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How can I deduce the power of a capacitor from its datasheet?

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1

$begingroup$

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

share|improve this question

asked 1 hour ago

user135172user135172

24029

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
  $endgroup$
  – Marko Buršič
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

share|improve this question

asked 1 hour ago

user135172user135172

24029

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The caps doesn't have power spec. They have capacitnce and ESR, and the voltage.
  $endgroup$
  – Marko Buršič
  1 hour ago
  

  
  
  
  

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

share|improve this question

asked 1 hour ago

user135172user135172

24029

$endgroup$

share|improve this question

asked 1 hour ago

user135172user135172

24029

share|improve this question

asked 1 hour ago

user135172user135172

24029

asked 1 hour ago

user135172user135172

24029

asked 1 hour ago

user135172user135172

24029

2 Answers
2

active

oldest

votes

3

$begingroup$

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.)

share|improve this answer

edited 50 mins ago

answered 1 hour ago

TransistorTransistor

85.3k784182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  58 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Regarding energy storage for the rail, consider that 1 farad when dischanged by 1 amp will sag at 1 volt/second.

Lets assume you have 50Hz 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 amp (assumed) * 0.01 second / 0.1volt= 0.1 Farad or

100,000 uF.

share|improve this answer

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

$endgroup$

add a comment | 

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3

$begingroup$

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.)

share|improve this answer

edited 50 mins ago

answered 1 hour ago

TransistorTransistor

85.3k784182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  58 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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.)

share|improve this answer

edited 50 mins ago

answered 1 hour ago

TransistorTransistor

85.3k784182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  58 mins ago
  

  
  
  
  

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.)

share|improve this answer

edited 50 mins ago

answered 1 hour ago

TransistorTransistor

85.3k784182

$endgroup$

share|improve this answer

edited 50 mins ago

answered 1 hour ago

TransistorTransistor

85.3k784182

answered 1 hour ago

TransistorTransistor

85.3k784182

answered 1 hour ago

TransistorTransistor

85.3k784182

0

$begingroup$

Regarding energy storage for the rail, consider that 1 farad when dischanged by 1 amp will sag at 1 volt/second.

Lets assume you have 50Hz 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 amp (assumed) * 0.01 second / 0.1volt= 0.1 Farad or

100,000 uF.

share|improve this answer

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

$endgroup$

add a comment | 

0

$begingroup$

Regarding energy storage for the rail, consider that 1 farad when dischanged by 1 amp will sag at 1 volt/second.

Lets assume you have 50Hz 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 amp (assumed) * 0.01 second / 0.1volt= 0.1 Farad or

100,000 uF.

share|improve this answer

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

$endgroup$

add a comment | 

$begingroup$

Regarding energy storage for the rail, consider that 1 farad when dischanged by 1 amp will sag at 1 volt/second.

Lets assume you have 50Hz 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 amp (assumed) * 0.01 second / 0.1volt= 0.1 Farad or

100,000 uF.

share|improve this answer

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

$endgroup$

share|improve this answer

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718

answered 23 mins ago

analogsystemsrfanalogsystemsrf

14.8k2718