Why are the input and the battery connected in a transistor amp?
$begingroup$
I'm trying to make a simple boost pedal for my guitar using the simplest schematic I found. I'm wondering why the input is connected to the +9v battery in this schematic through a 430kohm resistance. There should be no such connection from what I understood in this video about transistors.
Here is the schematic for the boost pedal I'm trying to make.
What happens current-wise where the two different voltages meet at R1?
I found another schematic here that I understand better according to the video.
transistors amplifier boost
asked 4 hours ago
ServietskyServietsky
1205
$endgroup$
add a comment |
$begingroup$
I'm trying to make a simple boost pedal for my guitar using the simplest schematic I found. I'm wondering why the input is connected to the +9v battery in this schematic through a 430kohm resistance. There should be no such connection from what I understood in this video about transistors.
Here is the schematic for the boost pedal I'm trying to make.
What happens current-wise where the two different voltages meet at R1?
I found another schematic here that I understand better according to the video.
transistors amplifier boost
asked 4 hours ago
ServietskyServietsky
1205
$endgroup$
add a comment |
$begingroup$
I'm trying to make a simple boost pedal for my guitar using the simplest schematic I found. I'm wondering why the input is connected to the +9v battery in this schematic through a 430kohm resistance. There should be no such connection from what I understood in this video about transistors.
Here is the schematic for the boost pedal I'm trying to make.
What happens current-wise where the two different voltages meet at R1?
I found another schematic here that I understand better according to the video.
transistors amplifier boost
asked 4 hours ago
ServietskyServietsky
1205
$endgroup$
I'm trying to make a simple boost pedal for my guitar using the simplest schematic I found. I'm wondering why the input is connected to the +9v battery in this schematic through a 430kohm resistance. There should be no such connection from what I understood in this video about transistors.
Here is the schematic for the boost pedal I'm trying to make.
What happens current-wise where the two different voltages meet at R1?
I found another schematic here that I understand better according to the video.
transistors amplifier boost
transistors amplifier boost
asked 4 hours ago
ServietskyServietsky
1205
asked 4 hours ago
ServietskyServietsky
1205
edited 13 mins ago
Servietsky
asked 4 hours ago
ServietskyServietsky
1205
asked 4 hours ago
ServietskyServietsky
1205
asked 4 hours ago
ServietskyServietsky
1205
1205
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.
It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.
In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.
C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k$Omega$, doing the math in my head), and the source impedance is pretty low (600$Omega$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.
answered 3 hours ago
TimWescottTimWescott
3,4841210
$endgroup$
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.
It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.
In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.
C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k$Omega$, doing the math in my head), and the source impedance is pretty low (600$Omega$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.
answered 3 hours ago
TimWescottTimWescott
3,4841210
$endgroup$
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
add a comment |
$begingroup$
You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.
It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.
In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.
C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k$Omega$, doing the math in my head), and the source impedance is pretty low (600$Omega$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.
answered 3 hours ago
TimWescottTimWescott
3,4841210
$endgroup$
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
add a comment |
$begingroup$
You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.
It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.
In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.
C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k$Omega$, doing the math in my head), and the source impedance is pretty low (600$Omega$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.
answered 3 hours ago
TimWescottTimWescott
3,4841210
$endgroup$
You have found two different schematics for circuits that do two different things. The bottom schematic is for making a beep with no volume control, and no real sound reproduction ability. The top schematic is an actual amplifier, which needs to faithfully reproduce the guitar's output voltage, only bigger.
It's what's known as a "single stage amplifier" -- try Googling for it to get a description of the circuit theory.
In short, R1, R2 and R4 are there to make sure that with no sound there's some standing current in Q1, because the only way that Q1 can make its collector voltage increase is by turning off a little bit. By biasing Q1 into the center of its operating range, the amplifier can accommodate large (ish) voltage swings from the guitar.
C1 is a DC blocking capacitor. It lets the transistor base voltage ride where it needs to (and the input voltage ditto), while allowing audio-frequency AC to pass. The impedance looking into the transistor base is pretty high (somewhere around 30k$Omega$, doing the math in my head), and the source impedance is pretty low (600$Omega$ for a guitar???). So the AC voltage at that node is pretty much set by the guitar.
answered 3 hours ago
TimWescottTimWescott
3,4841210
edited 1 hour ago
answered 3 hours ago
TimWescottTimWescott
3,4841210
answered 3 hours ago
TimWescottTimWescott
3,4841210
answered 3 hours ago
TimWescottTimWescott
3,4841210
3,4841210
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
add a comment |
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Without the bias current of R1 the sound would be distorted.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
$begingroup$
Thank you for the quick answer! I now understand that R1 is used for the bias. I have looked up single stage amplifier but I still don't understand what happens to the voltage where the end of C1 meets the end of R1 as we have two different voltages that meet. I also have another question if you don't mind, would it still work if I used a 10nf capacitor instead of a 100nf at C1?
$endgroup$
– Servietsky
1 hour ago
2
2
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
$begingroup$
You would be unhappy with a 10nF cap there. If I'm doing the math right (in my head, which means I'm probably screwing it up at least a little) that circuit starts ignoring low-frequency signals at about 60Hz. If you changed to the 10x smaller cap that'd be 600Hz, and your guitar would sound horribly tinny. If I were building that circuit I'd experiment with a bigger cap there (up to 470nF), especially if it were for a bass.
$endgroup$
– TimWescott
1 hour ago
add a comment |
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