Reverse Polarity Protection Bridge Rectifier for a latching Solenoid Valve Design Question

Thread Starter

stumpr84

Joined Dec 10, 2018
9
I am trying to design a bridge rectifier to be used to protect a latching solenoid from reverse polarity. Due to the application being a solenoid there is an issue of back EMF. On this particular 24V @.5A 47.24ohm latching solenoid I measure a -600V spike goin from On to off and a -0.75A spike going from Off to On. However, the voltage spike is less than 1ms in duration (see attached screen capture). So my question is what effect would this back EMF have on the design of the bridge rectifier diodes?
My first guess was that I should use diodes with a low forward voltage and are rated for a Vrm of 500V plus. However, out of curiosity my first designed used the 50V 1A MCC SS15E Schottky diodes and to my surprise the diodes survived and removed all of the Back EMF transients I was seeing. Can anyone explain why these diodes survived?
 

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panic mode

Joined Oct 10, 2011
1,881
draw a circuit and then we can discuss it.

if you draw a diode in anti-parallel with solenoid, it would do the job to erase back-EMF pulse.
but using full wave bridge rectifier, those rectifier diodes are in the same direction as the mentioned diode (although two in series, instead of just one we just mentioned) and therefore doing the same thing.
1605801954924.png
so bridge diodes also work to block EMF and separate diode for this is redundant.
what bridge diodes must endure is two things:
1. when reverse biased, they must tolerate max supply voltage (AC supply)
2. when forward biased, they must tolerate max load current.

since your supply is 24V which is less than 50V diode rating, 1st goal is met...
 
Last edited:

Thread Starter

stumpr84

Joined Dec 10, 2018
9
draw a circuit and then we can discuss it.

if you draw a diode in anti-parallel with solenoid, it would do the job to erase back-EMF pulse.
but using full wave bridge rectifier, those rectifier diodes are in the same direction as the mentioned diode (although two in series, instead of just one we just mentioned) and therefore doing the same thing.
View attachment 222737
so bridge diodes also work to block EMF and separate diode for this is redundant.
what bridge diodes must endure is two things:
1. when reverse biased, they must tolerate max supply voltage (AC supply)
2. when forward biased, they must tolerate max load current.

since your supply is 24V which is less than 50V diode rating, 1st goal is met...
Thank you for the response. I have attached a schematic of my circuit. To clarify, the circuit is straight DC not AC. The driving voltage is +24VDC with the bridge being used for polarity protection. I have used a flyback diode in previous designs to remove these type of transients. However, my end questions is not how to remove the transients but why in this application do diodes rated for a reverse voltage of 50VDC withstand spikes of 600VDC?
 

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LesJones

Joined Jan 8, 2017
3,137
The diode is only subject to a reverse voltage of 24 volts when the supply is on. When the supply switches off the current in the coil attempts to remain flowing in the same direction. For this to happen the voltage on it's terminals reverses direction so the diode across it's terminals is now forward biased. So the voltage across the diode is now about 0.7 volts. (Or in the case of a bridge providing back EMF protection the voltage across it's terminals is about 1.4 volts as there are two diodes in series.)

Les.
 

Thread Starter

stumpr84

Joined Dec 10, 2018
9
LesJones, thank you for your feedback. If I am understanding you correctly you are saying that when the power is removed that -600V spike (back EMF) forward biases D3 and D4 (See schmatic) in order for the current in the inductor to keep flowing in the same direction. So the voltage drop across D3 and D4 is around .7V each which is the diodes specified Vf. So why doesn't the remaining 598V damage/destroy D3/D4 since they are only rated for 50V? Is it due to the duration of the spike?
 

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ScottWang

Joined Aug 23, 2012
7,005
IMG_20201120_013955.jpg

If you upload the file as *.gif *.png or *.jpg then it's more easier to see the circuit, because some of our helpers that them don't like to open the .pdf file for security reason.
 

