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#1 |
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diyAudio Member
Join Date: Jul 2006
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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! |
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#2 |
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diyAudio Member
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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 ??
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#3 |
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diyAudio Member
Join Date: Sep 2007
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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".
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-Juha |
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#4 |
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diyAudio Member
Join Date: Jul 2006
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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/dy...hi_amp_sch.png note the 1M resistor to ground after the input. Thanks all! |
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#5 |
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diyAudio Member
Join Date: Sep 2007
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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
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-Juha |
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#6 | |
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diyAudio Member
Join Date: Jul 2006
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Quote:
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#7 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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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.
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regards Andrew T. |
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#8 | |
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diyAudio Member
Join Date: Jul 2006
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Quote:
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/showt...51#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? |
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#9 | |||
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Quote:
Quote:
Quote:
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.
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regards Andrew T. |
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#10 | |
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diyAudio Member
Join Date: Jul 2006
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Quote:
AndrewT: I forgot about this thread Thanks! |
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