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Old 16th April 2009, 08:54 AM   #1
mudihan is offline mudihan  United States
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Default Op amps: output impedance and load resistance

If the output stage of a CD player/pre-amp is a simple non-inverting op amp set up in the following ways:

1) Op amp output => 100 ohm series resistor => RCA jack

2) Op amp output => 10k ohm parallel resistor to the ground (not part of the loop) => RCA jack

If the op amp has zero output impedance (being perfect!):

What's the output impedance of the CD player/Pre-amp and what's the load resistance the op amp sees in set-up 1?

What's the output impedance of the CD player/Pre-amp in set-up 2?
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Old 16th April 2009, 09:03 AM   #2
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0 + 100 = 100
0 || 10k = 0
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Old 16th April 2009, 09:22 AM   #3
mudihan is offline mudihan  United States
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Quote:
Originally posted by pacificblue
0 + 100 = 100
0 || 10k = 0

Thanks for answering the output impedance questions!

If the next stage has a 10k parallel resistor to the ground (an input resistor), would 10K be the load resistance the op amp sees in set-up 1 and 5k in set-up 2?
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Old 16th April 2009, 10:59 AM   #4
AndrewT is online now AndrewT  Scotland
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the opamp sees the two 10k in parallel, so yes you are right, 5k is the load.

If you fit DC blocking caps at the output and input of your equipment then add a grounding resistor at the RCA of both the output and input.
This grounding resistor gives a load if the RCA is left open circuit.
It also takes any leakage through the blocking cap to ground. This helps stop pops and bangs if you ever connect/disconnect hot!

The grounding resistor can be any value from 100k to 2M2. Not 10k.
I prefer the higher end. I also prefer Zin~50k.
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Old 16th April 2009, 06:15 PM   #5
mudihan is offline mudihan  United States
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Quote:
Originally posted by AndrewT
the opamp sees the two 10k in parallel, so yes you are right, 5k is the load.

If you fit DC blocking caps at the output and input of your equipment then add a grounding resistor at the RCA of both the output and input.
This grounding resistor gives a load if the RCA is left open circuit.
It also takes any leakage through the blocking cap to ground. This helps stop pops and bangs if you ever connect/disconnect hot!

The grounding resistor can be any value from 100k to 2M2. Not 10k.
I prefer the higher end. I also prefer Zin~50k.

Thanks for the clear explanation. Yes, a grounding resistor in high values following the op amp output would enable me to use smaller caps (more choices and lower cost), which is very helpful.

Now, I would like to place a 24-step 10K ladder-type attenuator after the op amp output stage. According to the this calculator:
http://homepages.tcp.co.uk/~nroberts/atten.html
in the first 8 steps, the ground resistors (Ry) in the attenuator will have values less than 100 ohm. Even with high values grounding resistors elsewhere, those low value resistors in the attenuator will lead to very low load resistance for the op amp due to parallel resistance. Am I correct? If I am, why do people say that ladder-type attenuator provides constant load to the source at the specified value (10K attenuator = 10k load)?

Thanks!
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Old 16th April 2009, 06:24 PM   #6
mudihan is offline mudihan  United States
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And for the stage after the attenuator:

Let's say that it is a non-inverting LM1875/3875 amp with a 22x gain set by 22k/1k resistor and an 22k input grounding resistor. What would be the input resistance the amp would see? The 22k set by the input resistor, or the value calculated by paralleling all the grounding resistors in between the input of the LM chip and the output of the op amp?

Thanks!
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Old 17th April 2009, 08:35 AM   #7
AndrewT is online now AndrewT  Scotland
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you're last two posts show you are confused.
Draw a diagram of what you have and then look at the effective impedances/resistances.
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Old 17th April 2009, 09:13 AM   #8
mudihan is offline mudihan  United States
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Let's say:

Op amp output=>100k ohm parallel/grounding resistor => 9993 ohm series resistor Rx => 7 ohm parallel/grounding resistor Ry => 22K ohm input parallel/grounding resistor for LM1875 => the + input of LM1875

The 9993 ohm series resistor Rx and the 7 ohm grounding resistor Ry are taken from Step No. 4 (Attenuation = 63 dB, a random choice) of a 24-step 10k ladder type attenuator, calculated from here:

http://homepages.tcp.co.uk/~nroberts/atten.html

If the calculator for the ladder-type attenuator is correct, isn't it true that the 7ohm Ry would make the load resistance for the op amp around 6.99ohm (100k, 7, and 22k parallel), which is very low? The values of Ry also varies greatly throughout the 24 steps. Why do people say that a ladder-type attenuator gives a constant load to the source? And what is the input resistance LM1875 sees in this set up, 22k ohm or 6.99 ohm? Thanks!
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Old 17th April 2009, 09:28 AM   #9
AndrewT is online now AndrewT  Scotland
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I told you to draw yourself a diagram.

The 10k stepped attenuator has two legs the high resistance leg is in series with the signal line the low resistance leg goes to ground.

LOOK at it.
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Old 17th April 2009, 09:38 AM   #10
mudihan is offline mudihan  United States
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Quote:
Originally posted by AndrewT
I told you to draw yourself a diagram.

The 10k stepped attenuator has two legs the high resistance leg is in series with the signal line the low resistance leg goes to ground.

LOOK at it.

Yes, the low resistance one is the 7 ohm grounding resistor Ry I am talking about (each step in a ladder-type attenuator is constructed with one series resistor and one parallel resistor, am I right?) And this resistor is in parallel with the 100k at the op amp output and the 22k at LM1875's input. Is there something I am missing? I am really confused now.

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