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-   -   Do relay ratings apply for RF? (https://www.radiobanter.com/homebrew/22685-do-relay-ratings-apply-rf.html)

nick March 28th 04 02:52 PM

Do relay ratings apply for RF?
 
If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.

Tnx
Nick
n1pzy

Paul_Morphy March 28th 04 04:19 PM


"nick" wrote in message
...
If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


There are two parameters specified, but you are reducing them to one. The
first is the current, the second is the voltage. With a non-reactive 50-ohm
load, the rf current at 1000 watts is about 4.5 amps, and the rf voltage is
70.7 V. These formulae are in every edition of the ARRL Handbook. See also
the Handbook discussion of skin effect. At higher frequencies, ac tends to
travel near the surface of a conductor. If the relay contacts are silver
plated, this may not be a problem. The contacts in the old Dow-Key coax
relays weren't very large; it doesn't take a fat relay to switch
non-reactive 50-ohm loads. If you want to switch high-impedance or badly
mismatched antennas, you may exceed the capability of the relay. Another
consideration is whether you will actually be hot-switching the rf. If the
contacts are closed before rf is applied, there is less likelihood of arcing
and damaging the contacts.

Relays designed for rf use are mechanically constructed to minimize
impedance variation through the contacts, to maintain the non-reactive
50-ohm impedance of the system. At HF, a slight impedance 'bump' is not a
big deal. For amateur HF use, this relay will probably do the job.

"PM"



Paul_Morphy March 28th 04 04:19 PM


"nick" wrote in message
...
If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


There are two parameters specified, but you are reducing them to one. The
first is the current, the second is the voltage. With a non-reactive 50-ohm
load, the rf current at 1000 watts is about 4.5 amps, and the rf voltage is
70.7 V. These formulae are in every edition of the ARRL Handbook. See also
the Handbook discussion of skin effect. At higher frequencies, ac tends to
travel near the surface of a conductor. If the relay contacts are silver
plated, this may not be a problem. The contacts in the old Dow-Key coax
relays weren't very large; it doesn't take a fat relay to switch
non-reactive 50-ohm loads. If you want to switch high-impedance or badly
mismatched antennas, you may exceed the capability of the relay. Another
consideration is whether you will actually be hot-switching the rf. If the
contacts are closed before rf is applied, there is less likelihood of arcing
and damaging the contacts.

Relays designed for rf use are mechanically constructed to minimize
impedance variation through the contacts, to maintain the non-reactive
50-ohm impedance of the system. At HF, a slight impedance 'bump' is not a
big deal. For amateur HF use, this relay will probably do the job.

"PM"



Ralph Mowery March 28th 04 06:44 PM


"nick" wrote in message
...
If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


At a low impedance point in the line they could be ok. If you are going to
use it for switching in and out an amplifier after the rig probably no
problem. If you are depending on it for some isolation such as a low level
preamp then it will need to be sjpecified for how much isolation it has. If
used as an antenna switch you need to check the voltage rating so it will
not arc over in a high impedance point. The contact spacing and internal
wire spacing may give a problem in some situations. It all depends on the
relay.



Ralph Mowery March 28th 04 06:44 PM


"nick" wrote in message
...
If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


At a low impedance point in the line they could be ok. If you are going to
use it for switching in and out an amplifier after the rig probably no
problem. If you are depending on it for some isolation such as a low level
preamp then it will need to be sjpecified for how much isolation it has. If
used as an antenna switch you need to check the voltage rating so it will
not arc over in a high impedance point. The contact spacing and internal
wire spacing may give a problem in some situations. It all depends on the
relay.



Avery Fineman March 28th 04 10:34 PM

In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


Relay CONTACT ratings have little to with relay COIL ratings.

Contact ratings assume a specific environment such as AC
primary power circuits, 28 VDC aircraft distribution, etc. That
is for the overwhelming majority of relay applications. In the
case of contacts rated for 115 or 230 VAC applications, those
will have open-contact spacings larger than lower-voltage DC;
spacings are generally designed to withstand about 3 times
the AC RMS voltage specified in ratings. Generally, but
not always.

