How can I deduce the power of a capacitor from its datasheet?
$begingroup$
I am looking for suitable capacitors for a sound amplifier I have designed.
The speaker power should be at least 7 watt (for the sound to be loud enough).
(I attach a figure of the design)
So, I think I need capacitors which are suitable for these level of power.
But, in the datasheets, i don't see any specification for the power capability of the capacitors..
Any idea how can I know I have chosen capacitors with high enough power resum capabilities?
Thanks!
power capacitor energy
asked Feb 24 at 9:52
user135172user135172
26529
$endgroup$
add a comment |
$begingroup$
I am looking for suitable capacitors for a sound amplifier I have designed.
The speaker power should be at least 7 watt (for the sound to be loud enough).
(I attach a figure of the design)
So, I think I need capacitors which are suitable for these level of power.
But, in the datasheets, i don't see any specification for the power capability of the capacitors..
Any idea how can I know I have chosen capacitors with high enough power resum capabilities?
Thanks!
power capacitor energy
asked Feb 24 at 9:52
user135172user135172
26529
$endgroup$
3
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37
add a comment |
$begingroup$
I am looking for suitable capacitors for a sound amplifier I have designed.
The speaker power should be at least 7 watt (for the sound to be loud enough).
(I attach a figure of the design)
So, I think I need capacitors which are suitable for these level of power.
But, in the datasheets, i don't see any specification for the power capability of the capacitors..
Any idea how can I know I have chosen capacitors with high enough power resum capabilities?
Thanks!
power capacitor energy
asked Feb 24 at 9:52
user135172user135172
26529
$endgroup$
I am looking for suitable capacitors for a sound amplifier I have designed.
The speaker power should be at least 7 watt (for the sound to be loud enough).
(I attach a figure of the design)
So, I think I need capacitors which are suitable for these level of power.
But, in the datasheets, i don't see any specification for the power capability of the capacitors..
Any idea how can I know I have chosen capacitors with high enough power resum capabilities?
Thanks!
power capacitor energy
power capacitor energy
asked Feb 24 at 9:52
user135172user135172
26529
asked Feb 24 at 9:52
user135172user135172
26529
asked Feb 24 at 9:52
user135172user135172
26529
asked Feb 24 at 9:52
user135172user135172
26529
asked Feb 24 at 9:52
user135172user135172
26529
26529
3
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37
add a comment |
3
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37
3
3
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.
Instead, you need to consider the following:
- The voltage rating needs to be at least that of the maximum voltage they will see in service.
- For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
- The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $ Z = frac {1}{2 pi f C} $ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
$endgroup$
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
add a comment |
$begingroup$
Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.
Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.
Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?
dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.
Rearrange this, and C = I / (dV/dT) = I * T/ V
In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or
100,000 µF.
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.
Instead, you need to consider the following:
- The voltage rating needs to be at least that of the maximum voltage they will see in service.
- For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
- The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $ Z = frac {1}{2 pi f C} $ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
$endgroup$
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
add a comment |
$begingroup$
Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.
Instead, you need to consider the following:
- The voltage rating needs to be at least that of the maximum voltage they will see in service.
- For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
- The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $ Z = frac {1}{2 pi f C} $ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
$endgroup$
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
add a comment |
$begingroup$
Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.
Instead, you need to consider the following:
- The voltage rating needs to be at least that of the maximum voltage they will see in service.
- For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
- The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $ Z = frac {1}{2 pi f C} $ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
$endgroup$
Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.
Instead, you need to consider the following:
- The voltage rating needs to be at least that of the maximum voltage they will see in service.
- For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
- The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $ Z = frac {1}{2 pi f C} $ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
edited Feb 24 at 10:35
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
answered Feb 24 at 10:09
TransistorTransistor
85.8k784184
85.8k784184
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
add a comment |
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
1
1
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small).
$endgroup$
– Digiproc
Feb 24 at 10:27
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
$begingroup$
@Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
$endgroup$
– J...
Feb 24 at 21:09
1
1
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
$begingroup$
@J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem.
$endgroup$
– Transistor
Feb 24 at 21:20
add a comment |
$begingroup$
Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.
Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.
Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?
dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.
Rearrange this, and C = I / (dV/dT) = I * T/ V
In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or
100,000 µF.
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
$endgroup$
add a comment |
$begingroup$
Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.
Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.
Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?
dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.
Rearrange this, and C = I / (dV/dT) = I * T/ V
In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or
100,000 µF.
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
$endgroup$
add a comment |
$begingroup$
Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.
Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.
Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?
dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.
Rearrange this, and C = I / (dV/dT) = I * T/ V
In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or
100,000 µF.
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
$endgroup$
Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.
Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.
Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?
dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.
Rearrange this, and C = I / (dV/dT) = I * T/ V
In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or
100,000 µF.
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
edited Feb 25 at 2:39
Peter Mortensen
1,60031422
1,60031422
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
answered Feb 24 at 11:02
analogsystemsrfanalogsystemsrf
14.9k2718
14.9k2718
add a comment |
add a comment |
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3
$begingroup$
The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
$endgroup$
– Marko Buršič
Feb 24 at 10:09
$begingroup$
Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
$endgroup$
– CL22
Feb 24 at 13:37