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Q&A Switch-off time of relay with flyback diode

When switching a relay there should be a flyback diode. It could be just a diode or diode with resistor in series. I understand there is a difference in the speed of relay switch-off. But what is t...

1 answer  ·  posted 4y ago by Chupacabras‭  ·  last activity 4y ago by Olin Lathrop‭

Question relay
#3: Post edited by user avatar Chupacabras‭ · 2020-06-26T20:15:32Z (almost 4 years ago)
  • When switching a relay there should be a flyback diode.
  • It could be just a diode or diode with resistor in series.
  • I understand there is a difference in the speed of relay switch-off. But what is the difference really, how to calculate it?
  • Example in this answer:
  • https://electrical.codidact.com/questions/276116#answer-276117
  • Inductance is L=1 mH
  • Current is I=63 mA
  • Diode drop Vd=0.7 V
  • Resistor in series R=82 Ω
  • Is this calculation correct?
  • Energy stored in the coil is E=1/2*L*I=31.5 uJ
  • Power dissipated in diode:
  • P1=Vd*I=44.1 mW
  • Time needed to dissipate that energy:
  • T1=E/P1=714 us
  • Power dissipated in diode and resistor:
  • P2=(Vd+R*I)*I=369.6 mW
  • Time needed to dissipate that energy:
  • T2=E/P2=85.2 us
  • So it is 8.3x faster. Correct? Or it has to be calculated different way?
  • When switching a relay there should be a flyback diode.
  • It could be just a diode or diode with resistor in series.
  • I understand there is a difference in the speed of relay switch-off. But what is the difference really, how to calculate it?
  • Example in this answer:
  • https://electrical.codidact.com/questions/276116#answer-276117
  • Inductance is L=1 mH
  • Current is I=63 mA
  • Diode drop Vd=0.7 V
  • Resistor in series R=82 Ω
  • Is this calculation correct?
  • Energy stored in the coil is E=1/2*L*I^2=1.98 uJ
  • Power dissipated in diode:
  • P1=Vd*I=44.1 mW
  • Time needed to dissipate that energy:
  • T1=E/P1=44.9 us
  • Power dissipated in diode and resistor:
  • P2=(Vd+R*I)*I=369.6 mW
  • Time needed to dissipate that energy:
  • T2=E/P2=5.36 us
  • So it is 8.3x faster. Correct? Or it has to be calculated different way?
#2: Post edited by user avatar Chupacabras‭ · 2020-06-26T19:56:27Z (almost 4 years ago)
  • When switching a relay there should be a flyback diode.
  • It could be just a diode or diode with resistor in series.
  • I understand there is a difference in the speed of relay switch-off. But what is the difference really, how to calculate it?
  • Example in this answer:
  • https://electrical.codidact.com/questions/276116#answer-276117
  • Inductance is L=1 mH
  • Current is I=63 mA
  • Diode drop Vd=0.7 V
  • Resistor in series R=82 Ω
  • Is this calculation correct?
  • Energy stored in the coil is E=1/2*L*I=31.5 uJ
  • Power dissipated in diode:
  • P1=Vd*I=44.1 mW
  • Time needed to dissipate that energy:
  • T1=E/P1=714 us
  • Power dissipated in diode and resistor:
  • P2=(Vd+R*I)*I=369.6 mW
  • Time needed to dissipate that energy:
  • T2=E/P2=85.2 us
  • So it is 8.3x faster. Correct? Or it has to be calculated different way?
  • When switching a relay there should be a flyback diode.
  • It could be just a diode or diode with resistor in series.
  • I understand there is a difference in the speed of relay switch-off. But what is the difference really, how to calculate it?
  • Example in this answer:
  • https://electrical.codidact.com/questions/276116#answer-276117
  • Inductance is L=1 mH
  • Current is I=63 mA
  • Diode drop Vd=0.7 V
  • Resistor in series R=82 Ω
  • Is this calculation correct?
  • Energy stored in the coil is E=1/2*L*I=31.5 uJ
  • Power dissipated in diode:
  • P1=Vd*I=44.1 mW
  • Time needed to dissipate that energy:
  • T1=E/P1=714 us
  • Power dissipated in diode and resistor:
  • P2=(Vd+R*I)*I=369.6 mW
  • Time needed to dissipate that energy:
  • T2=E/P2=85.2 us
  • So it is 8.3x faster. Correct? Or it has to be calculated different way?
#1: Initial revision by user avatar Chupacabras‭ · 2020-06-26T19:55:24Z (almost 4 years ago)
When switching a relay there should be a flyback diode.  
It could be just a diode or diode with resistor in series.  
I understand there is a difference in the speed of relay switch-off. But what is the difference really, how to calculate it?  

Example in this answer:  
https://electrical.codidact.com/questions/276116#answer-276117
Inductance is L=1 mH  
Current is I=63 mA  
Diode drop Vd=0.7 V  
Resistor in series R=82 Ω

Is this calculation correct?  
Energy stored in the coil is E=1/2*L*I=31.5 uJ  

Power dissipated in diode:  
P1=Vd*I=44.1 mW  
Time needed to dissipate that energy:  
T1=E/P1=714 us  

Power dissipated in diode and resistor:  
P2=(Vd+R*I)*I=369.6 mW  
Time needed to dissipate that energy:  
T2=E/P2=85.2 us  

So it is 8.3x faster. Correct? Or it has to be calculated different way?