Are there any issues involved in putting (quality) attenuators in the line out of a phono pre amp ?
I ask this because my phono pre amp output has a bit too high of an output level .
I have also read conflicting opinions about if it will effect high frequencies.
I tend to believe it does not as it is no different than an amp's gain control .
I ask this because my phono pre amp output has a bit too high of an output level .
I have also read conflicting opinions about if it will effect high frequencies.
I tend to believe it does not as it is no different than an amp's gain control .
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A pair of resistors forming a ladder is all you need. Say 10k and 10k to ground, that will give you a 3dB loss, halving the signal and halving any noise.
No frequency degrading will happen.
No frequency degrading will happen.
Jons suggestion is very workable in the real world.
If you have a circuit for the preamp then we might be able to suggest a simple component change to lower the gain of the amp. Doing that would preserve the (presumably) low output impedance of the unit.
If you have a circuit for the preamp then we might be able to suggest a simple component change to lower the gain of the amp. Doing that would preserve the (presumably) low output impedance of the unit.
Are not all inline (RCA) attenuators just resistors? .
I prefer not modify any components if a simple inline attenuator will do.
I prefer not modify any components if a simple inline attenuator will do.
Yes, but they are a network of two resistors configured as a divider and not just a single resistor in series.
Two resistors of same size will do 6dB, not 3 (half the output).
Degradation of the highest frequencies may happen if the load
capacitance is high at the output.
Yes,an "inline attenuator" will do, better connect this at the line input.
Don't understand the bit too high outputlevel issue...
What do you need attenuate for? 🙄
Yes - you don't want to raise the output impedance too high or you'll create HF roll-off in conjunction with the cable capacitance as well as increase your noise susceptibility. You don't want to go too low or dynamics of the source will start to suffer.
Better to use a step down transformer.
Such is life on a forum 🙂
Are you happy with the answer/s or are you still not sure how to best proceed ?
Perhaps a simple line drawing of jonsnells' suggestion?
For those of us who know little to nothing.
For those of us who know little to nothing.
Not a bad idea. It also helps others researching this kind of thing.
Here we have R1 and R2 forming the attenuator. The output is 50% of the applied input voltage as we are using equal values for resistive components of the attenuator.
Cx is the unknown cable capacitance and input capacitance of whatever we connect the attenuator to. Typically this could be around 100pF to a worst case of perhaps 470pF or even 1nF.
So here we can see the effect of 100pf loading followed by 1nf (1000pf). Even here, the response is only 1.5db down at 20kHz.
The output impedance is 5k. In other words, if we load the output by 5k then our output voltage now becomes one quarter of the applied input voltage (-12db)
On a practical note... guessing the attenuation factor needed is impossible. Far better to initially make the lower resistor variable and get the level you are looking for. Then take the variable pot out and measure it and fit a fixed resistor in its place.
Here we have R1 and R2 forming the attenuator. The output is 50% of the applied input voltage as we are using equal values for resistive components of the attenuator.
Cx is the unknown cable capacitance and input capacitance of whatever we connect the attenuator to. Typically this could be around 100pF to a worst case of perhaps 470pF or even 1nF.
So here we can see the effect of 100pf loading followed by 1nf (1000pf). Even here, the response is only 1.5db down at 20kHz.
The output impedance is 5k. In other words, if we load the output by 5k then our output voltage now becomes one quarter of the applied input voltage (-12db)
On a practical note... guessing the attenuation factor needed is impossible. Far better to initially make the lower resistor variable and get the level you are looking for. Then take the variable pot out and measure it and fit a fixed resistor in its place.
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