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Help needed. Zero crossing with RC snubber problem

M

michael nikolaou

Jan 1, 1970
0
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the contact
of the driver relay i placed one RC snubber circut (27NF with 100 R
resisitor in series) to help with some spikes that were influencing the low
voltage driver circuits.
The driver circuit is able to detect mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform to 0
volt after 5 periods of 500 HZ but no overshoot. The relay presents no
arcing. If i remove the snubber and make the experiment the best place to
switch is zero crossing but i also see large SHARP spikes up to 500 Volts
Peak.
My question is
The switching with snubber must be made at zero crossing or at the peak of
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

Michael
 
P

petrus bitbyter

Jan 1, 1970
0
michael nikolaou said:
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the
contact of the driver relay i placed one RC snubber circut (27NF with 100
R resisitor in series) to help with some spikes that were influencing the
low voltage driver circuits.
The driver circuit is able to detect mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform to
0 volt after 5 periods of 500 HZ but no overshoot. The relay presents
no arcing. If i remove the snubber and make the experiment the best place
to switch is zero crossing but i also see large SHARP spikes up to 500
Volts Peak.
My question is
The switching with snubber must be made at zero crossing or at the peak of
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

Michael

The best moment for switching off, highly depends on the load. As long as
the load is resistive, the zero crossing point of the voltage is best as
switching is done at minimum voltage and current. As soon as the load has a
reactive component, zero crossing of the voltage differs from zero crossing
of the current. It is the breaking of the current that causes the voltage
spikes.

petrus bitbyter
 
C

Chris

Jan 1, 1970
0
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the contact
of the driver relay i placed one RC snubber circut (27NF with 100 R
resisitor in series) to help with some spikes that were influencing the low
voltage driver circuits.
The driver circuit is able to detect  mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point  to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV  at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform  to 0
volt after 5 periods of    500 HZ  but no overshoot. The relay presents no
arcing.  If i remove the snubber and make the experiment the best place to
switch is zero crossing but i also see large SHARP spikes up to 500 Volts
Peak.
My question is
 The switching with snubber must be made at zero crossing or at the peakof
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large  spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

 Michael

Hi, Michael. First off, you should spec a 12V relay that's made to
handle inductive loads (you can see a HP rating). This type of relay
has contacts which open more quickly, and are farther apart when
open. A small standard relay might not even open up far enough to
stop an inductive arc at line voltage.

Next, when using a relay to drive a relay, you have to be aware of the
delay-on-make, which can be several milliseconds, especially for
larger relays. That might help explain some of the curious results
you're getting. Turn-on delay can be affected by wear and aging. It
also varies from unit to unit, even in relays with the same part and
lot number. Trying to get this kind of timing accuracy might be the
wrong way to go.

It might be better to take a look at suppressing the arc, which you've
already started to do. Here's an intuitive way to start. First off,
remember your basic goal: you want the voltage across the contacts to
rise just slowly enough so the contacts can pull away without
sustaining an arc. That's all.

Remove the cover from the driving relay so you can see the contacts.
Next, find the current rating of those contacts, and use Ohm's law to
find a resistor which will result in about half that current.

For example, if you have a 220VAC load, and your relay can handle
10A,:

R = 220V / 5A = 44 ohms

Choose a 47 ohm, 1 watt resistor here (carbon comp is best). Now get
a selection of self-healing AC line-rated capacitors, and switch the
inductive load with the 47 ohm and C snubber directly across the load,
repeating and increasing the cap value until the contact arcing
disappears, or at least is minimized. Without knowing anything about
your 220VAC relay, it sounds like your 0.027uF cap might be on the
small side.

Note that ITW Paktron makes a series of Quencharc snubbers that you
can just plug in, which makes selection a snap. They're one-piece,
epoxy-encapsulated units, and are very easy to use.

http://www.paktron.com/pdf/Quencharc_QRL.pdf

If all else fails, remember that physical distance is also helpful.
Do what you can to place the arcing contact as far away as possible
from sensitive circuitry.

