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a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power. That means a coil current of 0.7 watts / 24 volts = ~30 mA. A relay might have a coil inducta...
Answer
#4: Post edited
by
Andy aka
·
2021-01-11T12:21:15Z (almost 4 years ago)
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
- What does R equal? It equals voltage squared divided by power (700 mW): -
$$P = \dfrac{V^2}{R}$$So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries.- -----
- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
- So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.
- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
- But, if in doubt, you could easily simulate this circuit.
- -----
- > _Q2: What about using TVS diodes as flyback?_
- When the relay is deactivated, the diode becomes forward biased so a TVS is ineffective in the normal position for a flyback diode. If, however, you wish to reverse the direction of the TVS in order to dissipate the power more quickly in order to turn the relay off more quickly, then you will still need a diode in addition to the TVS to avoid the TVS conducting when the relay is activated.
- This time, the power dissipated will be higher than the previously calculated 186 mW because, in the worst case scenario, the relay could be energized and rapidly deactivated nearly twice as fast. So assume 372 mW (worst case) and assume also that most of this will be "spent" in the TVS (if you want to be cautious).
- I'd still use a simulator because the time taken to set it up and get numbers is about the same length of time I took to write this answer.
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
- What does R equal? It equals voltage squared divided by power (700 mW): -
- $$R = \dfrac{V^2}{P}$$
- So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries based on L/R = 20 ms.
- -----
- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
- So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.
- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
- But, if in doubt, you could easily simulate this circuit.
- -----
- > _Q2: What about using TVS diodes as flyback?_
- When the relay is deactivated, the diode becomes forward biased so a TVS is ineffective in the normal position for a flyback diode. If, however, you wish to reverse the direction of the TVS in order to dissipate the power more quickly in order to turn the relay off more quickly, then you will still need a diode in addition to the TVS to avoid the TVS conducting when the relay is activated.
- This time, the power dissipated will be higher than the previously calculated 186 mW because, in the worst case scenario, the relay could be energized and rapidly deactivated nearly twice as fast. So assume 372 mW (worst case) and assume also that most of this will be "spent" in the TVS (if you want to be cautious).
- I'd still use a simulator because the time taken to set it up and get numbers is about the same length of time I took to write this answer.
#3: Post edited
by
Andy aka
·
2021-01-11T12:19:38Z (almost 4 years ago)
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
- What does R equal? It equals voltage squared divided by power (700 mW): -
- $$P = \dfrac{V^2}{R}$$
- So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries.
- -----
- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
- So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.
- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
But, if in doubt, you could easily simulate this circuit.
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
- What does R equal? It equals voltage squared divided by power (700 mW): -
- $$P = \dfrac{V^2}{R}$$
- So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries.
- -----
- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
- So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.
- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
- But, if in doubt, you could easily simulate this circuit.
- -----
- > _Q2: What about using TVS diodes as flyback?_
- When the relay is deactivated, the diode becomes forward biased so a TVS is ineffective in the normal position for a flyback diode. If, however, you wish to reverse the direction of the TVS in order to dissipate the power more quickly in order to turn the relay off more quickly, then you will still need a diode in addition to the TVS to avoid the TVS conducting when the relay is activated.
- This time, the power dissipated will be higher than the previously calculated 186 mW because, in the worst case scenario, the relay could be energized and rapidly deactivated nearly twice as fast. So assume 372 mW (worst case) and assume also that most of this will be "spent" in the TVS (if you want to be cautious).
- I'd still use a simulator because the time taken to set it up and get numbers is about the same length of time I took to write this answer.
#2: Post edited
by
Andy aka
·
2021-01-11T12:11:31Z (almost 4 years ago)
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
What does R equal? It equals power (700 mW) divided by voltage squared: -- $$P = \dfrac{V^2}{R}$$
So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals 16.5 henries.- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything is 186 mW.- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
- But, if in doubt, you could easily simulate this circuit.
- > _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
- That means a coil current of 0.7 watts / 24 volts = ~30 mA.
- A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
- What does R equal? It equals voltage squared divided by power (700 mW): -
- $$P = \dfrac{V^2}{R}$$
- So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals ~16.5 henries.
- -----
- Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
- So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything due to magnetic energy stored is 186 mW.
- -----
- It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
- But, if in doubt, you could easily simulate this circuit.
#1: Initial revision
by
Andy aka
·
2021-01-11T12:08:36Z (almost 4 years ago)
> _a coil with plain on/off functionality, for example a 24VDC relay coil with 700mW max coil power._
That means a coil current of 0.7 watts / 24 volts = ~30 mA.
A relay might have a coil inductance of anything from 1 henry upwards but this can be estimated by the relay activation time in the data sheet. It's approximate but you could assume L/R equals the activation time. Let's say it's 20 ms.
What does R equal? It equals power (700 mW) divided by voltage squared: -
$$P = \dfrac{V^2}{R}$$
So, with 700 mW and 24 volts, the resistance is 823 Ω. Hence inductance equals 16.5 henries.
Energy stored is half of \$0.03^2 \times 16.5\$ = 7.425 mJ.
So, if the relay was toggled on and off continuously (20 ms to activate and 20 ms to deactivate) that is a frequency of 25 Hz and the worst case power dissipated in everything is 186 mW.
-----
It's looking fairly trivial for virtually any diode that has a reverse voltage rating of 30 volts or above because most of that power would be eaten by the relay's internal resistance of 823 Ω.
But, if in doubt, you could easily simulate this circuit.