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Single supply non-inverting amplifier using op amp
My (single ended, class A) tube amp sounds great while it's shutting off. What's happening?How to control the amplification of a Transistor?Help Needed with Voltage Controlled Synth CircuitSoft diode clipping for 'controlling' amplifier levels and avoiding harsh distortionPhotodiode non-inverting amplifier circuitOp-Amp circuit: Check understandingCapacitor into resistor to groundReasons for choosing a transistor instead of an op-amp as a bufferSimple Non-Inverting Op-Amp ProblemHow to make a guitar amp switcher?
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$begingroup$
I'm working on building a distortion guitar effect pedal for an analog electronics class I'm taking. I am struggling with building an amplifying circuit using an op amp (we just began learning about them), and I'm hoping I could get some tips. Guitar pedals generally run on a single 9V supply. This is the circuit I attempted to build in lab today, however, it did not give any output.
simulate this circuit – Schematic created using CircuitLab
In particular, I am wondering if it is necessary for me to bias the non-inverting input at 4.5V, and if this circuit would give me the desired gain of +10 if I build it correctly. This is not the specific Op amp I'll be using, I'm not sure which I'll be using yet, I haven't gotten that far.
operational-amplifier amplifier gain single-supply-op-amp guitar-pedal
New contributor
$endgroup$
add a comment |
$begingroup$
I'm working on building a distortion guitar effect pedal for an analog electronics class I'm taking. I am struggling with building an amplifying circuit using an op amp (we just began learning about them), and I'm hoping I could get some tips. Guitar pedals generally run on a single 9V supply. This is the circuit I attempted to build in lab today, however, it did not give any output.
simulate this circuit – Schematic created using CircuitLab
In particular, I am wondering if it is necessary for me to bias the non-inverting input at 4.5V, and if this circuit would give me the desired gain of +10 if I build it correctly. This is not the specific Op amp I'll be using, I'm not sure which I'll be using yet, I haven't gotten that far.
operational-amplifier amplifier gain single-supply-op-amp guitar-pedal
New contributor
$endgroup$
1
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago
add a comment |
$begingroup$
I'm working on building a distortion guitar effect pedal for an analog electronics class I'm taking. I am struggling with building an amplifying circuit using an op amp (we just began learning about them), and I'm hoping I could get some tips. Guitar pedals generally run on a single 9V supply. This is the circuit I attempted to build in lab today, however, it did not give any output.
simulate this circuit – Schematic created using CircuitLab
In particular, I am wondering if it is necessary for me to bias the non-inverting input at 4.5V, and if this circuit would give me the desired gain of +10 if I build it correctly. This is not the specific Op amp I'll be using, I'm not sure which I'll be using yet, I haven't gotten that far.
operational-amplifier amplifier gain single-supply-op-amp guitar-pedal
New contributor
$endgroup$
I'm working on building a distortion guitar effect pedal for an analog electronics class I'm taking. I am struggling with building an amplifying circuit using an op amp (we just began learning about them), and I'm hoping I could get some tips. Guitar pedals generally run on a single 9V supply. This is the circuit I attempted to build in lab today, however, it did not give any output.
simulate this circuit – Schematic created using CircuitLab
In particular, I am wondering if it is necessary for me to bias the non-inverting input at 4.5V, and if this circuit would give me the desired gain of +10 if I build it correctly. This is not the specific Op amp I'll be using, I'm not sure which I'll be using yet, I haven't gotten that far.
operational-amplifier amplifier gain single-supply-op-amp guitar-pedal
operational-amplifier amplifier gain single-supply-op-amp guitar-pedal
New contributor
New contributor
New contributor
asked 1 hour ago
alexamvdoralexamvdor
112
112
New contributor
New contributor
1
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago
add a comment |
1
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago
1
1
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your schematic should be as following:
simulate this circuit – Schematic created using CircuitLab
Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally.
Now let's take a look at the capacitors in dashed rectangles:
If you don't put C2, the DC bias will be multiplied by 11 as well. Thus the output will saturate and you'll never get the amplified signal from output. Since C2 will be open in DC, the net gain in DC will be unity. Thus the DC bias will be multiplied by 1. So the amplified signal will have an offset of Vcc/2 instead of 11 x Vcc/2.
