Parallel LC filter

WBahn

Joined Mar 31, 2012
29,979
Can anyone explain that? I really get stuck.
Why we can ignore RL as it has a great impact in characteristic of the filter?
Who says you can ignore RL?

Any time you say that you can ignore something, you need to provide a justification for the claim (and I know it is exactly this justification that you are trying to get it, I'm just making a generic statement to establish a mental context).

That justification almost always ends up with qualifications attached, such as R5 can be ignored as long as R5 >> (R3||R2)

To identify these qualifiers, you start with the equation that counts -- the one that tells you the value of the quantity you are interested in -- and then manipulate it to see how sensitive it is to the parameter you are looking at.

In this case, the equation that counts is your transfer function, H(jw):

\(
H(j \omega ) \, = \, \frac{R_L}{R_S+R_L+R+j \omega R_L(R_S+R)C}
\)

What happens if we divide top and bottom by RL?

\(
H(j \omega ) \, = \, \frac{1}{ \( 1 + \frac{R_S+R}{R_L} \) +j \omega (R_S+R)C}
\)

Since RL now appears in only one place (the real part of the denominator), it becomes obvious what our qualification is. As long as RL is much larger than (Rs+R), the real term will be about 1.

The more formal way is to do a "sensitivity analysis" by taking the derivative of the transfer function with respect to RL and setting the magnitude of that less than some criterion Kmax. Then solve for RL and, in this case you would see that it has to be greater than (Rs+R), at least in the limit that Kmax gets arbitrarily small.
 

LvW

Joined Jun 13, 2013
1,752
Can anyone explain that? I really get stuck.
Why we can ignore RL as it has a great impact in characteristic of the filter?
If you think that it has "a great impact" it must not be neglected.
But WHY is that so?
The best method is to analyze the corresponding expression/formula and to verify if you can neglect any value or not.
Note: You must not neglect parts values only because the are small. You always have to verify: Small against what? (That means to compare two values which are summed or subtracted).
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
Well, that makes a lot of sense.

Now assuming that there is a case that RL is not much larger than Rs + R, then we have to use a buffer or something similar?
 

Jony130

Joined Feb 17, 2009
5,487
As long as RL is much larger than (Rs+R), the real term will be about 1.
Isn't this a basic rule for designs a circuit when the voltage is our signal? In real life we want RL to be at least 10 times larger than (Rs+R).
You did not know about it? You definitely need to read the art of electronics.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
Isn't this a basic rule for designs a circuit when the voltage is our signal? In real life we want RL to be at least 10 times larger than (Rs+R).
You did not know about it? You definitely need to read the art of electronics.
I read this but sometimes I forget!!!:D
 

WBahn

Joined Mar 31, 2012
29,979
Isn't this a basic rule for designs a circuit when the voltage is our signal? In real life we want RL to be at least 10 times larger than (Rs+R).
You did not know about it? You definitely need to read the art of electronics.
Who's the "you" you are referring to? :confused:
 

Jony130

Joined Feb 17, 2009
5,487
Who's the "you" you are referring to? :confused:
For sure not "you" I was referring to :). I was talking to anhnha.
I know that your (WBahn) knowledge is much greater than my. I'm not a EE, I'm just a hobbyist. Which thanks to his hobby works as a electrical maintenance technician in fridge factory.
 

WBahn

Joined Mar 31, 2012
29,979
For sure not "you" I was referring to :). I was talking to anhnha.
I know that your (WBahn) knowledge is much greater than my. I'm not a EE, I'm just a hobbyist. Which thanks to his hobby works as a electrical maintenance technician in fridge factory.
Oh, there are tons of things, including about electronics and electrics, that you know better than me. Be sure of it; I've learned more than a couple things from you.
 

LvW

Joined Jun 13, 2013
1,752
Isn't this a basic rule for designs a circuit when the voltage is our signal? In real life we want RL to be at least 10 times larger than (Rs+R).
You did not know about it? You definitely need to read the art of electronics.
Hi Joni130, may I ask you WHY in "real life we want RL to be at least 10 times larger than Rs+R" ?
Just because calculation is simplified? (I don`t know which page in Art of Electronics you are referring to, can you be specific?).

I can imagine some cases in which RL cannot or even shall not as large as possible. Don`t forget that RL also determines gain/attenuation and bandwidth.
 

Jony130

Joined Feb 17, 2009
5,487
Hi Joni130, may I ask you WHY in "real life we want RL to be at least 10 times larger than Rs+R" ?
Just because calculation is simplified? (I don`t know which page in Art of Electronics you are referring to, can you be specific?).
Yes, because calculation is simplified. Look at figure 1.12 and 2.7.
I almost always follow this rule when I design a audio frequency circuits.

I can imagine some cases in which RL cannot or even shall not as large as possible. Don`t forget that RL also determines gain/attenuation and bandwidth.
Can you give me an example?
 

JoeJester

Joined Apr 26, 2005
4,390
I'm 180 miles from my copy of the Art of Electronics. It would be nice when you cite a figure, you post a copy of it, for people like me and for people who don't own a copy of the book.
 

LvW

Joined Jun 13, 2013
1,752
Hi Jony,
you are lucky if you are able to follow this rule "almost always".
Of course, it is our obvious goal that any voltage transfer function should not (or only to a small extent) depend on load impedances - if this is possible!
In particular, this applies to voltage sources (as outlined in the referenced book, Fig. 1.12).

My only point was that it is not always possible to "follow this rule".
Think, for example, of a two-stage BJT amplifier. In most (if not in all) cases the finite input resistance of the load (second stage) must be considered for gain calculation.
In other cases - and also for the circuit in post #13 of this thread - the load resistor must not be neglected by calculating the bandwidth of a band pass output. Sometimes, this load is intentionally selected at medium values to allow a larger bandwidth. As you know, it is not always a high-Q circuit that is desired.
LvW
 

Jony130

Joined Feb 17, 2009
5,487

WBahn

Joined Mar 31, 2012
29,979
The exception that comes to my mind is RF in which you are often trying to transfer as much power from stage to stage as you can.
 
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