So, I've read a bit on the interweb about input impedance on amplifiers, preamplifers, etc. but I'm still a bit hazy on it. Can anyone give a concise explanation?
I see that people change the value of the pull up resistor, does this have anything to do with impedance matching between the devices?
How is the pull up resistor chosen on preamp / amp schematics and what are the implications of the choice?
Thanks!
I see that people change the value of the pull up resistor, does this have anything to do with impedance matching between the devices?
How is the pull up resistor chosen on preamp / amp schematics and what are the implications of the choice?
Thanks!
I don't know what circuit you are refering to using pull up resistors, however the input impedance e.g. of a power amp is the load that the output stage of the preamp sees.
Compare it to a power amp output stage and a speaker unit; the speaker is the load (input impedance) seen from the power amp output stage....
The input impedance of e.g. a power amp needs to be higher than the output impedance of the output stage of the preamp connected to it, in order for the preamp to be able deliver the Current needed to drive the power amp input stage.
A common input impedance on many commercial amps are from 47K to 100K Ohm, while the output impedance from preamps are in the order of anything from 2 to several 100s Ohms.
When choosing an input impedance, it's important that the Current in the input stage is sufficiant to drive the next stage of the amplifier.
Was it this type of answer you where looking for ??
Compare it to a power amp output stage and a speaker unit; the speaker is the load (input impedance) seen from the power amp output stage....
The input impedance of e.g. a power amp needs to be higher than the output impedance of the output stage of the preamp connected to it, in order for the preamp to be able deliver the Current needed to drive the power amp input stage.
A common input impedance on many commercial amps are from 47K to 100K Ohm, while the output impedance from preamps are in the order of anything from 2 to several 100s Ohms.
When choosing an input impedance, it's important that the Current in the input stage is sufficiant to drive the next stage of the amplifier.
Was it this type of answer you where looking for ??
When you have two equal impedances in series, there will be equal voltage across both of them. If the other impedance is much greater, most of the total voltage will be across that larger one.
That's the reason why amplifiers need to have very small output impedance (to be able to deliver maximum voltage across load - speaker) and large input impedance (so that most of the voltage signal would be across the input impedance and not the sources impedance).
In my language this fact is known as "voltage divider theorem".
That's the reason why amplifiers need to have very small output impedance (to be able to deliver maximum voltage across load - speaker) and large input impedance (so that most of the voltage signal would be across the input impedance and not the sources impedance).
In my language this fact is known as "voltage divider theorem".
excellent replys! The circuit I'm referring to is when you have an input, let's say from a CD player to a circuit. Next, there is a attenuator often. Then after this, there is a resistor that goes to ground. This is what I was referring to as a pull up (down?) resistor, perhaps incorrectly. I was wondering what the function of this resistor is and how its value is determined. Here's an example:
http://www.ibiblio.org/tkan/audio/dynahi/dynahi_amp_sch.png
note the 1M resistor to ground after the input.
Thanks all!
http://www.ibiblio.org/tkan/audio/dynahi/dynahi_amp_sch.png
note the 1M resistor to ground after the input.
Thanks all!
I've understood that pull-up resistor means the following: Using large resistor between positive voltage and some floating point (for example) causes virtually no current to flow through that large resistor. If no current flows between two points, then those points will be at the same potential. With that large resistor you can "pull" the floating point to the positive voltage.
In this case, that large resistor defines the input resistance of that amplifier. I presume you know the difference between resistance and impedance. So now that amplifier has input resistance of 1 M-ohm. To be exact, the real input impedance would be 1 M in parallel with the rest of the impedance between those fets and ground. And if you have two different impedances in parallel, the total impedance turns out to be quite close to the value of the smaller impedance.
This is why I wouldn't call that 1 Meg a "pull-down"-resistor, since the main purpose of it is not to pull the JFET drains to ground potential (it will essentially do that although). Instead, it makes almost all of the input voltage (signal) to appear across it and so the input signal will have greatest effect on the circuit (that is, the amplification will be largest). That attenuation resistor you mentioned means that a part of the input signal will stay across that attenuating resistor without having effect on the ampkifier circuit.
Hope you can get something out of this messy stuff
This explanation may include mistakes, so remember to read critically
In this case, that large resistor defines the input resistance of that amplifier. I presume you know the difference between resistance and impedance. So now that amplifier has input resistance of 1 M-ohm. To be exact, the real input impedance would be 1 M in parallel with the rest of the impedance between those fets and ground. And if you have two different impedances in parallel, the total impedance turns out to be quite close to the value of the smaller impedance.
This is why I wouldn't call that 1 Meg a "pull-down"-resistor, since the main purpose of it is not to pull the JFET drains to ground potential (it will essentially do that although). Instead, it makes almost all of the input voltage (signal) to appear across it and so the input signal will have greatest effect on the circuit (that is, the amplification will be largest). That attenuation resistor you mentioned means that a part of the input signal will stay across that attenuating resistor without having effect on the ampkifier circuit.
Hope you can get something out of this messy stuff
This explanation may include mistakes, so remember to read critically
Bootstrapper said:
thank you! yes, this helps. So, what are the effects of changing this 1M resistor, to say 100K and why would one do this? Doesn't this value depend on the input impedance of the input transistors? So, if you changed the input transistors, you might have to alter this resistor in this case as well?
Hi,
that 1M0 sets Zin. It also gives a route for the input offset current to go to ground.