Good luck
Chris
 
L

legg

Jan 1, 1970
0
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the contact
of the driver relay i placed one RC snubber circut (27NF with 100 R
resisitor in series) to help with some spikes that were influencing the low
voltage driver circuits.
The driver circuit is able to detect mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform to 0
volt after 5 periods of 500 HZ but no overshoot. The relay presents no
arcing. If i remove the snubber and make the experiment the best place to
switch is zero crossing but i also see large SHARP spikes up to 500 Volts
Peak.
My question is
The switching with snubber must be made at zero crossing or at the peak of
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

Michael
If your load is inductive, the current lags the voltage by 90 degrees.

Generally, to avoid switch stress, you would try to switch inductive
loads at zero current, capacitive loads at zero voltage.

RL
 
P

Pieter

Jan 1, 1970
0
What chris writes really makes sense. I am adding my extra's here too.

Hi, Michael. First off, you should spec a 12V relay that's made to
handle inductive loads (you can see a HP rating). This type of relay
has contacts which open more quickly, and are farther apart when
open. A small standard relay might not even open up far enough to
stop an inductive arc at line voltage.

Very true!
Next, when using a relay to drive a relay, you have to be aware of the
delay-on-make, which can be several milliseconds, especially for
larger relays. That might help explain some of the curious results
you're getting. Turn-on delay can be affected by wear and aging. It
also varies from unit to unit, even in relays with the same part and
lot number. Trying to get this kind of timing accuracy might be the
wrong way to go.

There also is a relay-off delay. And the RC snubber makes the turn-off
delay even worse as the current will run a little longer. Expect
something here between 5 to 20 milliseconds.
It might be better to take a look at suppressing the arc, which you've
already started to do. Here's an intuitive way to start. First off,
remember your basic goal: you want the voltage across the contacts to
rise just slowly enough so the contacts can pull away without
sustaining an arc. That's all.

What you need to know is the current running through the contacts, and
the voltage spike you want to allow. What a snubber does is store the
energy of the coil in the capacitor. The resistors "eats" this up as
the capacitor discharges over the load. The voltagespike at turnoff
(aasuming an empty capacitor at that moment) is the load current *
resistor. So if you have R= 100, C= 47n, I=2A, you get a voltage spike
of 100*2 = 200 volt even before your capacitor charges. The load
inductance with the current give the stored energy: Q=L*I=C*U. So the
capacitor size should match the load inductance, otherwise you get a
high voltage there too. The capacitor will for example be loaded to
100 volts with a Q = 27n*100V = 2,7 uCoulomb. An inductance that
contains this would be L=Q/I= 2,7uC/2A = 1,35 uH. But the resistor
already "eats" up a lot, so the voltage will be lower. When you turn
off larger motors, transformers, inductors (the coil of a large
relay!), the capacitor must match the load to keep voltages limited.

The prevent oscillations, resistance values must not be too low,
ususally between 30 to 100 Ohm is ok. For large inductive loads I
would not take 100 Ohm but 47 Ohm as normal value (see below what
Chris wrote), you contact must ve able to handle that. The capacitance
could be anything you want, for larger devices 220n, 470n etc.
Remove the cover from the driving relay so you can see the contacts.
Next, find the current rating of those contacts, and use Ohm's law to
find a resistor which will result in about half that current.

For example, if you have a 220VAC load, and your relay can handle
10A,:

R = 220V / 5A = 44 ohms

Choose a 47 ohm, 1 watt resistor here (carbon comp is best). Now get
a selection of self-healing AC line-rated capacitors, and switch the
inductive load with the 47 ohm and C snubber directly across the load,
repeating and increasing the cap value until the contact arcing
disappears, or at least is minimized. Without knowing anything about
your 220VAC relay, it sounds like your 0.027uF cap might be on the
small side.

