inductor switch protection diode

A

andy

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I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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Subject: inductor switch protection diode
From: andy news4@earthsong.free-online.co.uk
Date: 8/4/2004 11:04 AM Central Daylight Time
Message-id: <pan.2004.08.04.16.04.17.474838@earthsong.free-online.co.uk

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------
Click to expand...
A good rule of thumb is to make sure the current rating of the diode exceeds
the current going through the inductor when it's on. If you can, choose a
diode with a Peak Reverse Voltage (PRV) rating twice the voltage across the
inductive load when it's on. I would guess you're using 12 or 10AWG wire (30
amp) to the load. A similar size gauge going to the diode would definitely be
helpful, if you can.

I'd be a little more concerned about the turn-off for the FET. It looks like
you've got a driver that will turn on the FET quickly, but turn it off slowly
with the 10K pulldown. You might want to modify that.

Good luck
Chris
 
andy wrote:
I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.

12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------
Click to expand...
The diode across the inductor will see a peak current almost equal to
the current at the moment the mosfet turns off. However, the current
falls with something like an L/R time constant, with L being the
inductance of the coil and R being its resistance. So the diode does
not need to be rated for the continuous current in the coil, though it
should have a surge current rating well above the peak current it will
pass. A 3 amp rectifier should be fine.

Note that adding this diode will also lengthen the time that current
passes through the coil by the same L/R time constant. Is this a
problem?
--
John Popelish
 
andy wrote:
On Wed, 04 Aug 2004 17:04:18 +0100, andy wrote:

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

P.S. the way i'm thinking of constructing this is to put the BUZ10 in a
piece of stripboard, with all the terminals soldered just to hold it in
place, but leave the ends of the terminals sticking out so i can solder
the high current wire directly to them. I.e. the gate is using the
tracks on the board, but the source and drain are soldered directly to
the cable. Is this a good way to do it?
Click to expand...
I think so.

--
John Popelish
 
On Wed, 04 Aug 2004 17:04:18 +0100, andy wrote:

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
Click to expand...
P.S. the way i'm thinking of constructing this is to put the BUZ10 in a
piece of stripboard, with all the terminals soldered just to hold it in
place, but leave the ends of the terminals sticking out so i can solder
the high current wire directly to them. I.e. the gate is using the
tracks on the board, but the source and drain are soldered directly to
the cable. Is this a good way to do it?

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.
 
On Wed, 04 Aug 2004 11:49:34 -0400, John Popelish wrote:

andy wrote:

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.

12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------

The diode across the inductor will see a peak current almost equal to
the current at the moment the mosfet turns off. However, the current
falls with something like an L/R time constant, with L being the
inductance of the coil and R being its resistance. So the diode does
not need to be rated for the continuous current in the coil, though it
should have a surge current rating well above the peak current it will
pass. A 3 amp rectifier should be fine.
Click to expand...
thanks.

Note that adding this diode will also lengthen the time that current
passes through the coil by the same L/R time constant. Is this a
problem?
Click to expand...
no. The time just needs to be just long enough for the coil to reach its
peak (resistance determined) current, so as not to waste power keeping it
alive longer than needed.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
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On Wed, 04 Aug 2004 15:44:22 +0000, CFoley1064 wrote:

Subject: inductor switch protection diode
From: andy news4@earthsong.free-online.co.uk
Date: 8/4/2004 11:04 AM Central Daylight Time
Message-id: <pan.2004.08.04.16.04.17.474838@earthsong.free-online.co.uk

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------


A good rule of thumb is to make sure the current rating of the diode exceeds
the current going through the inductor when it's on. If you can, choose a
diode with a Peak Reverse Voltage (PRV) rating twice the voltage across the
inductive load when it's on. I would guess you're using 12 or 10AWG wire (30
amp) to the load. A similar size gauge going to the diode would definitely be
helpful, if you can.

I'd be a little more concerned about the turn-off for the FET. It looks like
you've got a driver that will turn on the FET quickly, but turn it off slowly
with the 10K pulldown. You might want to modify that.

Good luck
Chris
Click to expand...
why will it be a slow turn off?

Would decreasing the resistor to 1k or 200ohm be enough? The transistor is
being driven from the inverting output of a CMOS monostable.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with
HTML: 
 or [attachment] in the subject line.
 
andy wrote:
On Wed, 04 Aug 2004 15:44:22 +0000, CFoley1064 wrote:

Subject: inductor switch protection diode
From: andy news4@earthsong.free-online.co.uk
Date: 8/4/2004 11:04 AM Central Daylight Time
Message-id: <pan.2004.08.04.16.04.17.474838@earthsong.free-online.co.uk

I'm using this circuit to control a short (1s) pulse of 14A through an
electromagnet coil. The coil is 3 separate windings of 3 ohms each, which
should give about 13.5-14A when switched on. I'm putting in a protection
diode to stop the inductor destroying the mosfet when it's switched off.
The question is, how do i work out the current rating for the diode and
the wiring to it? I'm using heavy 30A wire for the wires between the
battery, magnet, and mosfet, wired through an electrician's terminal
block, but i'm not sure if i need to do the same with the diode.


12V ---------o------------o-------
| |
|< |
On --| ----|
|\ - C|
| ^ C| Electromagnet
----o | C|
| | -----
.-. | |
10k | | | ||-+
| | | ||<- BUZ10
'-' o-----||-+
| |
---------o------------o--------


A good rule of thumb is to make sure the current rating of the diode exceeds
the current going through the inductor when it's on. If you can, choose a
diode with a Peak Reverse Voltage (PRV) rating twice the voltage across the
inductive load when it's on. I would guess you're using 12 or 10AWG wire (30
amp) to the load. A similar size gauge going to the diode would definitely be
helpful, if you can.

I'd be a little more concerned about the turn-off for the FET. It looks like
you've got a driver that will turn on the FET quickly, but turn it off slowly
with the 10K pulldown. You might want to modify that.

Good luck
Chris

why will it be a slow turn off?

Would decreasing the resistor to 1k or 200ohm be enough? The transistor is
being driven from the inverting output of a CMOS monostable.
Click to expand...
The switch won't be slow to turn off. The current through the
inductor will be slow to reach zero, since it will free wheel through
the diode. The time constant of the current decay is roughly L/R of
the inductor. If you put a resistor in series with the diode, you
allow more peak reverse voltage to develop (the peak inductor current
times the resistance) but the decay time constant then becomes
L/(Rind+Rseries). And the resistor also has to handle the peak
current. Other voltage drop elements can also be used, like several
diodes in series, or a big zener in series with the diode. If a slow
release of the magnetic field is no problem, then just use one diode.

--
John Popelish
 
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