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Updated Schematic Updated Schematic DS1 was placed at the drain of Q1 instead of the source.
This circuit consists of two solenoids, L1 and L2. L1 is normally off. While L2 is always on unless SW1 opens. L2 does not have any voltage clamping because we want it to release as quickly as possible.

When SW1 opens, a large voltage spike develops on the NC terminal ~ 400v+ and in order not to have TVS1 clamps the voltage from L2 I added a 600V shotcky diode. I know this is not an ideal solution as it develops a 0.8V voltage drop when Q1 is ON.

Does anyone have any suggestions that blocks the inductive quick back voltage from reaching TVS1 while not exhibiting a large voltage drop of a diode?

Q1 Mosfet https://www.mouser.com/datasheet/2/427/sud90330e-1766527.pdf

Q2 Mosfet https://www.mouser.com/datasheet/2/308/1/FDS89141_D-2312937.pdf

TVS1 https://www.littelfuse.com/~/media/electronics/datasheets/tvs_diodes/littelfuse_tvs_diode_smbj_datasheet.pdf.pdf [![enter image description here][2]][2]

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  • \$\begingroup\$ You only show one terminal of the Gate driver output. Where does the other one go? Please show us the circuit of the Gate driver. \$\endgroup\$ Jun 2, 2021 at 21:50
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    \$\begingroup\$ Your circuit is seriously flawed. Assuming DS1 gets hundreds of volts across it when the switch turns off, Q1's Gate-Source junction is also get hundreds of volts across it. The FET will not survive. \$\endgroup\$ Jun 3, 2021 at 8:39
  • \$\begingroup\$ @bruce Abbott It's not seriously flawed you just have not analyzed it properly. The cathod of DS1 gets about 400 volts. DS1 blocks any current flowing through Q1. The gate of Q1 is protected by TVS2. I have already tried it. My issue is with the large voltage drop across DS1 when Q1 is on. \$\endgroup\$
    – Rocky79
    Jun 3, 2021 at 17:00
  • \$\begingroup\$ I did analyze it with LTspice, and wasn't happy with the result. How can you say that TVS2 protects the Gate-Source junction when it isn't connected across it? If Q2B is turned on (the 'normal' state when L1 is turned off) the Gate is connected to L2 while the Source (according to you) is 400V away from it. If it's turned off then TVS2 drops ~16V, so still nearly 400V difference. \$\endgroup\$ Jun 3, 2021 at 23:21
  • \$\begingroup\$ @BruceAbbott I updated the schematic and included the rest of the circuit per your request. DS1 was actually placed at the drain of Q1 instead of the source. TVS2 clamps the gate to source voltage of Q1. Are you still not convinced that it protects Q1? One solution I came up with is to upgrade Mosfet Q1 to a 600V mosfet and get rid of the diode DS1. I then clamp the voltage across L1 with a TVS breakdown voltage of 390V. Part#SMBJ35CA ( Link in description) What do you think? \$\endgroup\$
    – Rocky79
    Jun 4, 2021 at 1:10

3 Answers 3

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The essential problem with your circuit is that L1 and its entire driver circuit (rc delay, gate driver etc.) are connected across L2, and so get the full flyback voltage across them when SW1 opens. For fastest current decay in L2 this part of the circuit should be isolated from it. The obvious way to do this is to insert a diode between the 'ground' side of the L1 circuit and L2. This isolates not only L1 but also resistors R7, R4 and R1 etc., which are also shunting across L2.

The only problem with this solution is the ~0.8 V drop across the diode when L1 is turned on, which you deem unacceptable.

You can reduce this voltage drop by replacing the diode with an N-channel MOSFET wired 'backwards' so its body diode performs the same function, then tie its Gate to Q1's Gate to turn it on (bypassing the body diode) when Q1 is turned on. If Q1 is off then this isolation FET will be turned off when SW1 is turned off. If Q1 is on there will be a small delay before the Gate voltage drops enough to turn it off.

schematic

simulate this circuit – Schematic created using CircuitLab

I simulated this circuit in LTspice and it predicted current through L2 would drop to zero in ~180 μs when L1 is turned off, and 200 μs when L1 is turned on.

Here's another way that could also work. L2 is controlled by a FET which normally is turned on, but turns off when SW1 opens and removes power to the circuit. It is similar to the other circuit except the switch is connected to the Source side of the FET, which is turned on with a fixed voltage. The advantage of this configuration is that the FET doesn't have to pass current from L1, so its RDSON can be higher.

schematic

simulate this circuit

R1 and C1 help the FET to turn off faster. R4 and R5 quench possible oscillations in the coil caused by parasitic capacitance resonating with the inductance (I put 2 resistors in series to handle the high voltage).

