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'[PICLIST] How to control a BIG inductive load?'
2001\06\06@222134
by
Alexandre Domingos F. Souza
2001\06\06@223231
by
tims800
check out
http://www.teccor.com
look for alternistors.... I have been using them ..they work where other
devices have failed.
{Original Message removed}
2001\06\07@005403
by
Alexandre Domingos F. Souza
2001\06\07@015344
by
Robert A. LaBudde
|
At 11:21 PM 6/6/01 -0300, you wrote:
> Can I use a TRIAC to control an inductive load? I need 10 or more
> amps in 110V, any triac/solid state device can do that?
>
How much inductance?
If a few hundred microhenries, you can deal with back-emf by a snubber
network of reasonable size.
If more than a millihenry or two, you would need a varistor or zener to
prevent breakdown triggering of the triac.
In principle, a solid state relay or triac can be used for such an
application. In practice you need to know a little about snubbing design.
A relay would solve your problem with less knowledge required, although you
would still want a snubber network across the relay points and a diode
across the relay activation coil.
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral
KILLspamlcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
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2001\06\07@063559
by
tims800
in my circuitry I have it set up like this:
use a zero crossing opto triac like moc 3030 or moc 3031.
this isolates the line voltage (trigger signal to power line)
the trigger voltage is square wave +5vdc.@50% duty cycle.
0 volts led off-----+5vdc led fully on instantly current limit the led
portion with a 330 ohm resistor.
triac side of opto connects between gate and mt2 terminal of alternistor
(hot wire side).
look under application notes on thyristors.
an1007 and such..
just becareful to connect as shown to terminals mt1-mt2.
it blows up if the terminals are reversed with the triac what a mess <G>.
{Original Message removed}
'[EE]: Re: How to control a BIG inductive load?'
2001\06\07@131112
by
Robert A. LaBudde
|
At 06:39 AM 6/7/01 -0700, you wrote:
>in my circuitry I have it set up like this:
>use a zero crossing opto triac like moc 3030 or moc 3031.
> this isolates the line voltage (trigger signal to power line)
>the trigger voltage is square wave +5vdc.@50% duty cycle.
>0 volts led off-----+5vdc led fully on instantly current limit the led
>portion with a 330 ohm resistor.
>triac side of opto connects between gate and mt2 terminal of alternistor
>(hot wire side).
>look under application notes on thyristors.
>
>an1007 and such..
>just becareful to connect as shown to terminals mt1-mt2.
>it blows up if the terminals are reversed with the triac what a mess <G>.
I use a 820 ohm resistor through a MOC3011 LED to use the triac side to
switch a heavy duty triac. I've found that 5 mA reliably triggers the
MOC3011 device, so 820 ohm works for 5 V.
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: .....ralKILLspam
.....lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
================================================================
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2001\06\07@141452
by
Spehro Pefhany
|
At 01:09 PM 6/7/01 -0400, you wrote:
>
>I use a 820 ohm resistor through a MOC3011 LED to use the triac side to
>switch a heavy duty triac. I've found that 5 mA reliably triggers the
>MOC3011 device, so 820 ohm works for 5 V.
It may trigger the particular one you have, at the moment, but it's generally
quite poor design technique to do this sort of thing.
Best regards,
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2001\06\07@171358
by
Robert A. LaBudde
|
At 02:16 PM 6/7/01 -0400, Spehro wrote:
> >I use a 820 ohm resistor through a MOC3011 LED to use the triac side to
> >switch a heavy duty triac. I've found that 5 mA reliably triggers the
> >MOC3011 device, so 820 ohm works for 5 V.
>
>It may trigger the particular one you have, at the moment, but it's generally
>quite poor design technique to do this sort of thing.
Why do you say this?
The trigger current required is inversely proportional to pulse width.
When switching a triac at zero crossings (MOC3011->triac), there's no point
to switching faster than 100 Hz.
With pulse widths of 100 us or longer, the required trigger current is only
1-2 mA.
5-6 mA is a considerable margin of error.
So what's the problem?
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral
spam_OUTlcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
================================================================
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2001\06\07@233231
by
Spehro Pefhany
|
At 05:11 PM 6/7/01 -0400, you wrote:
>Why do you say this?
Because it is true, and I because don't want the other fellow to have
problems.
>The trigger current required is inversely proportional to pulse width.
No, it is not, there is a dependency only for very short pulses (<
100useconds);
normally one would use a longer pulse when an inductive load is involved.
>When switching a triac at zero crossings (MOC3011->triac), there's no point
>to switching faster than 100 Hz.
