Pre-amplifier input protection
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I am building the Velleman K1803 pre-amplifier kit. This amplifier has a maximum input signal of 40mv. The audio input to the pre-amplifier will be a piezo-electric sensor, and this can certainly exceed the specified maximum.
I believe that the input can be protected with a pair of diodes, but there is a huge range of diodes available.
It is some time since I have done any electronics, and so far my searches have not resulted in a suitable circuit design which could achieve the protection at the low signal voltage specified. For the record, the input is audio in the range 20Hz to 20kHz, and could possibly lie in the range +/- 0.5V.
I would appreciate some guidance on where to look for a suitable diode and circuit. I can of course supply a circuit diagram of the amplifier if needed
amplifier audio diodes protection
asked 7 hours ago
GeoffGeoff
61
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Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I am building the Velleman K1803 pre-amplifier kit. This amplifier has a maximum input signal of 40mv. The audio input to the pre-amplifier will be a piezo-electric sensor, and this can certainly exceed the specified maximum.
I believe that the input can be protected with a pair of diodes, but there is a huge range of diodes available.
It is some time since I have done any electronics, and so far my searches have not resulted in a suitable circuit design which could achieve the protection at the low signal voltage specified. For the record, the input is audio in the range 20Hz to 20kHz, and could possibly lie in the range +/- 0.5V.
I would appreciate some guidance on where to look for a suitable diode and circuit. I can of course supply a circuit diagram of the amplifier if needed
amplifier audio diodes protection
asked 7 hours ago
GeoffGeoff
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
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– Toor
7 hours ago
2
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago
add a comment |
$begingroup$
I am building the Velleman K1803 pre-amplifier kit. This amplifier has a maximum input signal of 40mv. The audio input to the pre-amplifier will be a piezo-electric sensor, and this can certainly exceed the specified maximum.
I believe that the input can be protected with a pair of diodes, but there is a huge range of diodes available.
It is some time since I have done any electronics, and so far my searches have not resulted in a suitable circuit design which could achieve the protection at the low signal voltage specified. For the record, the input is audio in the range 20Hz to 20kHz, and could possibly lie in the range +/- 0.5V.
I would appreciate some guidance on where to look for a suitable diode and circuit. I can of course supply a circuit diagram of the amplifier if needed
amplifier audio diodes protection
asked 7 hours ago
GeoffGeoff
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am building the Velleman K1803 pre-amplifier kit. This amplifier has a maximum input signal of 40mv. The audio input to the pre-amplifier will be a piezo-electric sensor, and this can certainly exceed the specified maximum.
I believe that the input can be protected with a pair of diodes, but there is a huge range of diodes available.
It is some time since I have done any electronics, and so far my searches have not resulted in a suitable circuit design which could achieve the protection at the low signal voltage specified. For the record, the input is audio in the range 20Hz to 20kHz, and could possibly lie in the range +/- 0.5V.
I would appreciate some guidance on where to look for a suitable diode and circuit. I can of course supply a circuit diagram of the amplifier if needed
amplifier audio diodes protection
amplifier audio diodes protection
asked 7 hours ago
GeoffGeoff
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
GeoffGeoff
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
GeoffGeoff
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
GeoffGeoff
61
asked 7 hours ago
GeoffGeoff
61
61
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Geoff is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
$endgroup$
– Toor
7 hours ago
2
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago
add a comment |
3
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
$endgroup$
– Toor
7 hours ago
2
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago
3
3
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
$endgroup$
– Toor
7 hours ago
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
$endgroup$
– Toor
7 hours ago
2
2
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 5 hours ago
JustmeJustme
2,0321413
$endgroup$
add a comment |
$begingroup$
A pair of inexpensive back-to-back silicon diodes across the input lines should be sufficient to limit input to 600 mV. Germanium diodes or Schottky diodes would keep the voltage lower yet, but they're generally more fragile and/or more expensive than ordinary Si iodes. Since the specifications limit response to 20 kHz, even Si rectifier diodes should not degrade performance noticeably.
Though the maximum rated signal for the Velleman K1803 is 40 mV, there is no DC path from input to IC1a, below, so a transient 600 mV should do no harm.