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  • \$\begingroup\$ Thank you for this, I appreciate it. I see an issue with this circuit. When L1 is active the body diode of Q3 will be forward biased with ~0.7V @ 2.1A. Q3 will be on so the current should bypass the body diode however the Rdson is 0.25Ω which yields ~0.5V drop drain to source. \$\endgroup\$
    – Rocky79
    Jun 4, 2021 at 15:22
  • \$\begingroup\$ If you look at the updated circuit on top. If I remove DS1 and replace Q1 with a 600V mosfet. Do you see an issue with the fly back voltage going through R1 and R4? I choose those resistors to be anti-surge 400V rating 1206 package. Even if the rest of the circuit is not isolated it shouldn't be an issue, correct? \$\endgroup\$
    – Rocky79
    Jun 4, 2021 at 15:27
  • \$\begingroup\$ "Q3 will be on so the current should bypass the body diode however the Rdson is 0.25Ω" - I only chose that FET because it was in LTspice already and had suitable ratings. If you can find one with lower Rdson then use that. So what is the maximum voltage drop acceptable to you? \$\endgroup\$ Jun 4, 2021 at 21:28
  • \$\begingroup\$ Even if you remove DS1, L1 is still across L2 (in series with Q1's body diode) so it shunts it strongly and greatly increases the 'release' time of L2. The other problem is you need a >400V FET, so you are back to having high Rdson and unacceptable voltage drop! \$\endgroup\$ Jun 4, 2021 at 21:44
  • \$\begingroup\$ I see. I got this 600V Mosfet from Infineon. Part# IPB60R055CFD7ATMA1. It has pretty low rds_On. ~ 60mΩ. I will be using a 400V TVS as well across L1. \$\endgroup\$
    – Rocky79
    Jun 4, 2021 at 22:41
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If you want L2 to turn off quickly, and still provide protection, then you can use a diode and zener pair. The voltage of the zener is picked to be around the supply voltage.

This will allow the current to cycle, but then cut off when it approaches the supply voltage.
If you have enough capacitance on your supply to absorb the power, you can increase the zener value, or put some in series to get the desired voltage, and get a faster turn off.
This is effectively what BLDC motor drivers do (use caps to absorb power from slowing down the motor)

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT Per the comments, an ideal diode is what is being requested. Here is an example circuit for that. The P-FET acts as the diode. Since you have measured about 400V during the flyback event, I would replace the transistors and FET with ~600V parts.

enter image description here

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  • \$\begingroup\$ Thanks Aaron. We have to let the voltage rise on L1. We can't do any voltage clamping. What I am really looking for is a way to block the release energy from L1 getting into TVS1. (TVS1 will clamp it ) and without the forward voltage drop of DS1 in the circuit above. \$\endgroup\$
    – Rocky79
    Jun 2, 2021 at 19:37
  • \$\begingroup\$ @Rocky79 That leads me to think of an ideal diode. But I don't know of any that go 400+V. Maybe make one discretely? \$\endgroup\$
    – Aaron
    Jun 2, 2021 at 20:00
  • \$\begingroup\$ thank you for the suggestion. I will have to study this circuit and how it works. The challenge like you said is making sure that high voltage will not kill any components in the way. \$\endgroup\$
    – Rocky79
    Jun 2, 2021 at 20:39
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TVS diode is replaced by a diode (Schottky) and Zener. This provides fastest turn off. However you can retain the TVS, but you have to place it also on other solenoid. The dump energy is about 0.5J for the 97mH solenoid at 2.1A.

The schottky diode is moved on high side of the MOSFET. It prevents the current re-circulation due to intrinsic diode of the MOSFET.

Higher the zener voltage, higher the spike, faster is the turn off.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you for your suggestions however I have tried different clamping voltages on L1 and none of them achieved the speed we are looking for. We have to let the voltage rise as quickly as possible on L1. If I add a Zener that will only conduct when the reverse voltage hits the zener breakdown. As soon as it conducts it will slow down the solenoid release. \$\endgroup\$
    – Rocky79
    Jun 2, 2021 at 19:31
  • \$\begingroup\$ @Rocky79 You have to make a compromise, leaving the circuit without a clamp it will destroy the circuit sooner or later. Now it's the R1 that takes the dump through transistor diode Q2B. So, you'd rather use a zener of 200V than nothing. \$\endgroup\$ Jun 2, 2021 at 20:02
  • \$\begingroup\$ How would it destroy the circuit? The -400V will go through the body diode of Q2 as you mentioned and through resistor R1. @400V that's a peak power of 6.4W dissipation but it's very brief in microseconds. What power rating for R1 would you use? \$\endgroup\$
    – Rocky79
    Jun 2, 2021 at 20:36
  • \$\begingroup\$ It's not clear to me why you need D4? Once you reach the zener breakdown of ZD1 it will start conducting and circulate the current within coil L1. \$\endgroup\$
    – Rocky79
    Jun 3, 2021 at 17:04
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    \$\begingroup\$ @Rocky79 But the zener would conduct as normal diode in forward direction and "kaabooom". \$\endgroup\$ Jun 3, 2021 at 18:02

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