Pre-amplifier input protection The Next CEO of Stack Overflow50 volts regulated power supplyAmplifiers: IC selection and determining its correct valuesMSP430 Low Voltage ADC Input ProtectionLimiting small signals to one direction (ideally with passive components)Designing a safe Amplifier to drive Sound Card Mic InputHow are the resistor values R1 and R2 calculated for a transistor amplifier that has a voltage divider biasSoft diode clipping for 'controlling' amplifier levels and avoiding harsh distortionSome questions regarding input stage of audio signals to the pre-amplifierCrossover distortion returns when load is appliedIs the ESD and overvoltage protection for my ADC circuit sufficient?
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Pre-amplifier input protection
The Next CEO of Stack Overflow50 volts regulated power supplyAmplifiers: IC selection and determining its correct valuesMSP430 Low Voltage ADC Input ProtectionLimiting small signals to one direction (ideally with passive components)Designing a safe Amplifier to drive Sound Card Mic InputHow are the resistor values R1 and R2 calculated for a transistor amplifier that has a voltage divider biasSoft diode clipping for 'controlling' amplifier levels and avoiding harsh distortionSome questions regarding input stage of audio signals to the pre-amplifierCrossover distortion returns when load is appliedIs the ESD and overvoltage protection for my ADC circuit sufficient?
$begingroup$
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
Added:
Following @DrMoishePippik comments on back-to-back schottky diodes, the lowest switch-on voltage I have found is 0.33 to 0.45 for the BAT43 small signal Schottky diode.
amplifier audio diodes protection
asked 2 days ago
GeoffGeoff
112
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.
$endgroup$
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
Added:
Following @DrMoishePippik comments on back-to-back schottky diodes, the lowest switch-on voltage I have found is 0.33 to 0.45 for the BAT43 small signal Schottky diode.
amplifier audio diodes protection
asked 2 days ago
GeoffGeoff
112
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).
$endgroup$
– Toor
2 days 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
2 days ago
$begingroup$
Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
$endgroup$
– Geoff
14 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
Added:
Following @DrMoishePippik comments on back-to-back schottky diodes, the lowest switch-on voltage I have found is 0.33 to 0.45 for the BAT43 small signal Schottky diode.
amplifier audio diodes protection
asked 2 days ago
GeoffGeoff
112
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
Added:
Following @DrMoishePippik comments on back-to-back schottky diodes, the lowest switch-on voltage I have found is 0.33 to 0.45 for the BAT43 small signal Schottky diode.
amplifier audio diodes protection
amplifier audio diodes protection
asked 2 days ago
GeoffGeoff
112
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 2 days ago
GeoffGeoff
112
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.
edited 13 hours ago
Geoff
asked 2 days ago
GeoffGeoff
112
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 2 days ago
GeoffGeoff
112
asked 2 days ago
GeoffGeoff
112
112
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
2 days 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
2 days ago
$begingroup$
Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
$endgroup$
– Geoff
14 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
2 days 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
2 days ago
$begingroup$
Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
$endgroup$
– Geoff
14 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
2 days 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
2 days 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
2 days 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
2 days ago
$begingroup$
Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
$endgroup$
– Geoff
14 hours ago
$begingroup$
Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
$endgroup$
– Geoff
14 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 2 days ago
JustmeJustme
2,0421413
$endgroup$
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 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
$endgroup$
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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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 2 days ago
JustmeJustme
2,0421413
$endgroup$
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 2 days ago
JustmeJustme
2,0421413
$endgroup$
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 2 days ago
JustmeJustme
2,0421413
$endgroup$
Just change one of the feedback resistors to have less gain so it can accept larger input voltages without clipping.
answered 2 days ago
JustmeJustme
2,0421413
answered 2 days ago
JustmeJustme
2,0421413
answered 2 days ago
JustmeJustme
2,0421413
answered 2 days ago
JustmeJustme
2,0421413
2,0421413
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 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
$endgroup$
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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days 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 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
$endgroup$
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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days 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 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
$endgroup$
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 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
answered 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
answered 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
answered 2 days ago
DrMoishe PippikDrMoishe Pippik
8967
8967
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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days 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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days 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
2 days 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
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days ago
$begingroup$
reducing R5 may cause oscillation; that OA is already at unity gain.
$endgroup$
– analogsystemsrf
2 days 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 days 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 days 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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$begingroup$
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
2 days ago
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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.
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– user207421
2 days ago
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Thank you for clarifying the reason for the low maximum input. If I have read the data sheet for the LM358 correctly, it confirms that the maximum input signal voltage range is -0.3 to +32 volt.
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– Geoff
14 hours ago