Chosing the right back EMF diodes

Thread Starter

Raymond175

Joined May 28, 2019
57
Hi,

For my project I need to add a back EMF diode to protect my MOSFETs from the back EMF from my solenoids, but I have no knowledge about deducing which diode I need (e.g. type, ratings, etc.). So I hope anyone can help me out here.

I currently have the following:
- DC power supply: 48V
- Solenoid: 50V, 133Ω, 19W@50V
- Solenoid: 50V, 3.7Ω, 675W@50V
- MOSFET: 28A, 100V, Enhancement type, logic-level, N-channel (IRL540PBF)
- Arduino: 5V
- Pull-down resistor: 10kΩ

One important detail is that the diode must influence the turn-on and release time as little as possible.

How can I chose the right diode, what search/filter parameters that I need to enter in the Mouser search, and do you have any recommendations for diodes to use here?

Thanks in advance,
Raymond
 

BobTPH

Joined Jun 5, 2013
2,993
A diode will definitely increase the turn off time, because it is clamping the voltage across the solenoid to about 0.7V and the change in current of an inductor is proportional to the voltage across it. A low voltage means a slow reduction in current.

To make the voltage higher, you can place a Zener diode in series with the flyback diode in reverse polarity to it. Then the voltage on the coil will be 0.7V + the Zener breakdown voltage.

Bob
 

crutschow

Joined Mar 14, 2008
26,726
To make the voltage higher, you can place a Zener diode in series with the flyback diode in reverse polarity to it. Then the voltage on the coil will be 0.7V + the Zener breakdown voltage.
That works well to speed the release time of the solenoid.
A Zener voltage equal to the supply voltage will make the solenoid current fall time approximately equal to the turn-on rise time.
Just note that the turn-off voltage across the transistor is now the supply voltage plus the Zener voltage, so you may have to go with a higher voltage rated transistor.
 

MisterBill2

Joined Jan 23, 2018
8,207
You can also put a resistor in series with the diode, which will turn part of the spike energy into heat. A 100 ohm half watt resistor will help quite a bit. And given the small price increment for diode voltages, you should go at the very least 3 times the solenoid voltage rating. The current rating for the diode on the 19 watt coil will not be high, while on the 675 watt coil it will need to be quite high.
 

Thread Starter

Raymond175

Joined May 28, 2019
57
I always use a diode thats rated at the same peak current as the continuous load current and the working voltage about 50% higher than DC supply voltage.
How do I know the continuous load current of the solenoids? My DC power supply outputs a max op 6.7 A, but 675 W on 50V is 13.5 A (how is that even possible).
 

Thread Starter

Raymond175

Joined May 28, 2019
57
A diode will definitely increase the turn off time, because it is clamping the voltage across the solenoid to about 0.7V and the change in current of an inductor is proportional to the voltage across it.
How do you know it is about 0.7 V? In the (near) future more solenoids will be added with different ratings, so I'd like to know how to calculate it, although the release times are of a less importance for those solenoids...

To make the voltage higher, you can place a Zener diode in series with the flyback diode in reverse polarity to it. Then the voltage on the coil will be 0.7V + the Zener breakdown voltage.
Very interesting, I've been reading up on Zener diodes! This is something I definitely want to try out and compare on my prototype PCB. But how do I decide on the Zener breakdown voltage? Does a higher value come with different cons?
 

Thread Starter

Raymond175

Joined May 28, 2019
57
A Zener voltage equal to the supply voltage will make the solenoid current fall time approximately equal to the turn-on rise time.
Just note that the turn-off voltage across the transistor is now the supply voltage plus the Zener voltage, so you may have to go with a higher voltage rated transistor.
Ah, this actually answers my question in the post above. But then I my "problem" shifts to finding a different MOSFET :)
 

ericgibbs

Joined Jan 29, 2010
12,200
How do I know the continuous load current of the solenoids? My DC power supply outputs a max op 6.7 A, but 675 W on 50V is 13.5 A (how is that even possible).
hi Ray,
The only way to be sure of the V and I is to measure the solenoids V and I, when powered from the supply,
The values posted don't add up.
E
 

Thread Starter

Raymond175

Joined May 28, 2019
57
You can also put a resistor in series with the diode, which will turn part of the spike energy into heat. A 100 ohm half watt resistor will help quite a bit. And given the small price increment for diode voltages, you should go at the very least 3 times the solenoid voltage rating. The current rating for the diode on the 19 watt coil will not be high, while on the 675 watt coil it will need to be quite high.
This is something that I also want to try and compare, but however it put me into thinking... 675 W on 50 V is 13.5 A, and since I wanted to incorporate this back EMF protection on the PCB, I have to be able to deal with this 13.5 A on my connectors as well? Currently I'm working with Molex Micro-Fit 3.0 connectors, but these only allow up to 10.5 A, which means I have to revisit my connectors as well?

Or am I missing something here?

PS: I must be missing something here, as 18 AWG wires don't even support 13.5 A...
 
Last edited:

Thread Starter

Raymond175

Joined May 28, 2019
57
The only way to be sure of the V and I is to measure the solenoids V and I, when powered from the supply,
This is quite hard for the 675 W solenoid, as it actually burns out when activated for more than 1-2 seconds. Using my microcontroller I make sure the solenoid is not activated for more than a few milliseconds (e.g. 500).
 

MisterBill2

Joined Jan 23, 2018
8,207
How do you know it is about 0.7 V? In the (near) future more solenoids will be added with different ratings, so I'd like to know how to calculate it, although the release times are of a less importance for those solenoids...



Very interesting, I've been reading up on Zener diodes! This is something I definitely want to try out and compare on my prototype PCB. But how do I decide on the Zener breakdown voltage? Does a higher value come with different cons?
The forward voltage drop of a silicon diode is always around 0.7 volts because of the property of the diode. All diodes have some forward direction voltage drop except perfect diodes in simulations.
 

crutschow

Joined Mar 14, 2008
26,726
At high currents the stray wiring inductance can alos generate a significant voltage spike, so it might be better if you put the diodes directly from the MOSFET drain pin to V+.
That way the diodes will then absorb the spike from both the wiring and the solenoid.
 

Thread Starter

Raymond175

Joined May 28, 2019
57
At high currents the stray wiring inductance can alos generate a significant voltage spike, so it might be better if you put the diodes directly from the MOSFET drain pin to V+. That way the diodes will then absorb the spike from both the wiring and the solenoid.
That was exactly my plan, as I already copied that from the original PCB :)

1613423834396.png
 
Top