To switch RF in a coaxial line system you have to allow for
the VSWR having a high value...choose at your own
convenience for expected conditions. The reason for that is
because, at a high VSWR, the relay contacts may be at a
maximum voltage point or a maximum current point.
Depending on the VSWR, either maximum may exceed the
contact ratings values. Location of either maximum depends
on frequency, length of transmission line, velocity of
propagation of the line used, antenna characteristics, and
the distance from the discontinuity (typically the antenna
feedpoint).

There are many tables and nomographs in ham literature to
determine the maximum voltage or maximum current with a
given SWR for a given characteristic impedance of a line.
A realistic worst-case value should be selected to avoid
having to replace a relay often due to arc-over or contact
sticking due to maxima. If an RF system is perfect (1:1
VSWR, line Z exact and unchanging) then compute the
required voltage and current from ordinary Ohm's Law.

I have yet to see such a perfect system but I know others
claim them as such... :-)

Len Anderson
retired (from regular hours) electronic engineer person

Avery Fineman March 28th 04 10:34 PM

In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.


Relay CONTACT ratings have little to with relay COIL ratings.

Contact ratings assume a specific environment such as AC
primary power circuits, 28 VDC aircraft distribution, etc. That
is for the overwhelming majority of relay applications. In the
case of contacts rated for 115 or 230 VAC applications, those
will have open-contact spacings larger than lower-voltage DC;
spacings are generally designed to withstand about 3 times
the AC RMS voltage specified in ratings. Generally, but
not always.

To switch RF in a coaxial line system you have to allow for
the VSWR having a high value...choose at your own
convenience for expected conditions. The reason for that is
because, at a high VSWR, the relay contacts may be at a
maximum voltage point or a maximum current point.
Depending on the VSWR, either maximum may exceed the
contact ratings values. Location of either maximum depends
on frequency, length of transmission line, velocity of
propagation of the line used, antenna characteristics, and
the distance from the discontinuity (typically the antenna
feedpoint).

There are many tables and nomographs in ham literature to
determine the maximum voltage or maximum current with a
given SWR for a given characteristic impedance of a line.
A realistic worst-case value should be selected to avoid
having to replace a relay often due to arc-over or contact
sticking due to maxima. If an RF system is perfect (1:1
VSWR, line Z exact and unchanging) then compute the
required voltage and current from ordinary Ohm's Law.

I have yet to see such a perfect system but I know others
claim them as such... :-)

Len Anderson
retired (from regular hours) electronic engineer person

N2EY March 29th 04 12:58 AM

In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts.


It will take 1100 watts at 110 volts. But if you lower the voltage to, say, 50
volts, and the contacts are still rated at 10 amps, it can only handle 500
watts. Etc.

Will it take 1000 watts of RF?

I would think not, but I don't know why.


Depends entirely on the RF characteristics - impedance, SWR, etc.

The trick is to exceed neither the voltage nor current ratings of the contacts.
In a 50 ohm RF application with low SWR (less than 2:1), that relay is probably
limited by its contact voltage rating, not the current rating. Which works out
to about 200 watts.

73 de Jim, N2EY


N2EY March 29th 04 12:58 AM

In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts.


It will take 1100 watts at 110 volts. But if you lower the voltage to, say, 50
volts, and the contacts are still rated at 10 amps, it can only handle 500
watts. Etc.

Will it take 1000 watts of RF?

I would think not, but I don't know why.


Depends entirely on the RF characteristics - impedance, SWR, etc.

The trick is to exceed neither the voltage nor current ratings of the contacts.
In a 50 ohm RF application with low SWR (less than 2:1), that relay is probably
limited by its contact voltage rating, not the current rating. Which works out
to about 200 watts.

73 de Jim, N2EY


Tom S March 29th 04 03:17 PM

Are these specs for hot switching? I would think that if you can figure out
a way to switch the relay THEN apply power, everything would be happy and
the contacts will last much longer.

Tom


"N2EY" wrote in message
...
In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts.


It will take 1100 watts at 110 volts. But if you lower the voltage to,

say, 50
volts, and the contacts are still rated at 10 amps, it can only handle 500
watts. Etc.

Will it take 1000 watts of RF?

I would think not, but I don't know why.


Depends entirely on the RF characteristics - impedance, SWR, etc.

The trick is to exceed neither the voltage nor current ratings of the

contacts.
In a 50 ohm RF application with low SWR (less than 2:1), that relay is

probably
limited by its contact voltage rating, not the current rating. Which works

out
to about 200 watts.

73 de Jim, N2EY





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