Relay against load dump ?

[kellogs]

It is pretty hard to cover all the nastiness from ISO 16750-2-2010E, and especially so when replacing a blown fuse is not desirable.

How about this approach ?

The relay armature will take some 5-ish ms to operate, which is less than the 24.4V TVS clamp can take.

I think some extra resistance will be needed ("possibly R") because i would like to use an automotive relay and these are rated for 12-16V, not 24.4V. I do want to drive it hard for a fast changeover; this relay will not have to operate more than maybe 10 times during the lifetime of a vehicle, so if it can take all those 24.4V I would not worry about # of cycles reduction.

Q1 will be operated by a MCU which interrupts on a threshold rising voltage, fast enough for the whole thing to cut off in much less than 10 ms and so the TVS clamp survives. When the surge disappears the MCU will eventually command Q1 again and operation is reestablished.

I expect this to work for a few years, but does it stand any chance to still work in 10 - 15 years ? Of course, the relay will spend 99.9999% of its life in its NC position. I wonder if an automotive relay will still be able to break the circuit after such a long time during which it has not clicked once.

 

[Diver300]

How big is the load? You can just use semiconductors to turn off the load if it's not too large. You can turn the semiconductors off fast enough that you don't need the TVS.

Is the idea of your circuit that the lamp loads the alternator enough to keep the voltage down? If your are protecting against load dump, you can just short the alternator out. Load dump only happens if the battery is missing, so you don't need to worry about shorting the alternator.

I wouldn't worry in the slightest about using a 12 V relay on your application. One of those will last for several minutes on 24 V, and they will not be damaged by the voltage or duration of a 24 V system load dump event.

 

[kellogs]

how do I short out the alternator, tie + to - ? What sort of amperage would that amount to through my tying element ? Nah!

Semiconductors for protecting other semiconductors - all fine as long as it is just a TVS diode

Ok, maybe i can get it to work. I am not very skilled.

What is going to happen when SW1 closes the circuit into an 150V voltage spike, Q2 is going to fry, yes ?
Because I know not how to protect its GS junction. I can only find 25V Vgs max good enough for my circuit, and the TVS clamping at 24.4V is too close for comfort.
Also can't find any other option than paralleling a bunch of smaller TVS diodes in order to get a lower clamping voltage. Not comfortable with that either.
Thought about throwing a Zener in there but these are notoriously slow, it will protect nothing when not already biased.

 

[Diver300]

how do I short out the alternator, tie + to - ? What sort of amperage would that amount to through my tying element ? Nah!

The reason that load dump is a thing is that alternators generate a certain current. That current depends mainly on the excitation current in the field, and doesn't really change much with the voltage on the terminals.

When a big load is disconnected from the alternator, if there is nowhere for the current to go, the voltage leaps up. If there is a battery connected, the alternator current goes into that, and there is no problem and only a small voltage rise.

If you short out an alternator, and there is no battery to provide additional current, you get much the same current that the alternator was producing before you shorted it out. It is completely unlike a battery, which is a constant voltage device, which will generate lots of current if shorted.

On motorcycle alternators and bicycle dynamos, which are very similar, permanent magnet alternators, it is common practice to short the alternator output. I owned a motorcycle (1983 Honda CG125) which had one alternator coil driving the ignition. When the key was turned off, that shorted out that alternator coil.

On car alternators, there is a voltage regulator which alters the excitation to alter the current, to try to keep the voltage constant. That works fine, but it takes time to respond, which is why load dump lasts for 1/2 second or so.

 

[Diver300]

You can easily protect the gate of a MOSFET with a zener to clamp the voltage, as long as you put a resistor to limit the current.

I would suggest that you have a resistor to turn on the main Q2, MOSFET, a zener to limit the gate voltage, and a transistor to short the gate to source. The transistor would be turned on when the voltage exceeds the cut-off voltage you want.

This is the circuit that I have used. It switches the negative not the positive, but could easily be reversed.

In normal use, R1 turns on Q1 and the output is energised. ZD2 protected Q1 gate from over-voltage.

If the voltage is too large, ZD1 conducts, which turns on Q2. R3 limits the current to avoid damage to ZD1 and Q2 from overcurrent. With Q2 turned on, Q1 turns off and the output is turned off.

R2 stops leakage in ZD1 from turning Q2 on until ZD1 conducts properly.

R4 provides a bit of hysteresis to stop the circuit oscillating if the input voltage is changing slowly.

