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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...
Question
relay
#3: Post edited
by
Chupacabras
·
2020-06-26T20:15:32Z (over 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
Chupacabras
·
2020-06-26T19:56:27Z (over 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
Chupacabras
·
2020-06-26T19:55:24Z (over 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?