# Bridge T attenuator equations

Discussion in 'Homework Help' started by Yuf, Sep 19, 2014.

1. ### Yuf Thread Starter New Member

Sep 11, 2014
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This particular attenuator is complicated and I could only get 1 equation via node voltage method, vo/vi in terms of R6, R7, Z. I need another equation of the resistors to derive to the equations but I could not get the right one.

Can anyone give me a tip of how to get the relationship of the 3 resistors? so the following derivations can be derived.

2. ### MrAl Distinguished Member

Jun 17, 2014
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791
Hi,

I have to ask where did you get this from?
I ask because it looks like a couple things are not right.

First thing, if you use the formulas for K and R6 and R7 you do not get the resistor values in the table, at least not for the 3db row.

Second, using those values do not result in the desired response unless we assume that there is either one of these two:
a. The source has zero impedance, or
b. The load has infinite impedance.
Since either of these contradicts the claim that "Z=source/load impedance (resistive)" this is the second problem.

Perhaps you can elaborate on where this problem came from or explain why these things are so.

In the usual circuit like this the source would have 50 ohms in series with it and the load would be 50 ohms to ground.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Mr Al,

When I apply the formulae shown I get the values shown for R6 & R7 on the 3dB line.

With an AC source having 50 ohms internal resistance and a 50 ohm load at the network output I also obtain 3dB loss from network input to output.

4. ### MrAl Distinguished Member

Jun 17, 2014
3,733
791
Hello again,

Ok, the reason they have those resistors shown in the 3db row is because they and yourself used exactly -3db. Usually when we say 3db attenuation we really mean -3.0103 db not -3.0000 db, but that's not that important as the resistor values come out almost the same anyway, so i'd be happy to go with the exact -3.0000 db instead and come out with the same value resistors they did. No worries there

So with a 1v test input generator voltage what do you get as the output voltage with the "3db" resistor values?
Remember we need 50 ohms of input series resistance and a 50 ohm load to ground too.
Free free to elaborate

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The network introduces an additional 3dB loss at the load when you compare the situation without the network. Without the network a 1V source would give 0.5V across the 50 ohm load. With the network in place there would be a load voltage of 0.3536V.

6. ### MrAl Distinguished Member

Jun 17, 2014
3,733
791
Hello again,

Yes, that's what seems strange here.

With the 50 ohm source series resistance and 50 ohm load, with the generator internal source set Vi=1v we get 0.35355v or approx 0.3536v as you said, but then look:
Vi/Vo=1/0.35355=2.828

Now i also see that they use Vi in the table but they use VI in the formula for K. If Vi is the open circuit voltage of the input source, then Vi is not equal to VI, but if i=I then it is.

So this is starting to look like it could have been written a lot better.

Also, when we do attenuators we dont have to assume that we get an additional cut as compared to the original cut, we can also assume that we have a signal and we want to see a certain cut, period.

The two different interpretations can lead to a problem as the signal may be too low for the application.

A quick formula for the output voltage is:
Vout=(Vin*(Z^2+2*R6*Z+R6*R7))/(4*(Z+R6)*(Z+R7))

where Vin is the open circuit source voltage.

Apr 26, 2005
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8. ### Yuf Thread Starter New Member

Sep 11, 2014
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this diagram is from "allaboutcircuits" online textbook semiconductor section. page 22. The resistor Z is equal to source/load impedance.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It would seem logical to me that one would view a matched attenuator in terms if its insertion loss.
In the old days when I worked in the telecommunications industry, we worked on 600 ohm systems in which each matched in-line device either had an insertion gain (amplifier) or an insertion loss. It was often difficult to define the actual originating signal source in such networks.
Usually one had a specific point in the signal path where there was a known signal level as a reference dB(x) level.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Re the OP's original question.
Replace R7 with a resistance n*Z, where n is any positive real value.
Analyze the circuit to find what value of R6 will ensure the input resistance is the matched impedance Z. Output side must be loaded with Z to obtain correct input resistance of Z.
After a lot of tedious algebra it is revealed that R6=Z/n.
The analysis is helped by use of the delta-wye transformation.
It is then a relatively simple matter to show for the value of K that K=n+1. Once repeated use of the voltage divider relationship helps with the latter.

Last edited: Sep 20, 2014
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