The "on" resistance of Q1, along with the capacitance of C1 provide a limit to how fast the output voltage can rise. This gives time for Q1 to be turned off so that even for very fast changes in the input voltage, the output voltage won't spike.

Q2 is turned on very hard if the circuit operates, so al the gate charge of Q1 can be removed quickly.

 

[kellogs]

Heeeey, a delay capacitor, oh yes!
There is already one cap in my design 1 uF @ 100V which allows none of the ISO 7637-2-2011 3b at the other end. That is an 150V @ 50 ohm, 0.2 us spike.

Now, what would it take to slow enough pulse 2a: 112V @ 2 ohm, ~50 us ? I am off to simulate this:

Why is your Q2 a BJT and not another FET, does it work faster ?

 

[Diver300]

On your circuit there is nothing to turn Q2 off when Q1 turns off. The gate - source voltage won't go down as all the devices connected to the gate of Q2 are high impedance if Q1 is off and the voltage is too small to make D2 conduct.

You might be able to have a resistor in parallel with D2. However, to get Q2 to turn off quickly, the gate charge needs to be discharged. I suspect that it would be difficult to make a circuit to work with a resistor to discharge the gate charge, and you would be better to have a transistor that it turned on discharge the gate and turn off Q2 quickly.

I don't think that C1 will help. It will slow down the turning off of the Q2.

The "On" resistance of Q2 is important to work out the speed that the output will change when the input voltage changes. I suspect that 1 μF won't be enough. You also need to calculate or measure how fast the transistor will turn off, to know how high the voltage will get, or how much energy will end up in the TVS.

In my circuit Q2 was a BJT because I used other identical devices elsewhere in the circuit and keeping the same device meant that there was no increase in the number of components in the Bill of Materials. I don't think that it's too important which is used.

 

[kellogs]

Err, right. I have taken q1 out to simplify the simulation. Does the simulation look right ?

blue - output voltage, drain pin
red - input voltage, source pin
green - gate pin voltage

PWL source:
0 0
10u 0
11u 112
61u 26
70u 15
200m 15

Without C22:

If I have not screwed up anything then it looks like the big TVS is actually fast enough not to allow MOSFET source pin to rise too much, in spite the slow Zener and starting from 0V. I find it a bit peculiar for Vds to be so low and for such a short amount of time (while Vgs is gaining some magnitude). Even ran a simulation starting directly in 112V, Vds was even lower (??). I suspect I am screwing up somewhere...

With C22 enabled in circuit, terrible:

Thank you!

 

[Diver300]

If that simulation is going up to 112V, and you are only getting 20 V anywhere, what I think is happening is that the TVS is keeping the voltage low.

What value for R14 are you using? Can you run the simulation and show the current in the TVS?

 

[kellogs]

Hey,

R14 is 2 ohm as the diagram shows. Yes, the TVS is indeed doing its job. But that is not the problem.

What is going to happen when SW1 closes the circuit into an 150V voltage spike, Q2 is going to fry, yes ?
Because I know not how to protect its GS junction

I have run a few dozen simulation changing various voltages and timings, disabling various components - all in order to get the above to happen. Even updated libngspice.so from v28 to v41. It has not!

Is my expectation plainly wrong that an unbiased Zener should not act fast enough to follow Drain pin voltage ? Or is the C_gs enough to contain any fast (and not so fast) spikes until the Zener eventually does its thing ?

 

[Diver300]

Is my expectation plainly wrong that an unbiased Zener should not act fast enough to follow Drain pin voltage ? Or is the C_gs enough to contain any fast (and not so fast) spikes until the Zener eventually does its thing ?

I don't think that zener diodes really have response time. They have capacitance, which is effectively in parallel with the MOSFET's gate-source capacitance. I don't think that you would get any measurable voltage overshoot of the gate-source voltage if you only supply the gate with a small enough current that the zener can handle it without overheating.

For instance, if you have a 12 V zener like a BZX84-12, it can handle about 20 mA. It has a capacitance of around 85 pF, so if you feed it 20 mA, the voltage will increase at 20 mA/85 pF = 235,000,000 V / s or 235 V/μs. Even if the zener took as long as 10 ns to turn on, which I think is unlikely, the voltage would only overshoot by 2.35 V.

The rate of change of voltage would be less if there is the gate-source capacitance of a MOSFET in parallel.

Unless you put a big capacitor between the gate and ground, an input spike will cause the source and gate to go up at the same time.

