Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

Thanks,

de KJ4UO

On 21 Dec 2004 14:23:10 GMT, (PDRUNEN) wrote:

Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

Hi OM,

That is the loss for using the network, and for this network it
appears to be Power lost IN the network (rather than Power not
delivered to the Load). However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

No matching network is required for this scenario.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

By your description, you are still not using the matching network. If
you think you are, then the math is incorrect.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

Of course not. You never accounted for your two matching resistors so
you cannot assign the 5.7dB loss to them.

I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

There are any number of reasons - accuracy and stability being
principal among them. On the other hand, the two resistor variety is
not the most suitable for accuracy (except under known loads).
Another virtue is that it will increase the input power specification
by that same 5.7dB. This very handy when you are putting a flame
thrower to its input. Burnt resistors are cheaper and easier to
replace than the input stage. One very useful 100 Ohm resistor is a
10mA glass fuse (Wollaston wire).

73's
Richard Clark, KB7QHC

In message , Richard Clark
writes
On 21 Dec 2004 14:23:10 GMT, (PDRUNEN) wrote:

Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

Hi OM,

Snips
However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.


Several manufacturers of precision matching pads might disagree!
Two resistors is enough, although you have to remember to add a rather
odd number (unless the measuring equipment can add it for you). If you
add a third resistor, you can make the correction a straight 6dB. That's
probably the only virtue.
Ian.
--

On Tue, 21 Dec 2004 20:27:54 +0000, Ian Jackson
wrote:

However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.
Several manufacturers of precision matching pads might disagree!

Hi Ian,

I seriously doubt it, but you are free to offer examples.

73's
Richard Clark, KB7QHC

In message , Richard Clark
writes
On Tue, 21 Dec 2004 20:27:54 +0000, Ian Jackson
wrote:

However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.
Several manufacturers of precision matching pads might disagree!

Hi Ian,

I seriously doubt it, but you are free to offer examples.

73's
Richard Clark, KB7QHC



http://www.maxim-ic.com/appnotes.cfm...mber/972/ln/en
http://www.maxim-ic.com/appnotes.cfm...te_number/3250
http://www.maxim-ic.com/appnotes.cfm...mber/972/ln/en
http://www.testmart.com/estore/produ...Fsearch%2Fspec.
cfm~~MICCOM~~AGILEN~~11852B~~%20~~%20|1.html
http://www.g4fgq.regp.btinternet.co.uk/padmatch.pas
http://www.g4fgq.regp.btinternet.co.uk/padmatch.pas
http://used-line.com/b2544p1pr0-Used-pads.htm
http://www.minicircuits.com/dg03-159.pdf#search='minimum%20loss%20pad'
+ many, many more!
Ian.

--

On Tue, 21 Dec 2004 22:51:30 +0000, Ian Jackson
wrote:

In message , Richard Clark
writes
On Tue, 21 Dec 2004 20:27:54 +0000, Ian Jackson
wrote:

However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.
Several manufacturers of precision matching pads might disagree!

Hi Ian,

I seriously doubt it, but you are free to offer examples.

73's
Richard Clark, KB7QHC
+ many, many more!
Ian.

Hi Ian,

Perhaps you can share from those many, many more, those which
disagree?

73's
Richard Clark, KB7QHC

On 21 Dec 2004 14:23:10 GMT, (PDRUNEN) wrote:

Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

This is correct, if the 100 V rms source is the Thevenin emf - not the
loaded terminal voltage.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

Under those condx the load voltage is 60 V rms and the load power is
48 watts, as you indicated.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

Since the original load voltage was 50 volts rms and now it's 60
volts, one could argue that the load voltage is up by 1.6 dB. However,
the load power at 48 watts is in fact down by 0.17 dB


I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

Interestingly enough, none of this has anything to do with your L-pad
matching network. All of your calculations above relate only to the
hypothetical situation in which you are actually "changing" the load
resistance and computing the new load power - nothing whatsoever to do
with the situation you would have when you actually insert an L-pad
between the real source and the original 50 ohm load.

A 75 to 50 ohm resistive matching pad is designed to work with a 75
ohm load and to present a 50 ohm load to the source, which it will do.
When you calculate the amount of power in the 75 ohm load before and
after inserting the L-pad you will get a significant loss, depending
upon the exact design of the two-resistor network. 5.7 dB sounds a bit
high, but it depends upon how the L-Pad losses are defined.

As to the reason for using the L-Pad, it's probably for the purpose of
minimizing the SWR on the cable between the L-Pad/Load and the source.
Since you haven't specified anything about the source, the load or
what the purpose of the circuit is, one can only guess.



Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

I was reviewing a 75 to 50 ohm resistive matching network using two
resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving
a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power
is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

Thanks,

de KJ4UO

In test devices the losses are not usually a problem as long as the levels
are accounted for if absolute numbers are needed. The impedances must match
if any tuning is done. As teh impedance of a receiver is not usaully 50
ohms over a wide frequency range , you use a 6 db pad or so to isolate the
generator from the receiver . If tuned circuits are involved, the impedance
missmatch (swr) can cause many problems . You tune a device for a 50 ohm
source or load and then replace it with a 70 ohm device and the tuning will
usually change . By using the pads, it is an attempt to keep the impedance
changes at a minimum when the test instruments are removed and the device
put back into normal operation.

On 21 Dec 2004 14:23:10 GMT, (PDRUNEN) wrote:

Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

Thanks,

de KJ4UO


Dear KJ4UO,

Some additional help with your L-pad -

The minimum loss L-Pad for matching 75 ohms to 50 ohms would indeed
have a loss of 5.7 dB. This is from the ITT handbook, pages 10-4
through 10-8. The design values for such an L-Pad would give the 5.7
dB loss based upon the following: Let P1 be the power delivered to a
matched load (50 ohms) from the 50 ohm source. Let P2 be the power
delivered to a 75 ohm load when fed through the 75 to 50 resistive
matching network. Then P1 / P2 will be approximately 3.715.
10 x Log(3.7) = 5.7 dB. You can work that out from the design values
for the L-Pad, which are 43 ohms in the series arm and 86.7 ohms in
the shunt arm.

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

PDRUNEN wrote:
Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

It appears that you forgot to put the resistive matching network
into the circuit. There must be a series resistor in there between
the source impedance and the load impedance to obtain that 5.7 dB
of isolation.
--
73, Cecil http://www.qsl.net/w5dxp


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