Also very true: the RC snubber also gives a current peak when you turn
ON the relay. Nice thing about inductive loads is that the current
does not rise fast, so the load and RC current do not occure exactly
at the same time (also depending on RC, where a smaller R gives a
higher rush-in current but also a better timing).

The peak where the snubber works is at turning the relay OFF again.

What I often use is 47 Ohm to 100 Ohm with 100 nF.
Note that ITW Paktron makes a series of Quencharc snubbers that you
can just plug in, which makes selection a snap. They're one-piece,
epoxy-encapsulated units, and are very easy to use.

http://www.paktron.com/pdf/Quencharc_QRL.pdf

If all else fails, remember that physical distance is also helpful.
Do what you can to place the arcing contact as far away as possible
from sensitive circuitry.

Good luck
Chris

You need TWO RC networks: one across the driving relay, one across the
large relay, and do not forget the diode (or also RC) across the coil
of the driving relay. So you have 3 things that need attention.

Do not forget that long cables (also with resistive loads like lamps)
act as inductance. 10 meters or more may also need a RC!

Regards,
Pieter
 
M

michael nikolaou

Jan 1, 1970
0
Thank guys for your replies .Some of them i have to study first

Let me make some things clear about the circuit and values chosen

1. I've measured turn-on , turn-off delay at 3.3 ms for the driver relay.All
results are
after calculating this delay .So what is see on the scope is at the moment
i'm explaining
2. the arc is across the driver relay .The power board is inside a control
unit box so
i have to leave with small distances and cpu disturbances.Its actually a
microcontroller
having the problem .Driver relay contact current rating is 5A at 220V
..Power relay coil is rated is 6
watts consumption at 220V.
3. Using large value capacitors over 33 nf was causing sometimes latch of
the power relay so i have
value limitation here
4. The capacitors i've chosen are X2 self healing 275VAC. With no ZC control
they are blown
to 0 nf value (some of them) after 10-12 months of operation.
5. I don't have the space or budget to use large sized capacitors rated at
higher voltages or SSR .
The idea was to use ZC to avoid using expensive and large size protection
snubber
So the question is .

Does the relay On/OFF time differs with time .If it's 10% it's not a
problem since again
the arc will not be so high .Since its the current break that causes the
arc i must switch off at Peak of the
ac voltage .This is what my reading confirmed .In this case switching a
resistive load must i change the driver
algorithm ???

Any helpful comments will be apreciated
 
L

legg

Jan 1, 1970
0
Thank guys for your replies .Some of them i have to study first

Let me make some things clear about the circuit and values chosen

1. I've measured turn-on , turn-off delay at 3.3 ms for the driver relay.All
results are
after calculating this delay .So what is see on the scope is at the moment
i'm explaining
2. the arc is across the driver relay .The power board is inside a control
unit box so
i have to leave with small distances and cpu disturbances.Its actually a
microcontroller
having the problem .Driver relay contact current rating is 5A at 220V
.Power relay coil is rated is 6
watts consumption at 220V.
3. Using large value capacitors over 33 nf was causing sometimes latch of
the power relay so i have
value limitation here
4. The capacitors i've chosen are X2 self healing 275VAC. With no ZC control
they are blown
to 0 nf value (some of them) after 10-12 months of operation.
5. I don't have the space or budget to use large sized capacitors rated at
higher voltages or SSR .
The idea was to use ZC to avoid using expensive and large size protection
snubber
So the question is .

Does the relay On/OFF time differs with time .If it's 10% it's not a
problem since again
the arc will not be so high .Since its the current break that causes the
arc i must switch off at Peak of the
ac voltage .This is what my reading confirmed .In this case switching a
resistive load must i change the driver
algorithm ???


However the timing is controlled, yes, that's probably what you need
to do.

The larger relay coil will be rated for power consumption in the
continuous active condition - with the armature closed and inductance
fairly high. The inductance limits the current flow that generates
power loss in the coil, if your phase angle observations are correct.

