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RC differentiator giving a higher output amplitude than input amplitude

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3

$begingroup$

This is the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

And this is the output:

I am unable to understand the increase in output amplitude as compared to the input's amplitude. Mathematically it makes sense - rate of change is quite high. But electronically I am having a hard time wrapping my head around it. How can additional voltage be generated with R and C?

high-pass-filter

share|improve this question

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

This is the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

And this is the output:

I am unable to understand the increase in output amplitude as compared to the input's amplitude. Mathematically it makes sense - rate of change is quite high. But electronically I am having a hard time wrapping my head around it. How can additional voltage be generated with R and C?

high-pass-filter

share|improve this question

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

$endgroup$

add a comment | 

$begingroup$

This is the circuit:

^{simulate this circuit – Schematic created using CircuitLab}

And this is the output:

I am unable to understand the increase in output amplitude as compared to the input's amplitude. Mathematically it makes sense - rate of change is quite high. But electronically I am having a hard time wrapping my head around it. How can additional voltage be generated with R and C?

high-pass-filter

share|improve this question

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

$endgroup$

share|improve this question

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

share|improve this question

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

asked 9 hours ago

WhiskeyjackWhiskeyjack

4,72422 gold badges2222 silver badges6969 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

Think of a capacitor as "liking to keep the voltage across it constant" - at least in the short term.

Figure 1. Voltage difference analysis.

Just prior to the squarewave step down at (1) we can see that the right hand side of C1 is at 0 V so there is -1 V across the capacitor. Immediately after the step down there is still -1 V across the capacitor. Because the left side jumped from +1 to -1 the right side is "kicked" from 0 to -2 V.

We get a similar but opposite effect at (2).

In both cases capacitor voltage is maintained at 2 V for the instant of the square transition and is followed by the RC discharge.

share|improve this answer

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Crystal clear now. :)
  $endgroup$
  – Whiskeyjack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good. You can similarly think of inductors as "liking to keep the current through them constant" (again in the short term).
  $endgroup$
  – Transistor
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

It's called "charge pumping", and it is sometimes used to create low-power boosted-voltage supplies for some applications.

At the end of each half-cycle, the capacitor is essentially charged to the same voltage as the source. When the next half-cycle starts, this voltage cannot change instantaneously, so the capacitor voltage is added (in series) to the source voltage. But this quickly discharges through the resistor, and by the end of that half-cycle, the capacitor is charged in the other direction.

share|improve this answer

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

$endgroup$

add a comment | 

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4

$begingroup$

Think of a capacitor as "liking to keep the voltage across it constant" - at least in the short term.

Figure 1. Voltage difference analysis.

Just prior to the squarewave step down at (1) we can see that the right hand side of C1 is at 0 V so there is -1 V across the capacitor. Immediately after the step down there is still -1 V across the capacitor. Because the left side jumped from +1 to -1 the right side is "kicked" from 0 to -2 V.

We get a similar but opposite effect at (2).

In both cases capacitor voltage is maintained at 2 V for the instant of the square transition and is followed by the RC discharge.

share|improve this answer

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Crystal clear now. :)
  $endgroup$
  – Whiskeyjack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good. You can similarly think of inductors as "liking to keep the current through them constant" (again in the short term).
  $endgroup$
  – Transistor
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Think of a capacitor as "liking to keep the voltage across it constant" - at least in the short term.

Figure 1. Voltage difference analysis.

Just prior to the squarewave step down at (1) we can see that the right hand side of C1 is at 0 V so there is -1 V across the capacitor. Immediately after the step down there is still -1 V across the capacitor. Because the left side jumped from +1 to -1 the right side is "kicked" from 0 to -2 V.

We get a similar but opposite effect at (2).

In both cases capacitor voltage is maintained at 2 V for the instant of the square transition and is followed by the RC discharge.

share|improve this answer

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Crystal clear now. :)
  $endgroup$
  – Whiskeyjack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good. You can similarly think of inductors as "liking to keep the current through them constant" (again in the short term).
  $endgroup$
  – Transistor
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Think of a capacitor as "liking to keep the voltage across it constant" - at least in the short term.

Figure 1. Voltage difference analysis.

Just prior to the squarewave step down at (1) we can see that the right hand side of C1 is at 0 V so there is -1 V across the capacitor. Immediately after the step down there is still -1 V across the capacitor. Because the left side jumped from +1 to -1 the right side is "kicked" from 0 to -2 V.

We get a similar but opposite effect at (2).

In both cases capacitor voltage is maintained at 2 V for the instant of the square transition and is followed by the RC discharge.

share|improve this answer

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

answered 8 hours ago

TransistorTransistor

96.1k88 gold badges9595 silver badges212212 bronze badges

6

$begingroup$

It's called "charge pumping", and it is sometimes used to create low-power boosted-voltage supplies for some applications.

At the end of each half-cycle, the capacitor is essentially charged to the same voltage as the source. When the next half-cycle starts, this voltage cannot change instantaneously, so the capacitor voltage is added (in series) to the source voltage. But this quickly discharges through the resistor, and by the end of that half-cycle, the capacitor is charged in the other direction.

share|improve this answer

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

It's called "charge pumping", and it is sometimes used to create low-power boosted-voltage supplies for some applications.

At the end of each half-cycle, the capacitor is essentially charged to the same voltage as the source. When the next half-cycle starts, this voltage cannot change instantaneously, so the capacitor voltage is added (in series) to the source voltage. But this quickly discharges through the resistor, and by the end of that half-cycle, the capacitor is charged in the other direction.

share|improve this answer

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

$endgroup$

add a comment | 

$begingroup$

It's called "charge pumping", and it is sometimes used to create low-power boosted-voltage supplies for some applications.

At the end of each half-cycle, the capacitor is essentially charged to the same voltage as the source. When the next half-cycle starts, this voltage cannot change instantaneously, so the capacitor voltage is added (in series) to the source voltage. But this quickly discharges through the resistor, and by the end of that half-cycle, the capacitor is charged in the other direction.

share|improve this answer

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

$endgroup$

share|improve this answer

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges

answered 9 hours ago

Dave Tweed♦Dave Tweed

131k1010 gold badges164164 silver badges281281 bronze badges