Do anyone know the output impedance of the Philips CD880, or can they advise how I can measure this myself with a multimeter.
Ian
Ian
measuring impedance
You can measure the impedance- but it will require a few extra things. You can do this by modeling the output stage with a Thevenin equivalent circuit.
First, you need a CD with test-tones of different frequencies on it. To measure the impedance at a given frequency follow these steps:
1. Using an oscilloscope, accurately measure the voltage output at an RCA jack while a test frequency is played. Record this voltage as Vs.
2. Now, get several resistors that are approximately the estimated output impedance (if you don't know, just get random resistors). Be sure to measure the resistances with an ohm-meter even if you know their values- you need to get a precise reading- record the value of the resistor you use as R.
3. Connect the resistor from the signal wire of the audio output to the ground. Play the same audio frequency as you used earlier.
4. Connect the oscilloscope across the resistor and measure the output voltage. Record this voltage as VL. (If VL is not much different (i.e. about 1/2) of Vs, then try a larger resistor.
5. Using ohm's law, the impedance and the resistor form a voltage divider. To find the output impedance, take the numbers you found earlier and plug them into this equation:
Zout = [ R * (Vs-VL) ] / VL
I attached a picture showing the the Thevenin model.
Good luck.
Brian
You can measure the impedance- but it will require a few extra things. You can do this by modeling the output stage with a Thevenin equivalent circuit.
First, you need a CD with test-tones of different frequencies on it. To measure the impedance at a given frequency follow these steps:
1. Using an oscilloscope, accurately measure the voltage output at an RCA jack while a test frequency is played. Record this voltage as Vs.
2. Now, get several resistors that are approximately the estimated output impedance (if you don't know, just get random resistors). Be sure to measure the resistances with an ohm-meter even if you know their values- you need to get a precise reading- record the value of the resistor you use as R.
3. Connect the resistor from the signal wire of the audio output to the ground. Play the same audio frequency as you used earlier.
4. Connect the oscilloscope across the resistor and measure the output voltage. Record this voltage as VL. (If VL is not much different (i.e. about 1/2) of Vs, then try a larger resistor.
5. Using ohm's law, the impedance and the resistor form a voltage divider. To find the output impedance, take the numbers you found earlier and plug them into this equation:
Zout = [ R * (Vs-VL) ] / VL
I attached a picture showing the the Thevenin model.
Good luck.
Brian
Attachments
May I ask why you wonder this? Modern gear can drive at least 10 kohm without any problem and I think even down to 2-1 kohms is also possible.ijhill said:
output impedance
The target for the NE5534 is to drive with ease 600 Ohms loads!
It is still one of my favorits ,besides the fashions!!
Regards
Jorge Santos
The target for the NE5534 is to drive with ease 600 Ohms loads!
It is still one of my favorits ,besides the fashions!!
Regards
Jorge Santos
Yes, Jorge,
But, the lower is the load, the higher is the distortion.
10 kohms is a good value for domestic hi-fi standard purposes.
Beware of loading such device with a transformer (passive preamp) , distortion can occurs at low frequencies because of the insufficient inductance of the transformer.
Regards, Pierre Lacombe.
But, the lower is the load, the higher is the distortion.
10 kohms is a good value for domestic hi-fi standard purposes.
Beware of loading such device with a transformer (passive preamp) , distortion can occurs at low frequencies because of the insufficient inductance of the transformer.
Regards, Pierre Lacombe.
Re: output impedance
The target for the NE5534 is to drive with ease 600 Ohms loads!
It is still one of my favorits ,besides the fashions!!
Jorge Santos [/QUOTE]
---------------------------------
The 5534 is the designer/pros favourite. Low noise and cheap. But l;ook at the circuitry, comples topography with lots of active components. It is my least favourites except for instruments. The thing sound precise and impressive but is hard and mechanical-the so called pro sound. Nothing to do with fashion exfcept for its use in pro gear and Japanese products.
The target for the NE5534 is to drive with ease 600 Ohms loads!
It is still one of my favorits ,besides the fashions!!
Jorge Santos [/QUOTE]
---------------------------------
The 5534 is the designer/pros favourite. Low noise and cheap. But l;ook at the circuitry, comples topography with lots of active components. It is my least favourites except for instruments. The thing sound precise and impressive but is hard and mechanical-the so called pro sound. Nothing to do with fashion exfcept for its use in pro gear and Japanese products.
My reason for asking is I am attempting to make some attenuating interconnects for my hi-fi system.
Using the following PI attenuator circuit
http://www.qsl.net/4f5aww/module5p.htm
I have using a reduction from my CD's 2V to 150mV giving a 11dB required reduction and the impedance of my Rotel RC-850 pre-amp of 18kOhms. I have established I require resitors of value R1=42K8 and R1=40K2 (nearest available).
This should give the required attenuator, but I have assumed that the required impedance is the impedance at the pre-amp and that the CD will be able to drive this (it should as it does normally)
Is this correct or should I look at a different method of attenuation.
Ian
Using the following PI attenuator circuit
http://www.qsl.net/4f5aww/module5p.htm
I have using a reduction from my CD's 2V to 150mV giving a 11dB required reduction and the impedance of my Rotel RC-850 pre-amp of 18kOhms. I have established I require resitors of value R1=42K8 and R1=40K2 (nearest available).
This should give the required attenuator, but I have assumed that the required impedance is the impedance at the pre-amp and that the CD will be able to drive this (it should as it does normally)
Is this correct or should I look at a different method of attenuation.
Ian
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