# Understanding resistors in and around a signal path in music amplifier

#### beginnersluke

Joined Aug 26, 2015
35
Hi all,

I'm trying to understand and learn some things dealing with musical instrument amps.

This started with a discussion about converting a headphone out of an old electric piano to a line out level. I found this schematic online, which raised some questions. (See first attachment.)

1. The 100 ohm resistor between signal and ground:

100 ohms is not that much resistance really. How does this not just basically short things out (the amp of the mp3 player in this case)?

Secondly, what does it do exactly? (As in, if you were designing a device to lower a headphone signal to a weaker line level, why would you include this?)

2. What's the effective difference between the parallel (signal to ground) and series (in the signal path resistor)?

I feel like I could explain how each of these lowers the strength of the signal that results from this circuit.

a. The parallel resistor "sinks" most, but not all of the signal to ground, so what's left is a weaker signal.
b. The series resistor turns some of the signal's energy into heat energy, so what's left is a bigger signal.

Are either of these right? Which is the correct approach (or in what cases is one approach the correct one, and in what cases is the other the correct one)?

I hope it's somewhat clear what I'm trying to understand, and I appreciate any help. To some extent, I realize I don't know enough to ask the right questions.

I also have some questions about the second attachment. This is the schematic to an amplifier of an electric piano.

I've circled two resistors in blue. Please let me know if I've understood the functions of these correctly (and what I've gotten wrong).

1. The one on the left is a resistor/capacitor combination. It filters out frequencies above a certain threshold. (Looking at an online calculator, it looks like it filters out frequencies above around 3,500 Hz. Is it correct to say that even though 12 ohms is not much resistance, it doesn't short the amp because only part of the signal (AC above 3500 Hz) is actually getting to ground?

2. The one on the right serves to lower the strength of the signal. It simply sends much of the signal to ground, and what's left over goes to the headphones.

Thanks so much!

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#### Ylli

Joined Nov 13, 2015
933
First circuit: The 100 ohm resistor is there to replace the load normally provided by the headphone. The amplifier driving this point may misbehave if it does not see the proper load. The 680 ohm series resistor is simply for isolation, and prevents an inadvertent short to the amplifier output when using the 'line input' port. Line input impedance is normally around 47K ohms, so the 680 ohm does not introduce much attenuation.

Second drawing. Somewhat similar in that the amplifier off to the left expects to see an 8 ohm load. When you plug in headphones, the speakers are disconnected and the 8 ohm resistor takes their place. The headphones themselves are usually a high impedance and do not load the output. The series RC is called a "Zobel Network". Its purpose is to improve the amplifier stability and prevent it from going into oscillation with an inductive load. You will almost always see it on the output of an amplifier that uses emitter follower outputs.

#### beginnersluke

Joined Aug 26, 2015
35
First circuit: The 100 ohm resistor is there to replace the load normally provided by the headphone.
So would it be correct to say that this is done to match the impedance the output is looking for?

So headphones may have an impedance of around 100 (I know they vary in practice), and because a line-in provides a lower impedance, this makes it so that the amplifier sees the impedance it's expecting.

The 680 ohm series resistor is simply for isolation, and prevents an inadvertent short to the amplifier output when using the 'line input' port. Line input impedance is normally around 47K ohms, so the 680 ohm does not introduce much attenuation.
Can you say more about how it prevents an inadvertent short? I don't really understand. Sorry and thanks.

Second drawing. Somewhat similar in that the amplifier off to the left expects to see an 8 ohm load. When you plug in headphones, the speakers are disconnected and the 8 ohm resistor takes their place. The headphones themselves are usually a high impedance and do not load the output.
I'm going to do some reading on this, because I'm just not that clear on the function of both the load and the imprudence in these cases.

Anyway, sorry I'm just trying to learn and understand, and I quite obviously don't grasp it yet.

Thanks!

#### beginnersluke

Joined Aug 26, 2015
35
The 100 ohm resistor is there to replace the load normally provided by the headphone.
The headphones themselves are usually a high impedance and do not load the output.
This also has me a bit confused. In what cases do the headphones provide a load, and when do they not?

Last edited:

#### Ylli

Joined Nov 13, 2015
933
The impedance of headphones can be all over the board, but in general they are a higher impedance that speakers. Maybe 30 ohms or higher. Without more specific information on the application, my best assumption is that the 100 ohm resistor is to provide a load the the preceding amplifier that is representative of a headphone load.

If the 680 resistor was not there and the voltage across the 100 ohm resistor was feed directly out, to perhaps an RCA plug. It would not be hard to accidentally short the center pin of that RCA connector to ground when plugging it in. The 680 ohm resistor prevents that short circuit from being seen at the driving amplifier and possibly causing damage. And since the input impedance of a line port is typically 47k, the 680 ohm resistor does not cause significant loss.

As for the higher power amplifier with the Zobel Network.... Headphones provide a load, but a higher impedance load than loudspeakers. As I said above, they can be almost anything but normally 30+ ohms. Switching in the 8 ohm load resistor when plugging in headphones keeps the impedance the amplifier is driving closer to what it expects.

#### beginnersluke

Joined Aug 26, 2015
35
So the first circuit does not really do much to attenuate the signal. It's just there to help the amp see the load/impedance it was built for, correct?

Thanks!

#### Ylli

Joined Nov 13, 2015
933
So the first circuit does not really do much to attenuate the signal. It's just there to help the amp see the load/impedance it was built for, correct?

Thanks!
Pretty much.

#### MisterBill2

Joined Jan 23, 2018
5,741
It looks like that circuit with the 2 resistors is drawn backwards. The source of the signal should be connected to the 680 ohm resistor and the earphone connected across the 100 ohm resistor. The purpose is to reduce the signal fed to the earphones, the added benefit is that the 680 ohm resistor would indeed protect the driving circuitry from damage by a short circuit. The circuit of the "Wurly" device is sort of similar, except that the series resistor is connected between the sleeve of the earphone jack and ground, while the 8 ohm resistor is switched into the circuit to replace the speakers which are disconnected when the phones are plugged in. That connection will not work properly with an external amplifier because it has no actual "common" available. If the external amplifier "ground" is somehow commoned with the "Wurly" ground the 470 ohm resistor will be bypassed.