It's therefor possible that current in the coil is higher than a
guestimate (using the 6W label) might produce. What's the DC
resistance of the bigger coil? What's the actual coil current with the
voltage applied?

Larger currents could account for the control contact arcing and large
voltages that you see with 100R snubber....unless the resistor is open
circuit or intermittent.

RL
 
C

cadman

Jan 1, 1970
0
I have used a triac assisted relay before.

The triac shorts out the relay contacts just before the relay is
switched on and off.
Because the relay is no longer swicthing current they pretty much last
forever.
You can get away with a small triac as its not on for long.
 
J

Jamie

Jan 1, 1970
0
michael said:
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the contact
of the driver relay i placed one RC snubber circut (27NF with 100 R
resisitor in series) to help with some spikes that were influencing the low
voltage driver circuits.
The driver circuit is able to detect mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform to 0
volt after 5 periods of 500 HZ but no overshoot. The relay presents no
arcing. If i remove the snubber and make the experiment the best place to
switch is zero crossing but i also see large SHARP spikes up to 500 Volts
Peak.
My question is
The switching with snubber must be made at zero crossing or at the peak of
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

Michael
WHen you say relay, I assume you mean a mechanical contact?

if so, It takes time for the contacts the release. If you turn it
off at what you detect as the zero crossing point, the contacts most
likely will not actually release until some where in some mid point .
Many contactors are fast but not fast enough to open before current
can get a charge going.
That's just my evaluation of what you're doing.

By you signaling to turn off the relay at a peak, the contactor will
most likely not open until it gets near the zero crossing point.


--
 
C

Chris

Jan 1, 1970
0
Use a solid state releay and dont worry about it.

Bob- Hide quoted text -

- Show quoted text -

Right. If you have a resistive load for the final 220VAC load, a
plain old solid state relay will work fine (be sure to heat-sink the
SSR at about 1.5 watts per amp load). If you have an inductive load,
you may want to spec a SSR made to switch these loads, which have back-
to-back SCRs to eliminate the possibility of not being able to turn
off the SSR).

If this is a class project, you won't lose any points by going for the
simple solution, as long as it also happens to be the best one.

Good luck
Chris
 
WHen you say relay, I assume you mean a mechanical contact?

  if so, It takes time for the contacts the release. If you turn it
  off at what you detect as the zero crossing point, the contacts most
likely will not actually release until some where in some mid point .
   Many contactors are fast but not fast enough to open before current
can get a charge going.
    That's just my evaluation of what you're doing.

   By you signaling to turn off the relay at a peak, the contactor will
most likely not open until it gets near the zero crossing point.

--- Áðüêñõøç êåéìÝíïõ óå ðáñÜèåóç -

- ÅìöÜíéóç êåéìÝíïõ óå ðáñÜèåóç -

I observe all signals with a scope . I see actually the "driver relay"
signal .So all my observations
about timing are correct . My notion was different though .I expected
to turn off at zero crossing
voltage and have no arc .Actualy when the transition happents at the
peak of voltage then i see only a
decaying waveform ,no overshoot, for which i suspect since i know the
capacitor and measure the
frequency one could calculate the total inductance value (cables+relay
coil). Since its clear that V(emf)=-L*di/Dt
as pointed out by your emails also then the correct point to switch
off the driver relay is at peak voltage
since current lags voltage by 90 deg in any inductor.
Now i only have to find out how much can a relay type deviate from the
measurements i have made.
Thank's for all you help guys
 
O

Oppie

Jan 1, 1970
0
I have had some occasions using a solid state relay to switch a large
contactor. We found that a random turn-on relay was as good as a zero
crossing turn-on (and cost a bit less too). Turn-off with a solid state
relay will be at the next zero *current* crossing regardless of the type of
SSR used. As to snubbing, I found that a snubber across the contactor's coil
was more significant than one across the SSR. A secondary snubber can be
applied across the SSR if still needed.

