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Resistive load of infinity

by Idyllic
Tags: infinity, load, resistive
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Idyllic
#1
Jun10-11, 05:12 AM
P: 14
Hello,

I have a question. How do you design a buffered band pass filter with a resistive load of infinity?

I have a feeling that a resistor with a resistance of infinity is an open circuit. Although how can this be implemented in a type of filter seen in the attachment?


Thank you.
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bufferedbandpassfilter.jpg  
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uart
#2
Jun10-11, 06:15 AM
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P: 2,751
The assumption of an infinite resistance load (also know as "no load") is purely to make the analysis of the circuit easier, as it means that the low pass time constant is unaffected by the load resistance. You can achieve this by simply placing a second buffer (voltage follower) between the output and the load.
sophiecentaur
#3
Jun10-11, 07:06 AM
Sci Advisor
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PF Gold
sophiecentaur's Avatar
P: 12,186
One point of 'buffering' a filter is to isolate the resistances and capactiances that constitute the filter from other elements in the circuit which may not be defined and which might affect the frequency response. A load resistance of 'infinity' is what a good buffer will provide (no current will be taken). There is nothing more to it than that, I think.

Carl Pugh
#4
Jun10-11, 11:47 AM
P: 384
Resistive load of infinity

Your circuit will not work without a load.
C2 will charge up to a voltage and stay at this voltage forever.
berkeman
#5
Jun10-11, 11:55 AM
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P: 41,109
Quote Quote by Carl Pugh View Post
Your circuit will not work without a load.
C2 will charge up to a voltage and stay at this voltage forever.
His drawing is not very clear, but I think that's a buffer amp in the middle, and not a diode. If it's an amp, it will work okay. If it were a diode, it would have the problem that you mention.
sophiecentaur
#6
Jun10-11, 12:25 PM
Sci Advisor
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PF Gold
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P: 12,186
The left hand bit is the high pass (working into a high impedance) and the right hand bit is the low pass (fed from a low impedance) the buffer (in between them) ensures that there are only two time constants involved. Without it, R1 and R2 would both have an effect on the charge rates of both C2 and C1.
Idyllic
#7
Jun10-11, 06:53 PM
P: 14
Yeh sorry about that it is a buffer. So that buffer makes have a resistive load of infinity? Theres nothing else to add?
Antiphon
#8
Jun10-11, 09:51 PM
P: 1,781
The circuit is fine. It works as is.

All it means is that the designed cutoff will change if you attach a load. Simply adjust R2 once you have a finite load impedance and all is well.
sophiecentaur
#9
Jun11-11, 09:40 AM
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PF Gold
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P: 12,186
Or follow with another buffer. Many packages have more than one amp in them, in any case.
uart
#10
Jun11-11, 10:39 AM
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P: 2,751
Quote Quote by Idyllic View Post
Yeh sorry about that it is a buffer. So that buffer makes have a resistive load of infinity?
No that is not correct. That is not what people have been telling you.

The thing (in your given circuit) that makes the load resistance infinity is the absence of a load resistance (connected in parallel with c2).

If you were to place a load resistor in that location then it would alter both the gain and the frequency response of your filter. The load resistance effectively becomes an integral part of the filter circuit and modifies it's behavior.

In such cases where there is a load resistance present (that is, load resistance is not infinity) then we have two choices.

1. If the load resistance is well know and consistent enough then we can design the filter around the load resistance, so that it achieves it's a specific gain and frequency response with the load resistance included as part of the filter specifications (effectively as an integral part of the filter itself).

2. We can use a second buffer connected at the output of the filter (after c2) to isolate the load from the filter.


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