I have a 100W MOSFET power amp (the object was reliability not diy knowledge). Its to use as in a powered stage monitor. Having use these power sections before, I remember that the source input is sometimes insufficient to drive it to the max. Also no local control.
Obvious answer:- make a simple preamp with a gain variable between (say) -9db and +9db . An inverting amp where the gain is R2/R1. R1 = 10K R2 = 2K7 + 100K log pot. Would that work, or does it need to be more complicated?
Obvious answer:- make a simple preamp with a gain variable between (say) -9db and +9db . An inverting amp where the gain is R2/R1. R1 = 10K R2 = 2K7 + 100K log pot. Would that work, or does it need to be more complicated?
If you want to use op-amp in inverting mode remember it should work with gain higher than 1, furthermore it may be noisy to have a pot within feedback loop, if it fails gain jumps to huge value. If you want to still have option to attenuate or amplify better use pot at input creating voltage divider and then amplify using inverting opamp with constant (and high) gain.
best regards
best regards
darkfenriz is right about the noise and potential for overload under fault. However, what you propose I have seen in professional use in various bits of kit for over 15 years without problems. The way to do it is fixed-pot-fixed connected at one side to the input source and the other side to the output and the take off for the inverting input is the pot wiper. Thus the middle of the pot gives unity gain and sweeping either way gives you cut or boost.
Pbassred said:
richie00boy said:
So that would be no 🙂 As Leolabs said you need a resistor at the input side of the pot, but you also need one at the output side, too. If you made the resistors 10k and the pot 10k you would get +/-6dB adjustment.
Actually what I was trying to describe in the initial post was like this but with R1 in the more obvious position.
vAD, doesn't this have huge gain (unless R4 does something I don't understand)? The purpose of the preamp is to adjust a line level input 0.5 - 0.775V not boost a mike level.
I was only using inverting design because i thought I would be simpler. Maybe not.
vAD, doesn't this have huge gain (unless R4 does something I don't understand)? The purpose of the preamp is to adjust a line level input 0.5 - 0.775V not boost a mike level.
I was only using inverting design because i thought I would be simpler. Maybe not.
Attachments
That schematic is now correct except that the dashed resistor is redundant, the input connects to the left resistor on the pot.
I must be stupid.
vAD, If I was sure, I would not have started the post. I saw something that I did not understand so I asked the question.
However things have moved on. I decided to build the non inverting amp in the WIKI from this forum. I used a TL072 ( I know the pinout is different from a TL071). Power is from a 9V battery. input is from a minidisc player. Output is to a mixing desk. I't does not work! I can just hear a crunch as the MD overloads the input.
I measured the voltage getween -9Vcc, and non invert input. 1V curiously, this passed a low volume music but the 100K pot IS controlling the amplifire - i.e. low resistence caused higher volome.
What am I doing wrong.? It must be obvious.
R1 = 10K
VR1 = 100
C1 = 0.1 MF
R2 ( not shown)across C1 and earth is 470K
vAD, If I was sure, I would not have started the post. I saw something that I did not understand so I asked the question.
However things have moved on. I decided to build the non inverting amp in the WIKI from this forum. I used a TL072 ( I know the pinout is different from a TL071). Power is from a 9V battery. input is from a minidisc player. Output is to a mixing desk. I't does not work! I can just hear a crunch as the MD overloads the input.
I measured the voltage getween -9Vcc, and non invert input. 1V curiously, this passed a low volume music but the 100K pot IS controlling the amplifire - i.e. low resistence caused higher volome.
What am I doing wrong.? It must be obvious.
R1 = 10K
VR1 = 100
C1 = 0.1 MF
R2 ( not shown)across C1 and earth is 470K
Attachments
Use a good quality environmentally sealed multiwiper pot. Allen-Bradley or Bourns have conductive plastic sealed types that are reasonable. Modpots?
Cheers,
Greg
Cheers,
Greg
Hi,
I agree with the first few posts. Make a voltage divider from the pot at the input of the chip and give the amp a fixed gain. Come in on a capacitor (D.C. blocking) then into one side of the pot, the wiper goes to the (-) input of the chip, the other leg of the pot goes to say a 1K resistor to ground. Pot should be 100K or so. Then just make a regular inverting amplifier out of it, gain of 10. Putting the pot in the feedback loop is asking for trouble, along with non- linear gain. Also, if you want to get fancy, tie 2X 9 V batteries to form a +/- supply for the chip. For a stage monitor of only 100W, a little noise won't matter much, so even a 741 would do. Good luck!!!
With respect,
Chris
I agree with the first few posts. Make a voltage divider from the pot at the input of the chip and give the amp a fixed gain. Come in on a capacitor (D.C. blocking) then into one side of the pot, the wiper goes to the (-) input of the chip, the other leg of the pot goes to say a 1K resistor to ground. Pot should be 100K or so. Then just make a regular inverting amplifier out of it, gain of 10. Putting the pot in the feedback loop is asking for trouble, along with non- linear gain. Also, if you want to get fancy, tie 2X 9 V batteries to form a +/- supply for the chip. For a stage monitor of only 100W, a little noise won't matter much, so even a 741 would do. Good luck!!!
With respect,
Chris
Re: I must be stupid.
Which side of C1 is this R2? It should be on the op-amp side. If on the input side there will be no way for the op-amp input to get DC bias so it will go crazy.
Pbassred said:
Which side of C1 is this R2? It should be on the op-amp side. If on the input side there will be no way for the op-amp input to get DC bias so it will go crazy.
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