What's this surge protection component?

[DaveC]

On Tue, 18 Nov 2003 9:10:43 -0800, John Fields wrote
(in message <6bkkrvg8fbrr9r2i8eutpo5uok0c9m8aj9@4ax.com&gt:

Don't use a light bulb since it will do exactly the opposite of what you
want!

Hmm... so it seems. But why do many techs use it as a troubleshooting tool to
limit max current if the initial surge will fry/blow the fuse/component?

Want a schematic for the resistor and relay solution?

Sure!

Thanks,
--
DaveC
me@privacy.net
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[John Fields]

On Tue, 18 Nov 2003 9:15:27 -0800, DaveC <me@privacy.net> wrote:

On Tue, 18 Nov 2003 9:10:43 -0800, John Fields wrote
(in message <6bkkrvg8fbrr9r2i8eutpo5uok0c9m8aj9@4ax.com&gt:

Don't use a light bulb since it will do exactly the opposite of what you
want!

Hmm... so it seems. But why do many techs use it as a troubleshooting tool to
limit max current if the initial surge will fry/blow the fuse/component?

---
Prob'ly because they don't have a VARIAC.
It's used to limit gross steady-state current into a load, not to
prevent an initial surge from damaging the load.
---

Want a schematic for the resistor and relay solution?

Sure!

---
OK.

1. What are the surge spec's for the rectifiers?

2. What are the rail voltages when the caps are fully charged?

3. Are there any rail voltage maximum rise time spec's which need to
be adhered to?

4. What are the peak and continuous power ratings for the amp?

--
John Fields

 

[John Woodgate]

I read in sci.electronics.design that DaveC <me@privacy.net> wrote (in
<0001HW.BBDF873F005109B1F0080600@news.individual.net&gt about 'OK, so how
do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

the resistor-and-cutout relay solution, now.

I recommend that solution. It's much more flexible in design. The
appropriate lamp may not exist.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[John Woodgate]

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <6bkkrvg8fbrr9r2i8eutpo5uok0c9m8aj9@4ax.com&gt about
'OK, so how do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

Don't use a light bulb since it will do exactly the opposite of what you
want!

That was my initial reaction, but it's not true. The cold resistance of
the lamp needs to be high enough that initially almost all the available
voltage appears across it, so it heats up very rapidly and its
increasing resistance limits the inrush. When the caps are charged, the
lamp goes cool and its resistance falls.

A lamp that will do that for a given design may not actually exist.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[DaveC]

On Tue, 18 Nov 2003 10:35:59 -0800, John Fields wrote
(in message <f3okrvoip23letf9imel24496or59i7p95@4ax.com&gt:

1. What are the surge spec's for the rectifiers?

Don't know. I think I used 10 A bridges when I upgraded them. Not sure...

2. What are the rail voltages when the caps are fully charged?

Plus and minus 65vdc per side (one *big transformer, two bridge rectifiers,
four large caps). Caps are four 54,000 uF @ 75 vdc. I've e-mailed the cap
mfr. for ESR and max surge specs, but not a peep, yet.

3. Are there any rail voltage maximum rise time spec's which need to
be adhered to?

Nope. Pretty crude power supply, and MOSFET output with driver stages. No
fancy "muting for x seconds" relays or such features.

4. What are the peak and continuous power ratings for the amp?

Hmm... it's been awhile since I've seen the specs for this 20-year-old amp.
The MOSFET output stage is three 10 A devices per supply rail (12 total).
Since only three are operating per channel at any one time (conducting + or -
current through the load), isn't this P = I^2 / R = (10 x 3)^2 / 8 = 112.5 W
per channel max? I'm not sure how to rate the peak power, but I've noticed
that MOSFET spec sheets seem to give a "pulsed current" figure that is
usually 3 to 5 times the continuous.

Thanks,
--
DaveC
me@privacy.net
This is an invalid return address
Please reply in the news group

 

[John Woodgate]

I read in sci.electronics.design that DaveC <me@privacy.net> wrote (in
<0001HW.BBDFBB04005D2C1BF0080600@news.individual.net&gt about 'OK, so how
do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

4. What are the peak and continuous power ratings for the amp?

Hmm... it's been awhile since I've seen the specs for this 20-year-old
amp. The MOSFET output stage is three 10 A devices per supply rail (12
total). Since only three are operating per channel at any one time
(conducting + or - current through the load), isn't this P = I^2 / R =
(10 x 3)^2 / 8 = 112.5 W per channel max? I'm not sure how to rate the
peak power, but I've noticed that MOSFET spec sheets seem to give a
"pulsed current" figure that is usually 3 to 5 times the continuous.