The main purpose of snubbing with a SSR is to control the slew rate and peak
voltage across the SSR's switching element (traic or scr) to keep it below
the point of causing a spurrious trigger/conduction cycle. You should also
be mindful that triacs are more sensitive to triggering in certain quadrants
and may lead to an asymetrical output conduction (aka a net DC component)
which can cause inductive loads to saturate.

Oppie
 
J

Jamie

Jan 1, 1970
0
I observe all signals with a scope . I see actually the "driver relay"
signal .So all my observations
about timing are correct . My notion was different though .I expected
to turn off at zero crossing
voltage and have no arc .Actualy when the transition happents at the
peak of voltage then i see only a
decaying waveform ,no overshoot, for which i suspect since i know the
capacitor and measure the
frequency one could calculate the total inductance value (cables+relay
coil). Since its clear that V(emf)=-L*di/Dt
as pointed out by your emails also then the correct point to switch
off the driver relay is at peak voltage
since current lags voltage by 90 deg in any inductor.
Now i only have to find out how much can a relay type deviate from the
measurements i have made.
Thank's for all you help guys
if you're trying to save the contact life you can assist it with a SSR
across the terminals.
The SSR will conduct just prior before the contacts close, this will
create a shunt on the contacts. when the contacts finally make, they
will remove the load from the SSR.
When opening the contacts, the SSR will switch to save the day and
unlatch at the base line. If you decide to employ this, you need to have
a snubber tide across the SSR because of slight delays of the SSR, you
could damage it when the contacts open on a high peak.
Just select a SSR with the same turn on voltage as the contactor and
tie them to the same control voltage.
 
L

legg

Jan 1, 1970
0
The circuit below simulates fairly well. You don't want to use a
conventional snubber across the contacts because on opening the relay
coil voltage reverses and adds to the 220VAC source. Placing a snubber
in shunt with the coil with peak current limiting resistor as shown
increases operating power by about 10% but tends to maintain the contact
voltage and results in a very slew rate limited 0.5V/us contact voltage
peaking in the 450V range. There should be no arc at all with this
circuit, with or without zero crossing logic. K1 are the 12V relay
contacts and K2 is the 220VAC coil. I did not consider contact bounce on
closure, will leave that to you.
View in a fixed-width font such
as Courier.


.
.
.
.
. 220VAC
. o o
. | |
. | |
. | - K1
. | -
. | |
. | |
. | R 100R, 1.5W
. | |
. | |
. | +-----.
. | | |
. | - |
. | |\| |
. | K2|\| ===
. | |\| | 47nF
. | |\| |
. | - |
. | | |
. | | |
. '------+-----'
.
.
Placing impedance in series with the working solenoid could produce a
reduction in speed/dropout performance in the armature of the relay
switching the main working load. (not shown in the above drawing)

I'm not sure how you modelled the relay coil, but if it used a linear
inductor, it will not likely reflect actual performance. A relay
drive coil is coupled to a mechanically changing magnetic circuit.

As the OP already has a cost-free solution involving programmed timing
adjustments, perhaps it's best to let the issue drop?

RL
 
L

legg

Jan 1, 1970
0
He was working with a largely resistive 8KR coil, the 100R has no effect
on pull-in or hold-in. The programmed timing can be dropped, what
happens at line loss, does his controller lock up because things didn't
go exactly as planned, you tell me.

I've asked for this information, but still see only the reference to a
6W AC coil in the relay doing the work. Larger AC working relay coil
current is seldom determined by the DC impedance of the coil. You can
refer to the Leach tutorial on this issue, and their catalog, though
they don't seem to supply relays or contactors with low frequency AC
coils, at present.

For example, a 115V 400Hz coil in a 4W series, is listed as consuming
90mA. This indicates that the DC resistance of the coil must be less
than half an RDC value that would produce the specified current, in
order to keep combined power loss from coil and armature poles to the
value expected.

