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#1 |
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diyAudio Member
Join Date: Jun 2007
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I am about to assemble a Leach amp (ver 4.5). The Leach amp appears to have an input Z of 20k.
For various reasons I will be using a 10k resistor pot before the amp (conventionally wired, ie, no shunt) rather than using a pre-amp. So I would like to raise the input Z of the Leach amp. There is plenty of gain, the source Z is low, & the cabling will be short. http://users.ece.gatech.edu/~mleach/lowtim/instage.html Is this as simple as changing R1 from 20k to perhaps 47k? Am I missing something? Thanks, -Tom |
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#2 |
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diyAudio Member
Join Date: Apr 2004
Location: Norwich, UK
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Really, the input impedance resistor should match the feedback resistor to avoid an imbalance, however in this circuit it might work fine. Change it, but if you get an offset voltage at the output, you may need to rethink this.
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#3 | |
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diyAudio Member
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Quote:
__________________
The devil is not so terrible as his mathematical model! Wavebourn: We Create Creativity! |
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#4 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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The Leach has AC coupling on the negative feedback path (NFB).
This presents 22k//capacitor as the impedance/resistance on the inverting input. The schematic shows 20k+2k0 as the impedance on the non-inverting input. If you add a DC blocking cap to the input then the input becomes AC coupled to match the NFB route. If you add a pot directly, then the input becomes DC coupled and with it variable source impedance and variable RF attenuation. Consider the implications carefully before deciding on whether to adopt mixed AC & DC coupling for your amplifier. Impedance matching is usually OK if the RS ~<10% of the Zin. The 10k pot presents 0ohm to 2500ohms to the input terminal. The Zin=20k & 10% is ~2k so the 10k pot is an acceptable source in terms of impedance matching. I prefer Rs<=5% of Zin, but I would NEVER use mixed AC & DC coupling.
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regards Andrew T. |
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#5 |
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diyAudio Member
Join Date: Jun 2007
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Andrew, Thanks for clarifying things. I had forgotten that the 10k pot is presenting an impedance of 1/4 (2.5k) .
I think I will probably be okay. I am assuming that if the source Z and the input Z are not sufficiently different (too low of a ratio) then the the amp would be trying to draw too much current form the source (in this case an opamp). I am further assuming this would result in distortion at the low end of the spectrum. we shall see.... Thanks everyone, -Tom |
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#6 |
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diyAudio Member
Join Date: Oct 2008
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You can use boot-strapping to raise Zin to around 5X its current value. Let me make the following comments about the enclosed file:
1. AR1 represents all the amp past the input components this post deals with. It is an admittedly crude model. R1 is a crude stand in for all the existing feedback components. These components remain unchanged. I have tried to match component designations for the other components with the Leach's original schematic. 2. R1a and R1b replace R1 in the original. 3. Rnew and Cnew are the added bootstrapping components. Cnew should be non polar. 4. C1 shrinks from 390 pF to 68 pF to preserve the high impedance (110 kOhms) out to around 18 kHz, where it drops to 77 k Ohms. 5. you can pull the most of the mod in and out by disconnecting the left end of Cnew where it touches the midpoint of R1a and R1b. 6. Warning: this is a thought experiment verified by simulations, but I think it should work just fine. Comments invited. |
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