You do the calculations this way. You have +/-65 V rails, and you
probably lose 3 V at each end in saturation voltages, source resistors
and strays. So you have 124 V peak-to peak signal available, which is
43.8 V r.m.s. If the load is 8 ohms, you get 43.8^2/8 = 240 W. This is
the short-term average power, where 'short' isn't so short with your BIG
caps in place. We can't say what the continuous average power is,
because we don't know how much the 65 V rails drop on sustained load.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

John Woodgate wrote:

I read in sci.electronics.design that ehsjr@bellatlantic.net wrote (in
3FB97087.F335B6B4@bellatlantic.net&gt about 'OK, so how do I choose a
replacement thermistor?', on Tue, 18 Nov 2003:

Can you go into a little more detail on this? "Will not work" could mean
many things. For example: Say you add an identical NTC in parallel with
an existing one. Do you degrade the function, improve the function, or
have no effect on the function of the NTC?

The one with the lower initial resistance, or the one with the steeper
fall with temperature, will hog nearly all the current.

Why? These things don't operate like diodes, as far as I
know. I didn't see a Vf rating (which may or may not mean
something) but their time constant is relatively long, from
30 to 130 seconds. They start out as very high resistance
so there would have to be current sharing at the beginning.
The limiting R would be lower than that of a single thermistor,
which would obviously reduce the desired effect - inrush
limiting. Even if not perfectly equal, at the beginning, with
two in parallel, both would heat and each would lower its
internal resistance, but neither would reduce so quickly as
to hog all the current, leaving the other in a cold resistance
state. If one was faster than the other, conceivably there
could be a point in the ~100 second time constant when one
hogs the current to the extent that the other no longer
decreases resistance - but for inrush, we are interested in
the beginning part.

So, if the above paragraph is correct NTC's in parallel would
work, but not as effectively as a single NTC. The hogging, if
it does occur, would happen after inrush.

All of the above is said to show my thinking on the subject -
please correct me if I am wrong. I claim no expertise on this
subject - just what I've learned reading about it. You may
have extensive experience on this and the ability to correct
any misconceptions I may have about it.

--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[DaveC]

I've got a 20 ohm, 30 watt wirewound power resistor. How do I calculate how
this would effect inrush current? If it's now 48 A, can we presume that the
turn-on load is about 3.5 ohms? Inserting the 20 ohm resistor would limit it
to 8.5 A (using 170 V peak in calculations)?

Could I then use some passive components in-line with the relay's coil to
delay the closing of the (N.O.) contacts which would short circuit the
resistor?

Am I on the right track, here?
--
DaveC
me@privacy.net
This is an invalid return address
Please reply in the news group

 

[John Woodgate]

I read in sci.electronics.design that ehsjr@bellatlantic.net wrote (in
<3FBAEB10.430F96C3@bellatlantic.net&gt about 'OK, so how do I choose a
replacement thermistor?', on Wed, 19 Nov 2003:

All of the above is said to show my thinking on the subject - please
correct me if I am wrong. I claim no expertise on this subject - just
what I've learned reading about it. You may have extensive experience on
this and the ability to correct any misconceptions I may have about it.

You need to re-think what happens at switch on. Suppose one thermistor
is 100 ohms and the other 110 ohms, and they both vary with temperature
in the same way (which is not normal: the 'law' varies from sample to
sample). The 100 ohm one will take more current, get hotter, lower
resistance, take more current, get hotter...

The same thing happens if they have equal cold resistances but the
resistance of one falls more steeply with temperature.