Tonyr1084

Joined Sep 24, 2015
5,865
What I do is open the drawing using "Preview" or whatever program you have similar to this. Then drag a copy field around the part of the drawing you want to upload. Adjust the sides and top and bottoms to get the most compact copy field you want. Then copy the field in whatever preferred method you like. Then go into your text on AAC and do a simple paste. Quickest way I found to add drawings to the threads.
 

Thread Starter

stumpr84

Joined Dec 10, 2018
9
The two green diodes do the same thing as the (now) grey diode. Therefore the grey diode can be removed from circuit. Thank you @panic mode for the initial drawing.
View attachment 222765
Hi Tonyr1084 thank you for the feedback. I understand how the diodes are clamping the transients. What I dont understand is why a 600VDC transient is not damaging the or causing the 50VDC 1A rated Schottky diodes to fail.
 

BobTPH

Joined Jun 5, 2013
3,112
There is no 600V transient when there is a reverse biased diode. This is why we use them. It does not absorb or block the 600V, the 600V never happens as long as it is there.

You have to understand how an inductor works. The voltage across it L di/dt. The back EMF voltage occurs because the current is drops rapidly when power is disconnected. The diode allows the current to continue flowing, thus no rapid drop in current, thus no spike in voltage.

The reverse rating of the diode needs only to be above the driving voltage, not above the spike that would occur if the diode was not in the circuit!

Bob
 

LesJones

Joined Jan 8, 2017
3,137
In the explanation I simplified the circuit on of the pairs of diodes D2 and D4 (Which are in series.) D1 and D3 (Also in series.) are in parallel with D2 and D4 so the current through the coil is shared between the pairs of diodes. The high voltage only occurs when there is only a high resistance path between the coil terminals. So if the original coil current was 1 amp and the resistance connected across it was 100 ohms the initial voltage across the 100 ohm resistor would be 100 volts. It gets more complicated if a very value resistor on no resistor is connected across the coil (Infinite resistance.) due to the self capacity of the winding.

Les.
 

Thread Starter

stumpr84

Joined Dec 10, 2018
9
There is no 600V transient when there is a reverse biased diode. This is why we use them. It does not absorb or block the 600V, the 600V never happens as long as it is there.

You have to understand how an inductor works. The voltage across it L di/dt. The back EMF voltage occurs because the current is drops rapidly when power is disconnected. The diode allows the current to continue flowing, thus no rapid drop in current, thus no spike in voltage.

The reverse rating of the diode needs only to be above the driving voltage, not above the spike that would occur if the diode was not in the circuit!

Bob
Hi Bob, Thank you for the explanation, that makes sense.
 

crutschow

Joined Mar 14, 2008
27,010
Here's an LTspice simulation of an inductor with a diode to show the suppression of the inductive spike.
I have 1A going into an arbitrary 1H, 5Ω, inductor, giving 5V across the inductor when energized.
When the source current goes to zero (simulating a switch opening) at 1 second, the inductor current (yellow trace) is now carried by the diode (red trace) until it returns to zero.
The inductor voltage (blue trace) goes to near -1V maximum, which is the forward conduction voltage of the diode at 1A.

1605814997576.png
 

Thread Starter

stumpr84

Joined Dec 10, 2018
9
Here's an LTspice simulation of an inductor with a diode to show the suppression of the inductive spike.
I have 1A going into an arbitrary 1H, 5Ω, inductor, giving 5V across the inductor when energized.
When the source current goes to zero (simulating a switch opening) at 1 second, the inductor current (yellow trace) is now carried by the diode (red trace) until it returns to zero.
The inductor voltage (blue trace) goes to near -1V maximum, which is the forward conduction voltage of the diode at 1A.

View attachment 222781
Curtschow will you please share a screen capture showing the setting of your current supply? I am trying to run the same simulation in LTspice and it is not working.
1605901085484.png
 

crutschow

Joined Mar 14, 2008
27,010
You must be running LTspice XVII which is apparently different than IV, which I'm using (XVII has some features I don't like, this being one of them).

So set I1 as follows and see if that works:

1605911244384.png
 
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