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
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1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
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reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
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At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 5 hours ago
JustmeJustme
2,0321413
$endgroup$
add a comment |
$begingroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 5 hours ago
JustmeJustme
2,0321413
$endgroup$
add a comment |
$begingroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 5 hours ago
JustmeJustme
2,0321413
$endgroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 5 hours ago
JustmeJustme
2,0321413
answered 5 hours ago
JustmeJustme
2,0321413
answered 5 hours ago
JustmeJustme
2,0321413
answered 5 hours ago
JustmeJustme
2,0321413
2,0321413
add a comment |
add a comment |
$begingroup$
A pair of inexpensive back-to-back silicon diodes across the input lines should be sufficient to limit input to 600 mV. Germanium diodes or Schottky diodes would keep the voltage lower yet, but they're generally more fragile and/or more expensive than ordinary Si iodes. Since the specifications limit response to 20 kHz, even Si rectifier diodes should not degrade performance noticeably.
Though the maximum rated signal for the Velleman K1803 is 40 mV, there is no DC path from input to IC1a, below, so a transient 600 mV should do no harm.
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
$endgroup$
1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
$begingroup$
A pair of inexpensive back-to-back silicon diodes across the input lines should be sufficient to limit input to 600 mV. Germanium diodes or Schottky diodes would keep the voltage lower yet, but they're generally more fragile and/or more expensive than ordinary Si iodes. Since the specifications limit response to 20 kHz, even Si rectifier diodes should not degrade performance noticeably.
Though the maximum rated signal for the Velleman K1803 is 40 mV, there is no DC path from input to IC1a, below, so a transient 600 mV should do no harm.
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
$endgroup$
1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
$begingroup$
A pair of inexpensive back-to-back silicon diodes across the input lines should be sufficient to limit input to 600 mV. Germanium diodes or Schottky diodes would keep the voltage lower yet, but they're generally more fragile and/or more expensive than ordinary Si iodes. Since the specifications limit response to 20 kHz, even Si rectifier diodes should not degrade performance noticeably.
Though the maximum rated signal for the Velleman K1803 is 40 mV, there is no DC path from input to IC1a, below, so a transient 600 mV should do no harm.
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
$endgroup$
A pair of inexpensive back-to-back silicon diodes across the input lines should be sufficient to limit input to 600 mV. Germanium diodes or Schottky diodes would keep the voltage lower yet, but they're generally more fragile and/or more expensive than ordinary Si iodes. Since the specifications limit response to 20 kHz, even Si rectifier diodes should not degrade performance noticeably.
Though the maximum rated signal for the Velleman K1803 is 40 mV, there is no DC path from input to IC1a, below, so a transient 600 mV should do no harm.
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
answered 6 hours ago
DrMoishe PippikDrMoishe Pippik
8867
8867
1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
1
1
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
R2 puts a severe limit on transient current into the IC anyways. Not sure the OP has a transient problem to fix. Reducing R5 to reduce gain may be a better choice.
$endgroup$
– Sparky256
4 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
At max (100X) gain, this "preamp" will have 40nanoVolts/rtHz * sqrt(20,000Hz) * sqrt(2 res of 100K) * pi/2 * Av = 220/2.2 == 15uVrms * 100 = 1.5 milliVolts rms random noise, provided by R2 and R5.
$endgroup$
– analogsystemsrf
2 hours ago
add a comment |
Geoff is a new contributor. Be nice, and check out our Code of Conduct.
Geoff is a new contributor. Be nice, and check out our Code of Conduct.
Geoff is a new contributor. Be nice, and check out our Code of Conduct.
Geoff is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Normally you would clamp to the maximum that the input can take, not the expected maximum of the signal source. It is likely that the 40mV maximum is the maximum that the amp can take and still work properly...but you're not worried about that. You're worried about the maximum it can take and not have damage occur. There's a difference. You can either use a TVS diode that clamp in reverse-breakdown or "regular" sufficiently fast diodes that clamp in forward bias to clamp the voltage to the rail supply (but this requires a rail supply to be present).
$endgroup$
– Toor
7 hours ago
2
$begingroup$
The maximum input is specified as 40mV because the gain is up to 100 and the minimum Vcc is specified as 10V, giving you 5V peak output, or 3.5VRMS, so 40mVRMS roughly defines a clipping point rather than the damage point. A much larger input won't damage the first op-amp, up to at least the voltage rail.
$endgroup$
– user207421
3 hours ago