It isn't quite *inevitable* that hogging will occur. There is a limited
range of initial conditions under which the initial drop in resistance
is sufficient to reduce the power dissipated in that thermistor to less
than that in the other one. But this condition is almost impossible to
design for.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[John Woodgate]

I read in sci.electronics.design that DaveC <me@privacy.net> wrote (in
<0001HW.BBE04F46007FEBC0F0407600@news.individual.net&gt about 'OK, so how
do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

've got a 20 ohm, 30 watt wirewound power resistor. How do I calculate
how this would effect inrush current? If it's now 48 A, can we presume
that the turn-on load is about 3.5 ohms? Inserting the 20 ohm resistor
would limit it to 8.5 A (using 170 V peak in calculations)?

Not quite. You would have 23.5 ohms in the circuit, so the current would
be 7.23 A peak. It doesn't need to be as low as that, but it probably
won't do any harm.

Could I then use some passive components in-line with the relay's coil
to delay the closing of the (N.O.) contacts which would short circuit
the resistor?

You need to put a resistor in series with the coil and a large capacitor
in parallel with the coil. So that you can use an electrolytic
capacitor, you put a diode in series with the resistor. This diode needs
to be adequately rated, and a 1N4007 is a good choice. The R and C
values depend on the relay characteristics, and you will have to
experiment, with just the diode resistor, relay and capacitor, not the
whole amplifier. Start with a resistor of about twice the resistance of
the coil. You need to get the relay to close firmly. If it doesn't,
reduce the resistor value.

Am I on the right track, here?

Without more data that you give, it's not possible to say for sure, but
generally, yes.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[El Meda]

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that DaveC <me@privacy.net> wrote (in
0001HW.BBDF873F005109B1F0080600@news.individual.net&gt about 'OK, so how
do I choose a replacement thermistor?', on Tue, 18 Nov 2003:

the resistor-and-cutout relay solution, now.

I recommend that solution. It's much more flexible in design. The
appropriate lamp may not exist.

He can use the lamp as the resistor in the resistor-and-cutout relay
solution. The lamp can be an ordinary 100 W bulb. There is a power
amplifier made by Crest Audio that uses a thermistor-and-cutout relay.
When the termistor goes bad, i've used a 100W bulb in its place, and
it works OK.
---
Ing. Remberto Gomez-Meda <gomerem@hotmail.com>
http://ingemeda.tripod.com/
INGE - Ingenieria Electronica.
Puerto Vallarta, Jalisco, Mexico.

 

[John Woodgate]

I read in sci.electronics.design that El Meda <gomerem@hotmail.com>
wrote (in <c81nrvsgtg2tfokkblkpv4gaq9g8n3h1rs@4ax.com&gt about 'OK, so
how do I choose a replacement thermistor?', on Wed, 19 Nov 2003:

He can use the lamp as the resistor in the resistor-and-cutout relay
solution. The lamp can be an ordinary 100 W bulb. There is a power
amplifier made by Crest Audio that uses a thermistor-and-cutout relay.
When the termistor goes bad, i've used a 100W bulb in its place, and
it works OK.

Maybe it did in your case, but it won't work in all cases. You can
finish up with the lamp half on, the relay not energised yet and the amp
trying to work from half voltage or less. Remember that the amplifier's
no-signal current is not very large, even for a 240 W unit.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

John Woodgate wrote:

I read in sci.electronics.design that ehsjr@bellatlantic.net wrote (in
3FBAEB10.430F96C3@bellatlantic.net&gt about 'OK, so how do I choose a
replacement thermistor?', on Wed, 19 Nov 2003:
All of the above is said to show my thinking on the subject - please
correct me if I am wrong. I claim no expertise on this subject - just
what I've learned reading about it. You may have extensive experience on
this and the ability to correct any misconceptions I may have about it.

You need to re-think what happens at switch on. Suppose one thermistor
is 100 ohms and the other 110 ohms, and they both vary with temperature
in the same way (which is not normal: the 'law' varies from sample to
sample). The 100 ohm one will take more current, get hotter, lower
resistance, take more current, get hotter...

Right. And while the 100 ohm is doing that, the 110 ohm is
also taking current, getting hotter, lower resistance, take
more current, get hotter, etc. The heating in the higher
resistance unit will be less than the heating in the lower
resistance unit, but it will occur. Conceivably you can
arrive at the point where the first NTC has lowered resistance
enough so that the inrush voltage can no longer force enough
current through the other NTC to cause it to heat, but that
takes appreciable time.