>
>With pulse widths of 100 us or longer, the required trigger current is only
>1-2 mA.
No, it is not. I think perhaps you are misreading the data sheet's
'normalized'
curves. The typical is 5mA (what you say you are supplying; 1.0 on the
normalized curve) the maximum is 10mA, for the MOC3011. That's at 25'C (it
takes more current at low Ta). That's before aging. Plenty of app notes on
this,
you need a *lot* of margin to do this right, probably 15-20mA nominal.
I didn't work out your resistor value tolerance, Vf range and power supply
tolerance and Vol range but those are factors too. MOC3012's have 50% lower
guaranteed trigger current.
>So what's the problem?
By eliminating a 3 cent transistor you make the whole thing unreliable, it
will work with only some MOC3011's, at only some temperatures, and perhaps
only for a few months before it may quit. When it fails to trigger it can
fail to trigger on only half the cycles, feeding DC to the inductive load.
If the fuse wasn't left out, the fuse should blow, protecting the load, if
it is a motor.
IMO, this is far worse than a circuit that doesn't work at all, but, hey,
feel free to criticize me for being an overly conservative engineer.
Best regards,
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Spehro Pefhany --"it's the network..." "The Journey is the reward"
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2001\06\08@010310
by
Russell McMahon
|
> curves. The typical is 5mA (what you say you are supplying; 1.0 on the
> normalized curve) the maximum is 10mA, for the MOC3011. That's at 25'C (it
> takes more current at low Ta). That's before aging. Plenty of app notes on
> this,
> you need a *lot* of margin to do this right, probably 15-20mA nominal.
> I didn't work out your resistor value tolerance, Vf range and power supply
> tolerance and Vol range but those are factors too. MOC3012's have 50%
lower
> guaranteed trigger current.
Things you all already knew, (didn't you?) :-)
- You MUST always design using worst case specs for all components and
parameters . (unless you don't care whether the equipment always works).
- Typical values are fine for getting eg probable current drain of a batch
of IC's BUT typical values MUST NEVER be used for design.
- When multiple component and parameters are involved you MUST use the worst
case values of all of them simultaneously when designing (unless you don't
care ... etc).
For instance, as Spehro suggests -
Say you have
- a PORT that has a 4.9 volt typical and 4.5 volt guaranteed high output
level.
- a 470r resistor +/- 10%
- An opto coupler with typical / guaranteed trigger currents of 5 and 8 mA.
- Optocoupler diode voltage of typical guaranteed 2.0/2.5 volts at rated
working current.
- etc , but that will do
(Haven't got to load drive requirements, and more)
Using typical values, opto drive current = V/R
= (Vport - Vopto drop)/R
= (4.9-2)/470 = 6.2 mA
This exceeds the typical 5 mA drive requirement.
It has a safety margin (apparently of 6.2/5 ~= 1.25
The design will work (sometimes)(maybe).
Using worst case values we get -
(4.5-2.5)/(470*1.1) = 3.87 mA
This will not meet the 5 mA typical drive requirement, let alone the worst
case 8 mA worst case requirement.
It fails by a factor of 8/3.87 =~ 2:1
The "real" design and the typical design are different by a factor of about
2 x 1.25 = 2.5 times.
While these figures are cooked it shows how far off the mark one can be with
figures which are apparently OK to use.
NEVER USE TYPICAL FIGURES FOR SPECIFIC DESIGN DECISIONS.
(with the proviso that it may be OK to do so for statistically based aspects
of a design if you understand the implications).
Russell McMahon
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'[PICLIST] How to control a BIG inductive load?'
2001\06\09@021902
by
Peter L. Peres
|
Alexandre, please keep in mind that motors are big inductors that appear
almost as dead shorts when turned on. Unlike real big inductors. And the
kind of motor that is used in air conditioner compressors definitely
refuses to start if not supplied enough current. Plan for about as much
starting current as I = 1.5 * Uac / Rin, where Uac is your ac mains
voltage and Rin is the resistance you measure at the motor terminals with
the motor off. Multiply the result by 2 if the motor has a solenoid type
starter (because it does not appear in the circuit when measuring ohms).
If you get to do this and rub your eyes, and check the meter, it's ok for
a 2kW motor to show 2 to 10 ohms Rdc ;-), and sometimes less. 50 or 100A
start current is not uncommon for a motor that normally draws 10A. There
is a good reason for virtually all airconditioner makers using a relay to
switch the load in despite of fancy electronics.
Peter
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