 

[kellogs]

Ok then. Already over my head

The circuit you have shared works by virtue of collector-emitter dynamic resistance being much less than that of the Zener, correct ? It would have never crossed my mind to short-circuit a Zener with a transistor!

I suppose it works with a MOSFET too when Rds_on is 100x lower than Rzener. Sort of like this ?

Do I still need C12 with that D3 TVS on the output ? How about R12 ?

Thanks!

 

[danadak]

Zener switching time, the avalanche process :

https://www.onsemi.com/pub/Collateral/HBD854-D.PDF Page 27 on....


Regards, Dana.

 

[Diver300]

I think you should be swapping D1 and R1.

When the voltage is too large, D1 conducts, turning on Q1. That turns off Q2.

I've just realised one advantage of a bipolar transistor for Q1. It would not need a zener to limit the base voltage like a MOSFET needs a zener to limit the gate voltage.

C12 is useful to absorb the current spike until Q2 it off. You shouldn't need the TVS and you should run your simulations without it.

 

[kellogs]

When the voltage is too large, D1 conducts, turning on Q1. That turns off Q2.

Isn't that what happens in the current arrangement ?

Zener switching time, the avalanche process :

https://www.onsemi.com/pub/Collateral/HBD854-D.PDF Page 27 on....


Regards, Dana.

Thank you.
Figures 29 - 31 shows some capacitive spikes that my simulator would not show no matter what...

 

[danadak]

It all depends how good the spice model is and your simulator time step. In the
examples shown you reference I would set mine up to 1 nS time step. When
you do this the sim takes a loooong time to run depending on what time period
you are looking at / graphing.


Regards, Dana.

 

[Diver300]

Isn't that what happens in the current arrangement ?

In your arrangement, as soon as the input voltage gets over the gate-source threshold of Q1, which will be about 3 V, Q1 turns on, turning off Q2, so the circuit will block all voltages. Your simulation should start at 12 V or so, and then have a spike to 120 V. Your circuit needs to conduct at 12 V to make the load work.

In your arrangement D1 is limiting the gate source voltage of Q1.

BTW, you can't just swap D1 and R1. You need a resistor in parallel with D1 to discharge the gate-source capacitance of Q1. You need a 15 V zener in series with R1. When the input voltage gets to around 18 V, there will be about 3 V between the gate and source of Q1, and it will turn on, removing the gate-source voltage from Q2, which will turn off.

R1 is still needed so that the gate-source voltage of Q1 can be limited if there is a large voltage spike, without getting excessive current though D1 and the new 15 V zener.

I think the circuit would work better with Q1 as a bipolar transistor. The advantage is that you don't need a zener to protect the base-emitter voltage, and the turn-on is at a more predictable voltage. The fact that you need a base current for a bipolar transistor is no disadvantage in a situation where you have maybe 100 A to deal with.

 

[kellogs]

Oh, snap! It is that R12 causing this behaviour. Maybe I can remove it and be done with it? I don't understand how it works and how much it helps.

R4 provides a bit of hysteresis to stop the circuit oscillating if the input voltage is changing slowly.

 

[kellogs]

It all depends how good the spice model is and your simulator time step. In the
examples shown you reference I would set mine up to 1 nS time step. When
you do this the sim takes a loooong time to run depending on what time period
you are looking at / graphing.


Regards, Dana.

It is not happening... Tried to replicate the circuit on page 28:

V0 Source: pwl(0 0 10n 0 50n 18 90n 0) r=0

Tried with 6 spice models from 5 or 6 vendors. Best i could get was a fast voltage rise to 10V followed by slow voltage rise to 10.24V at around 50ns - not a spike. This was for just one of the models...

I must be doing something wrong. Using libngspice v4.1, Kicad 5.1. From a long ago time i remember simple rule "TVS fast, Zener slow". So I am expecting voltage overshoot just like in the document you have linked.

Diver300 says not necessarily:

I don't think that zener diodes really have response time. They have capacitance, which is effectively in parallel with the MOSFET's gate-source capacitance. I don't think that you would get any measurable voltage overshoot of the gate-source voltage if you only supply the gate with a small enough current that the zener can handle it without overheating.

 

[Diver300]

Oh, snap! It is that R12 causing this behaviour. Maybe I can remove it and be done with it? I don't understand how it works and how much it helps.

The circuit should work without R12. Worst case without R12 is that the circuit can't decide whether to turn on or off when the voltage input is near the cut-off point.

In a real load dump, the input voltage will never spend much time near that voltage.