Let me do some numbers. At the start, both thermistors are
cold. One is 110 ohms, the other is 100 ohms. Say the load
is 20 ohms. Ok, with only the 100 ohm thermistor and the
load in series, with a 120 volt source, the initial current
will be 1 amp, and the thermistor will drop 100 volts. Add
the 110 ohm thermistor. The total resistance of the two
thermistors in parallel and the load in series is ~72.38 ohms.
At turn on, the initial current will be ~1.67 amps, and the
thermistor pair will drop 87.3 volts. The current through
the 110 ohm thermistor will be ~.79 amps and the current
through the 100 ohm thermistor will be ~.87 amps. Thus both
will heat, but there will be more heating in the 100 ohm
thermistor. If by "current hogging" you are describing
the above scenario, then we are in agreement. The question
becomes one of rate of change of resistance over time in
a thermistor, but initially, the parallel NTC will reduce
the current that the first NTC has to handle, at the expense
of more total current (less limiting) to the load.
I haven't seen any curves of delta R vs time for these
limiters. If you have a url that shows the curves, that
would be helpful.

The same thing happens if they have equal cold resistances but the
resistance of one falls more steeply with temperature.

It isn't quite *inevitable* that hogging will occur. There is a limited
range of initial conditions under which the initial drop in resistance
is sufficient to reduce the power dissipated in that thermistor to less
than that in the other one. But this condition is almost impossible to
design for.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

DaveC wrote:

I've got a 20 ohm, 30 watt wirewound power resistor. How do I calculate how
this would effect inrush current? If it's now 48 A, can we presume that the
turn-on load is about 3.5 ohms? Inserting the 20 ohm resistor would limit it
to 8.5 A (using 170 V peak in calculations)?

Could I then use some passive components in-line with the relay's coil to
delay the closing of the (N.O.) contacts which would short circuit the
resistor?

Am I on the right track, here?
--
DaveC
me@privacy.net
This is an invalid return address
Please reply in the news group

Looks like you're on the right track. Your 8.5 A should be
lower, as John said, but you have the concept.

You can connect the relay coil across the power supply
through a resistor, as you suggest, but you may have a
problem. As the capacitors charge, the supply voltage
rises. When it is high enough, the relay operates and
its contacts bypass the limiting resistor. But the
voltage will continue to rise, possibly to the point
where it will force too much current through the relay
coil. So you have to be judicious in component selection,
such that the relay is protected from too much current,
yet gets enough to operate at the proper time.

An alternative is to use the power supply voltage level
as a signal, rather than a source - run it in to a
comparator, or use a zener & transistor, or whatever,
which in turn operates the relay from a fixed supply.
Or, you could use a turn on delay timing circuit, independent
of the supply you are limiting. After N seconds, it operates
the relay without regard to the voltage on the power supply.
The point is that there are a number of possibilities to
choose from - there are even time delay relays that contain
all the necessary circuitry that are made for this purpose.

 

[John Woodgate]

I read in sci.electronics.design that ehsjr@bellatlantic.net wrote (in
<3FBC465F.567BDC52@bellatlantic.net&gt about 'OK, so how do I choose a
replacement thermistor?', on Thu, 20 Nov 2003:

Let me do some numbers. At the start, both thermistors are cold. One is
110 ohms, the other is 100 ohms. Say the load is 20 ohms. Ok, with only
the 100 ohm thermistor and the load in series, with a 120 volt source,
the initial current will be 1 amp, and the thermistor will drop 100
volts. Add the 110 ohm thermistor. The total resistance of the two
thermistors in parallel and the load in series is ~72.38 ohms. At turn
on, the initial current will be ~1.67 amps, and the thermistor pair will
drop 87.3 volts. The current through the 110 ohm thermistor will be
~.79 amps and the current through the 100 ohm thermistor will be ~.87
amps.

Ratio of currents = 1.1

Thus both will heat, but there will be more heating in the 100 ohm
thermistor. If by "current hogging" you are describing the above
scenario, then we are in agreement.

You need to look at the next step. The thermistor that WAS 100 ohms gets
hotter due to the greater current, so its resistance goes down more.
Suppose it is 50 ohms at a certain time, and the other thermistor is 60
ohms. We get 47 ohms total, 2.55 A from the 120 V source and 69 V across
the thermistors. This gives 1.38 A through the 50 ohm and 1.15 A through
the 60 ohms.