The losses in AC activated coils is typically 5 times that for DC
rated ones, in simpler commodity forms like OMRON MGN.

http://components.omron.com/components/web/webfiles.nsf$FILES/family.html?ID=CNEN-6TJQPU

But that wasn't always the case.

Older SquareD parts anticipate coil current ratios of more tha 4:1
between closed and open armature, with different values expected for
50 and 60Hz dedicated parts..

http://ecatalog.squared.com/catalog/173/html/sections/21/17321026.html#1013844

The same relationship shows up in lower-powered parts:

http://catalog.tycoelectronics.com/...T&P=86596,80517&U=&BML=10576,16354,16453&LG=1

RL
 
L

legg

Jan 1, 1970
0
I'm aware of that. Generally AC coils with predominantly reactive
impedance are rated in VA and the so-called impedance limited coils with
impedance dominated by coil wire resistance are rated in Watts. The OP
is working with a 6W impedance limited coil.

Sorry, but I don't see a resistance limited coil described in any
correspondence from the OP.

The most common ( old P&B now Tyco ) contactor for high current that
still offers AC coils and a fair description of their impedance and
wattage expectations:

http://tinyurl.com/3845jx

RL
 
L

legg

Jan 1, 1970
0
That link proves my point, the AC coils are specified in Volt-Amps and
the listed DC resistance of those coils is ~20% of the reactance. The
fact that the OP describes his coil as AC and 6W means it's impedance
limited.

Could you post a link to a data sheet for any device meeting this
description? I am unable to find a relay in this coil power range that
even has an AC-operated rating specified in W, never mind one
operating at this power level resistively.

I see some smaller ones that come close to 45 degrees, but are still
on the inductive side. I guess it's hard to avoid, being a magnetic
component.

I appreciate that there's some confusion here, but I have a suspicion
that it is most likely to originate with poor characterization by the
OP. I see no reason to carve the misunderstanding into electronic
stone on the news server.

RL
 
L

LVMarc

Jan 1, 1970
0
michael said:
Hi

I have a 12 v relay driving an large 220 volt AC relay . Across the contact
of the driver relay i placed one RC snubber circut (27NF with 100 R
resisitor in series) to help with some spikes that were influencing the low
voltage driver circuits.
The driver circuit is able to detect mains zero crossing and fire the
driver relay at an angle i choose .
From what i read the best point to switch off the power relay is at zero
crossing . I did that and i show a large spike up to 1 KV at the relay
contact followed by a decaying 500hz waveform to 0 volts . After some
experimentation the best point came exactly when switching off at the peak
of the mains voltage .At this point there is smooth decaying waveform to 0
volt after 5 periods of 500 HZ but no overshoot. The relay presents no
arcing. If i remove the snubber and make the experiment the best place to
switch is zero crossing but i also see large SHARP spikes up to 500 Volts
Peak.
My question is
The switching with snubber must be made at zero crossing or at the peak of
an ac voltage waveform ?
What is the behaviour of the circuit ?.
As i understand any large spikes can harm the X2 capacitor i'm using so
what is the best operating practise ?.

Any help will be appreciated

Michael
Hello,


Two comments:


If the load is not purely resistive, there will be a voltage current
shift, aka Eli ICE Man... therefore in this case zero voltage crossing
is not at all zero current crossing and you may be creating more problem
by switching at the worst (or just a bad) time) .

BY looking at the signal that the rc snubber is trying to "tame" , you
adjust RC unitl you get a "critically damped response. you can dampen
more, but at the cost of higher stand by leakage via the RC, as it
becomes a part of the load too!

Good luck, this is an often asked problem, and the ability to visualizes
on scope and make changes and observations, will help you forever. This
is a re-occcuring problem and it is the variations in the load that
cause engineers to have to re-visit the solutions

Best regards

Marco
 
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