Voltage divider/attenuator

I have a 10V audio source that I want to bring it to line level 1V. That’s pretty easy: I just need to use a voltage attenuator 10:1 like image below. I used resistor values of 1k ohm (R1) and 100 ohm (R2) and it worked great!

@batata004

It’s all depends on the resistance / impedance of the "Source" and the "Load".

As a rule of thumb, input impedance of the load should be ten times greater than the source output impedance. So the load will not load down the source, or else frequency response will suffer.

In your case, I assume your 10V audio source is speaker level source, and its output impedance usually is less than 1Ω. The load for this source is R1+R2 (to simplify, we ignores the connected output device here) , which is 1000Ω + 100Ω = 1100Ω. It’s 1100 times greater than the source impedance and will not caused any problem.

Now, on the output side, R2 (100Ω) becomes the source impedance, the device that connected to it is the load, as long as this device has an input impedance more than 1000Ω (10 times greater than the source impedance) and it should be fine.

BUT I have a question: what if I had used 100k and 10k ohm resistors? The ratio is still 10:1 but I think it would not produce a good result (I have no idea why I think so).

Same as above, on the output side, as long as the connected device has an input impedance more than 100kΩ (10 times greater than R2), it should works as well.

If so... should I use 100 ohm (R1) and 10 ohm (R2) resistors instead of 1k ohm and 100 ohm that I am currently using? Would it be a good idea?

If the source is an amplifier speaker output with less than 1Ω output impedance, it should drive the load of 100Ω (R1) and 10Ω (R2) resistors without problem.

However, unless the connected device on the output side is about 100Ω, it is not advisable to use such a low value resistance R2.

Why? Because it waste power!

P = V² / R

In my case, if I use a potentiometer of 1k with a 100ohm (R2), what would be the resulting load when the potentiometer is at 1k ohm and when the potentiometer is at 0 ohm? I think the load value will be exactly the potentiometer value, right? So I should add a 100ohm resistor in series with it to make sure I never get below 100 ohm?

No, the load value = R1+R2. It is better to use a fix value R1(1kΩ) and a 1kΩ variable resistor as R2. So the source will see a load of a Max. value of 2kΩ to Min. of 1kΩ (variable resistor is at 0Ω).

It's important to consider the total, final impedance. 

You can use a 9:1 resistor ratio, but the total should be higher than the minimum recommended output impedance of the source.  Consider 9 Ohms and 1 Ohms.  That's a total R of 10 Ohms.

With a 10V source that's 1 Amp that you are going to waste in just the voltage divider.  The final circuit impedance will include whatever is connected across R2. If  your load (not shown) is 10x R2 no problem. :)

@rodman99999  thanks a lot!!! You left me with a question: I took a look at the specification of line output, and it says it should produce a load between 100 and 600 ohm. In my case, if I use a potentiometer of 1k with a 100ohm (R2), what would be the resulting load when the potentiometer is at 1k ohm and when the potentiometer is at 0 ohm? I think the load value will be exactly the potentiometer value, right? So I should add a 100ohm resistor in series with it to make sure I never get below 100 ohm?

     Regardless of the examples you’ve mentioned; with 10V in, you’ll end up with .909V out.

     Seems to me: the use of 50 (R1) and 12 (R2) Ohm resistors, giving a 1.94V output, would yield better results (sonically), 1V being kind of wimpy, far as line level.

     Multiplying those two values by 10, 100, 1000, etc, will still yield 1.94V.    Will that affect sonics?    That’s why we experiment.

     Use either wire-wounds or metal films, for the least coloration.    ie: Vishay makes some very nice ones