Ratio of currents = 1.2

Eventually, one thermistor gets to carry almost all the current, which
of course decreases with time as the inrush falls away, while the other
gets cooler. This actually happens experimentally, so there is no point
in arguing that it doesn't.

The question becomes one of rate of
change of resistance over time in a thermistor,

Not with time, primarily, but with body temperature.

but initially, the
parallel NTC will reduce the current that the first NTC has to handle,
at the expense of more total current (less limiting) to the load. I
haven't seen any curves of delta R vs time for these limiters. If you
have a url that shows the curves, that would be helpful.

Time is not significant if both thermistors have the same thermal mass.
Since they have to be nominally identical (otherwise current hogging is
instantaneous), they do have that property.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

John Woodgate wrote:

I read in sci.electronics.design that ehsjr@bellatlantic.net wrote (in
3FBC465F.567BDC52@bellatlantic.net&gt about 'OK, so how do I choose a
replacement thermistor?', on Thu, 20 Nov 2003:

Let me do some numbers. At the start, both thermistors are cold. One is
110 ohms, the other is 100 ohms. Say the load is 20 ohms. Ok, with only
the 100 ohm thermistor and the load in series, with a 120 volt source,
the initial current will be 1 amp, and the thermistor will drop 100
volts. Add the 110 ohm thermistor. The total resistance of the two
thermistors in parallel and the load in series is ~72.38 ohms. At turn
on, the initial current will be ~1.67 amps, and the thermistor pair will
drop 87.3 volts. The current through the 110 ohm thermistor will be
~.79 amps and the current through the 100 ohm thermistor will be ~.87
amps.

Ratio of currents = 1.1

Thus both will heat, but there will be more heating in the 100 ohm
thermistor. If by "current hogging" you are describing the above
scenario, then we are in agreement.

You need to look at the next step. The thermistor that WAS 100 ohms gets
hotter due to the greater current, so its resistance goes down more.
Suppose it is 50 ohms at a certain time, and the other thermistor is 60
ohms. We get 47 ohms total, 2.55 A from the 120 V source and 69 V across
the thermistors. This gives 1.38 A through the 50 ohm and 1.15 A through
the 60 ohms.

Ratio of currents = 1.2

Eventually, one thermistor gets to carry almost all the current, which
of course decreases with time as the inrush falls away, while the other
gets cooler.

This actually happens experimentally, so there is no point
in arguing that it doesn't.

Now I'm getting somewhere! This is the part where I could only
speculate. I've never experimented with them in parallel.

Thank You!

The question becomes one of rate of
change of resistance over time in a thermistor,

Not with time, primarily, but with body temperature.

but initially, the
parallel NTC will reduce the current that the first NTC has to handle,
at the expense of more total current (less limiting) to the load. I
haven't seen any curves of delta R vs time for these limiters. If you
have a url that shows the curves, that would be helpful.

Time is not significant if both thermistors have the same thermal mass.
Since they have to be nominally identical (otherwise current hogging is
instantaneous), they do have that property.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!

 

[El Meda]

John Woodgate <jmw@jmwa.demon.contraspam.yuk> wrote:

There is a power
amplifier made by Crest Audio that uses a thermistor-and-cutout relay.
When the termistor goes bad, i've used a 100W bulb in its place, and
it works OK.

Maybe it did in your case, but it won't work in all cases. You can
finish up with the lamp half on, the relay not energised yet and the amp
trying to work from half voltage or less. Remember that the amplifier's
no-signal current is not very large, even for a 240 W unit.

You can choose a proper wattage lamp by trial and error (that´s what I
did). The Crest amp was a 1200 W unit, with big filter capacitors and
with my first choice -a 60 W lamp- it did exactly what you describe.
With a 100 W bulb the relay energized correctly.

Plus, every time the amp was turned on, the bulb glowed beautifully
inside, not unlike an old tube unit. :')
---
Ing. Remberto Gomez-Meda <gomerem@hotmail.com>
http://ingemeda.tripod.com/
INGE - Ingenieria Electronica.
Puerto Vallarta, Jalisco, Mexico.