The maximum source impedance that can connect to this 1M0 is about 200k but would be better if it were <50k.
Stray capacitance on the line between Rs and Zin will severely curtail the high frequency response using these very high values.
The pot used as a volume control has a maximum source impedance of 50k/4 + 475r <=13k.
This will allow the amp Zin to be set to >=65k.
I would recommend that the 1M0 be changed to 100k.
I would also recommend that an RF filter be added to the input terminals. 1k0 in series to the pot and 680pF to ground after the resistor.
The route between the pot and the amplfier input must be kept VERY short to minimise the stray capacitance on the input.
that 1M0 sets Zin. It also gives a route for the input offset current to go to ground.
The maximum source impedance that can connect to this 1M0 is about 200k but would be better if it were <50k.
Stray capacitance on the line between Rs and Zin will severely curtail the high frequency response using these very high values.
The pot used as a volume control has a maximum source impedance of 50k/4 + 475r <=13k.
This will allow the amp Zin to be set to >=65k.
I would recommend that the 1M0 be changed to 100k.
I would also recommend that an RF filter be added to the input terminals. 1k0 in series to the pot and 680pF to ground after the resistor.
The route between the pot and the amplfier input must be kept VERY short to minimise the stray capacitance on the input.
AndrewT said:
AndrewT:
Thanks again for an excellent reply. This makes sense to me know. I have changed the 1M0 to 100K and trust me, the input is VERY short, probably 1" or so. However, I'm starting to second guess one thing. I have been planning on using a Twisted Pear Joshua Tree relay attenuator, and Russ White just posted that the input impedance ranges from approximately 2.2K to 10K and the output impedance is 750R:
http://www.diyaudio.com/forums/showthread.php?s=&postid=1341651#post1341651
Russ mentions the possible need for a buffer. So, perhaps my understanding is a bit lacking, but I should be okay, right?
As for the RF filter, I could stick it directly on the XLR jack p2p style. Is it okay if it comes before the attenuator?
With an output impedance of 750r, you can use Zin between 3k9 and 15k. I would go for the higher value. You can go higher still and retain the 100k but you would have the option to bring it down substantially, if you think noise or output offset is becoming a problem.luvdunhill said:
I don't think he meant a buffer on the output, 750r is considered lowish for output impedance although low tends to be thought of as between 10r and 200r. I suspect he may have been referring to an input buffer since 2k2 is a pretty severe load for many source components. They are liable to run out of current ability for high voltage outputs. If any of your source components cannot easily feed 2k2 with a signal of +10db above the maximum you are likely to need, then an input buffer will probably be required. Some builders would aim for near 20db of headroom for high level transient signals through the preamp/attenuator. i.e. if an amplifier needs 1Vrms=1Vac=1.4Vpk to go to maximum output power then the pre-amp/buffer should allow about 14Vpk into Zin, or about 7mApk for a source feeding the Twisted Pear.luvdunhill said:
Remember that the RF filter is a R+C combination. If it is located at the input socket then it must take account of the Rs of the source and any added resistance you put into the signal line. Then you can calculate (or listen for an effect) the appropriate filtering capacitor.luvdunhill said:
If you are using XLR with balanced impedances then a filter is required on each pole and these filters must be exactly matched to maintain that impedance balance, otherwise you throw away your interference rejection capability.
Impadance doesn't matter sp much these days...
Back in the early days of pro audio, most equipment used transformers. Even power amps did. The most efficient power transfer occurrs when impedances are matched. This is why older transformer output amplifiers had different output taps. That way, you could match your 8 ohm speaker to the 8 ohm output tap, etc.
Most equipment also used 600 ohm inputs and outputs, transformer balanced. In other words, the input and output impedances were "matched".
Today however, most audio equipment uses the "voltage distribution" system, where output impedances are very low (in the case of preamps, 100 ohms or less) and load impedances are very high (5-100K ohms) Basically, this means that the load is "bridged" across the source (as opposed to being matched to it). Much of this comes from the fact that the average opamp likes to see a 2000 ohm load or higher.
Solid state power amps work this way as well-typical output impedances are a fraction of an ohm, and they are bridged by the speaker load (8 ohms nominally).
Finally, most balanced inputs and outputs these days are done with amplifiers running differentially. In outputs, this means that each output is 180 degrees out of phase with the other. In the case of inputs, it means that the balanced input responds to the difference between the two signals on each input wire.
Back in the early days of pro audio, most equipment used transformers. Even power amps did. The most efficient power transfer occurrs when impedances are matched. This is why older transformer output amplifiers had different output taps. That way, you could match your 8 ohm speaker to the 8 ohm output tap, etc.
Most equipment also used 600 ohm inputs and outputs, transformer balanced. In other words, the input and output impedances were "matched".
Today however, most audio equipment uses the "voltage distribution" system, where output impedances are very low (in the case of preamps, 100 ohms or less) and load impedances are very high (5-100K ohms) Basically, this means that the load is "bridged" across the source (as opposed to being matched to it). Much of this comes from the fact that the average opamp likes to see a 2000 ohm load or higher.
Solid state power amps work this way as well-typical output impedances are a fraction of an ohm, and they are bridged by the speaker load (8 ohms nominally).
Finally, most balanced inputs and outputs these days are done with amplifiers running differentially. In outputs, this means that each output is 180 degrees out of phase with the other. In the case of inputs, it means that the balanced input responds to the difference between the two signals on each input wire.
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