C3 is just a DC-blocking capacitor. It removes the DC bias so you can get only the amplified AC signal.
$endgroup$
add a comment |
$begingroup$
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?)
The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
And you might also want to use a "bulk capacitor". (This is a short-time energy provider for your op-amp.)
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your schematic should be as following:
simulate this circuit – Schematic created using CircuitLab
Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally.
Now let's take a look at the capacitors in dashed rectangles:
If you don't put C2, the DC bias will be multiplied by 11 as well. Thus the output will saturate and you'll never get the amplified signal from output. Since C2 will be open in DC, the net gain in DC will be unity. Thus the DC bias will be multiplied by 1. So the amplified signal will have an offset of Vcc/2 instead of 11 x Vcc/2.
C3 is just a DC-blocking capacitor. It removes the DC bias so you can get only the amplified AC signal.
$endgroup$
add a comment |
$begingroup$
Your schematic should be as following:
simulate this circuit – Schematic created using CircuitLab
Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally.
Now let's take a look at the capacitors in dashed rectangles:
If you don't put C2, the DC bias will be multiplied by 11 as well. Thus the output will saturate and you'll never get the amplified signal from output. Since C2 will be open in DC, the net gain in DC will be unity. Thus the DC bias will be multiplied by 1. So the amplified signal will have an offset of Vcc/2 instead of 11 x Vcc/2.
C3 is just a DC-blocking capacitor. It removes the DC bias so you can get only the amplified AC signal.
$endgroup$
add a comment |
$begingroup$
Your schematic should be as following:
simulate this circuit – Schematic created using CircuitLab
Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally.
Now let's take a look at the capacitors in dashed rectangles:
If you don't put C2, the DC bias will be multiplied by 11 as well. Thus the output will saturate and you'll never get the amplified signal from output. Since C2 will be open in DC, the net gain in DC will be unity. Thus the DC bias will be multiplied by 1. So the amplified signal will have an offset of Vcc/2 instead of 11 x Vcc/2.
C3 is just a DC-blocking capacitor. It removes the DC bias so you can get only the amplified AC signal.
$endgroup$
Your schematic should be as following:
simulate this circuit – Schematic created using CircuitLab
Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally.
Now let's take a look at the capacitors in dashed rectangles:
If you don't put C2, the DC bias will be multiplied by 11 as well. Thus the output will saturate and you'll never get the amplified signal from output. Since C2 will be open in DC, the net gain in DC will be unity. Thus the DC bias will be multiplied by 1. So the amplified signal will have an offset of Vcc/2 instead of 11 x Vcc/2.
C3 is just a DC-blocking capacitor. It removes the DC bias so you can get only the amplified AC signal.
answered 38 mins ago
Rohat KılıçRohat Kılıç
4,5542926
4,5542926
add a comment |
add a comment |
$begingroup$
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?)
The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
And you might also want to use a "bulk capacitor". (This is a short-time energy provider for your op-amp.)
$endgroup$
add a comment |
$begingroup$
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?)
The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
And you might also want to use a "bulk capacitor". (This is a short-time energy provider for your op-amp.)
$endgroup$
add a comment |
$begingroup$
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?)
The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
And you might also want to use a "bulk capacitor". (This is a short-time energy provider for your op-amp.)
$endgroup$
Because you have a single supply op-amp configuration, you would also need to bias the inverting input of your op-amp to +4.5V (Where did you connect R3?)
The gain will be (R3+R4)/R3 which is +11 for the resistor values shown here.
And you might also want to use a "bulk capacitor". (This is a short-time energy provider for your op-amp.)
edited 1 hour ago
answered 1 hour ago
Stefan WyssStefan Wyss
2,3321414
2,3321414
add a comment |
add a comment |
alexamvdor is a new contributor. Be nice, and check out our Code of Conduct.
alexamvdor is a new contributor. Be nice, and check out our Code of Conduct.
alexamvdor is a new contributor. Be nice, and check out our Code of Conduct.
alexamvdor is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Where does the bottom end of R3 go? You're showing it as unconnected and that's rather important.
$endgroup$
– Finbarr
1 hour ago
$begingroup$
put a large cap in series with R3, to ground.
$endgroup$
– analogsystemsrf
33 mins ago