a safe path to dissipate in. Your proposal doesn't do that.

Check out this comparison simulation...

BackEMF_schem.png

BackEMF_plots.png

As you can see that diode across the transistor doesn't do anything... the correct position is across the inductor.

If you want to clamp across the transistor you use a zener.
#5
08-29-2010, 05:30 AM
 marshallf3 Senior Member Join Date: Jul 2010 Location: Oklahoma City, OK. Large enough to be modern but plenty of country left around it. Posts: 2,358

My mistake, all my original post would do is feed it back into the power supply.
__________________
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The very first course in engineering school should be how to use Google and Wikipedia
-
I very often misspell or miss things while making posts so come back and double check a few times.
I also have a full time job and often 10 projects going on at the same time so I'm not always online.
#6
08-29-2010, 12:39 PM
 EngIntoHW Senior Member Join Date: Apr 2010 Posts: 128

Hi Ghar,
You did a wonderful job here, I'm glad to be part of this thread and learn from forum members here.

Quote:
 Originally Posted by Ghar Can you justify this? The entire point of a diode is give the inductor current a safe path to dissipate in. Your proposal doesn't do that. Check out this comparison simulation... Attachment 22175 Attachment 22176 As you can see that diode across the transistor doesn't do anything... the correct position is across the inductor.
How come you were applying 5V on the base?
Wouldn't it damage the BE junction of the BJT?

Quote:
 Originally Posted by Ghar If you want to clamp across the transistor you use a zener.
If you were to use a zener diode across the CE junction (Anode to Emitter, Cathode to Collector), would you need its Zener Voltage to be larger than V1 (25V)? (So the Zener wouldn't conduct when the BJT is in cutoff, right?).

What Zener Voltage would you want in such application?

And, could you give up on the Rectifier (across the relay) if you use a Zener diode across the CE junction?

Thank you very much
#7
08-29-2010, 01:23 PM
 marshallf3 Senior Member Join Date: Jul 2010 Location: Oklahoma City, OK. Large enough to be modern but plenty of country left around it. Posts: 2,358

You will absolutely have to have the rectifier across the ignition coil as that's where the back EMF is coming from and it's best to quench an offending signal at its source.

I was just half asleep with another half dozen problems in my head when I spit that first post out, guess I should go edit it as not to accidentally mislead someone.

What you're doing is shunting out the back EMF generated when the transistor is switched back off. When the current going through an inductance is abruptly interrupted it generates a back EMF (voltage in reverse) that can easily damage other things in the circuit. The principle of using a rectifier to catch it is even used in circuits that contain absolutely no active components at all. (such as relay banks)

Herre's a little reading on the subject:
http://www.ehow.com/how_5593373_calculate-back-emf.html
__________________
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The very first course in engineering school should be how to use Google and Wikipedia
-
I very often misspell or miss things while making posts so come back and double check a few times.
I also have a full time job and often 10 projects going on at the same time so I'm not always online.
#8
08-29-2010, 03:01 PM
 rjenkins Senior Member Join Date: Nov 2005 Location: Sheffield, England Posts: 1,015

Putting a diode directly across an inductor results in a slow turn-off as the current only decays due to circuit resistance.

In circuits that need a fast switch-off, you should add a resistor in series with the flywheel diode. The resistor value depends on the inductor current and the voltage the switching device can safely withstand, above the supply voltage.

eg. if the inductor is taking 2A on a 12V supply and the transistor is rated at 60V, a 22 Ohm resistor is about right; it will drop 44V @ 2A, adding to the 12V supply so the peak collector voltage will be around 56V (plus diode drop).

The resistor is dissipating a peak power of 44(V) x 2(A) = 88W for a short time at each switch-off, so a power resistor of some form is essential.

For a fast turn-off, the higher the resistor the better - so also use a transistor with a high Vce rating.
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#9
08-29-2010, 03:25 PM
 Ghar Senior Member Join Date: Mar 2010 Location: Toronto Canada Posts: 655

Quote:
 Originally Posted by EngIntoHW How come you were applying 5V on the base? Wouldn't it damage the BE junction of the BJT?
I just wanted to switch the inductor current in the simulation, this wouldn't have been a real circuit.

Quote:
 If you were to use a zener diode across the CE junction (Anode to Emitter, Cathode to Collector), would you need its Zener Voltage to be larger than V1 (25V)? (So the Zener wouldn't conduct when the BJT is in cutoff, right?). What Zener Voltage would you want in such application? And, could you give up on the Rectifier (across the relay) if you use a Zener diode across the CE junction? Thank you very much
You're right, the zener voltage would need to be higher than the supply or else it would always conduct. The value you pick depends on the transistor, you want it to be quite a bit lower than the maximum rating to protect it.

I agree with marshallf though, in that it's probably better to keep the diode across the coil. Since the back EMF is really the inductor current, that zener would dissipate Vz*IL, while the diode would dissipate Vf*IL. Vz we already agree would be larger than 25V, while Vf is about a volt. However, that zener would reduce the current much faster than the diode, for the reasons that rjenkins mentioned. The inductor current will decay according to the simple equation:
$\frac{dI}{dt} = \frac{V}{L} = \frac{IR + V_f}{L}$

If you ignore the resistance you can see that a 25V zener would make the current decay 25 times faster than a 1V diode.

Which is best depends on the application; the required decay time, the power levels, the level of protection needed...
This kind of problem is almost an entire field which goes by the name of 'contact protection'.

Also just as a general note I find that by far the best way to understand back EMF is to realize that it is the inductor current that's the problem. The inductor current will want to remain at the same value and will raise the voltage if impeded. By providing an easy path it can't complain and you will not have a voltage spike beyond those caused by the new current path's impedance.

Edit:
On another note, I do not follow that eHow article...
They say calculate di/dt = V/L, but don't specify what V they're talking about.
The next line says to calculate maximum V by multiplying di/dt*L.
...but you already needed V to calculate di/dt, since that's the exact same equation?

Last edited by Ghar; 08-29-2010 at 03:33 PM.
#10
08-29-2010, 04:03 PM
 The Electrician Senior Member Join Date: Oct 2007 Posts: 1,722

Quote:
 Originally Posted by EAI Hello When connecting a power transistor to a ignition coil how is it possible to stop back emf. Thanks EAI
If you "stop" the back EMF, you won't get very good performance.

Your post describes this as an ignition coil driver. You haven't said what the application is, but assuming that it's an ignition coil for a gasoline engine, such coils have two windings, a primary and a secondary.

They are a kind of transformer, and the voltages across primary and secondary are approximately in the same ratio as the turns ratio, typically something like 1:100. So, if you want to get 30,000 volts out of the secondary, you must have 300 volts across the primary.

In other words, you must allow the back EMF to rise to several hundred volts. If you clamp it to a single diode drop by putting a diode across the primary directly, you won't get a spark out of the secondary.

If you clamp the primary "back EMF" to around 60 volts, to protect a 2N3055, you will get very poor performance.

What you must do is to use a transistor that can withstand several hundred volts when it's turned off.

There are special transistors made just for this purpose; see, for example:

http://www.newark.com/on-semiconduct...tor/dp/26K4435

These transistors have built-in protection and don't need another zener diode or other protective device, and they're easy to drive since they have a MOSFET-like input impedance.

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