In a relay contact RC snubber application like the one shown , why doesn't the snubber conduct the ac through the resistor and capacitor ?
I realize I should already know this but truthfully I have never thought about it before and was wondering why the
load doesn't remain powered through the snubber when the relay disengages and the contacts open ?
Yes, it's a dumb question but I have to ask because somehow the answer isn't coming to me...
It says here there is no current flow through an RC snubber across the contacts when the switch is open but it
doesn't say why.
If the snubber is connected in parallel to the load, the current flow is also stopped through the switch and the snubber circuit becomes active. In contrast to the first method there is no current flow when the switch is open but there is a leakage current when the switch is closed.
I think when the capacitor becomes fully charged it will stop conducting because even though it has 0V on it at
zero crossing , that is not the same as discharging a cap to ground because it is not being connected to ground so
the initial state is the current conducts through the contacts, bypassing the higher resistance of the resistor, then
the second state is contacts open and current must find a path through the resistor which then charges the cap .
But when the cap is fully charged and the negative ac transition doesn't the negative voltage do the same thing
in reverse ? Also, it would seem obvious that only a non-polarize cap would work in this application. By Ohm's
Law the current through the resistor is a function of the resistor value and if , for example, the load is a 100W
incandescent lightbulb. If the resistor is large enough it might not conduct enough current to illuminate the bulb
or the bulb might be dim. I'm still not seeing the factor that prevents it from powering the load to some degree.
What am I missing ?
for example, the load is a 100W
incandescent lightbulb. If the resistor is large enough it might not conduct enough current to illuminate the bulb
or the bulb might be dim. I'm still not seeing the factor that prevents it from powering the load to some degree.
What am I missing ?
Probably less than one volt, but without a load you'll get the full voltage.
Then, for a 0.047uF cap at 60 hz , that would make
XC = 1/(2Pi60Hz*0.000000047 F)
= 56,437.9 ohms
Xc = 56,437.9 ohms
P=I x V> I=P/V
I = 100W120V= 0.8333A
If my logic and math is correct, a capacitor with an
Xc of 56,437.9 would conduct 0.0000177A
I= V/R
= 120V / 56,437.9 Ohms
= 0.0000177A (17.7uA), which , conducting through the filament equivalent resistance of (120V/[(100W/120V)]A)
[(120V/0.833A) = 144 ohms)] V=I x R
= 0.0000177A x 144 Ohms
= 0.0025V. (2.5mV)
I think what I forgot was that for AC the cap has reactance which allows the application of ohm' law to calculate
what , if any ac passes through the RC snubber.
raschemmel:
It says here there is no current flow through an RC snubber across the contacts when the switch is open but it
doesn't say why.
If the snubber is connected in parallel to the load, the current flow is also stopped through the switch and the snubber circuit becomes active. In contrast to the first method there is no current flow when the switch is open but there is a leakage current when the switch is closed.
I think you are simply misreading it. The lower diagram on page 1 of the application note applies.
The previous paragraph also states:
If the snubber is
connected in parallel to the switch, the current flows through the
protection circuit at switch-off and then decays. Because of the
connection in parallel to the switch, there is a constant current
flow through the snubber when the switch is open. To keep this
current flow to an acceptable level, a well designed circuit is
needed.
Because of the connection in parallel to the switch, there is a constant current flow through the snubber when the switch is open. To keep this current flow to an acceptable level, a well designed circuit is needed.
I think that means the capacitor value needs to be chosen for a sufficiently high XC to limit the current to an acceptable value. I am still not clear on what the design criteria is for the resistor value. I originally thought
it was a function of the load current but now I am not so sure.
I also understand that for a properly designed RC snubber , the energy in joules (J) of the surge or spike to be
suppressed should be known, and then the cap value is
2*E(J)
C = ---------------
[VSWITCH([sub]MAX)[/sub]][sup]2[/sup]
In general, the current through the snubber is much less than the required operating current of the load, such as a solenoid or motor, so it does not move.
In terms of safety however, specifically personal safety, it is extremely dangerous.
Well, if the load is inductive, then at the very instant the contacts open, the original current will now be flowing through the resistor. So you decide what voltage you can tolerate and determine the resistor according to that current and that voltage.
Ohm's law - clearly the higher the resistance, the higher the instantaneous voltage that will appear before it decays.
Next you will want the answer to the difficult question.
Next you will want the answer to the difficult question(s).
From Reply #9:
"I originally thought it was a function of the load current but now I am not so sure."
Maybe we could start by listing the applications
for which an RC Snubber of the type I asked about
is used ?
i.e.: RC SNUBBER APPLICATIONS
Contactor for AC motor
Flyback Converter
Protecting switching devices (IGBT modules)
?
I found this :
RC snubbers are mostly used in low power and medium power applications. In high power applications, these snubbers exhibit excessive power losses making them unsuitable for such applications. In comparison, RCD snubbers are suitable for medium current and high current applications. They are commonly used for protecting various switching devices including IGBT modules. In high current applications, RC snubbers are mostly used for secondary damping. Snubber capacitors
and list our Design Parameters
Max switch contact to contact value
Load Voltage
Load current
AC Load frequency (50Hz or 60Hz)
Load Power (in W) (#3 x #2)
Back-EMF Pulse width (t) in seconds
Back-EMF Surge/Spike max amplitude/voltage
Energy (in joules(J)) of inductive Back-EMF surge/spike to be surpressed (calculated from #5 & #6)
