[SOLVED] False MOSFET triggering in buck converter.

[mrinalmani]

A MOSFET has been inserted in a buck converter. It has nothing to do with the converter itself. (Please refer to the circuit attached). The gate of the MOSFET is connected to the source, hence the MOSFET should ideally never turn on.
But referring to the simulation (also attached along with the circuit), it is evident that each time the buck converter is triggered, there is a large inrush current, indicating that the floating MOSFET also triggers for a brief duration. However when this "extra" MOSFET is not connected, the inrush disappears and the converter works normally.

1. Why is the "floating" MOSFET triggering even though the gate is shorted with the source?
2. How to overcome this spurious triggering?

Thanks

 

[KlausST]

Hi,

I can't see this in the diagram.

Could it be reverse recovery time of the fet's body diode, or
Reverse recovery of the external diode?


Klaus

 

[mrinalmani]

No... not of the external diode. Because the circuit without the MOSFET also has the diode, but does not glitch.
Perhaps the internal diode recovery time may be the culprit. Notice carefully that the current does not spike during the first pulse. The first pulse is absolutely clean. However, from the second pulse onward, it begins to spike. This may be due to the fact that after the first pulse the internal diode carries current during the free-wheeling period and is not able to recover before the subsequent pulses. What do you think?

 

[KlausST]

Hi.

What do you think?

Sounds good.

Can you place a current probe into the source line of the fet. Lets see what it shows.

If the fet's diode is the problem, then use a externsl schotky diode with lower forward voltage.....

Klaus

 

[mrinalmani]

Notice the polarity of current through the MOSFET. It has reversed.
By the way, the idea of an external schottky with lower voltage seems impressive!
I will actually be using a different MOSFET and may be that does not spike like this one. But the aim here is to rectify what is causing the problem, and not use another MOSFET with enhanced capability.

See the attachments below. Can you now figure out something from the simulations?

 

[FvM]

The output capacitance of a MOSFET (is it IRF840?) is considerably higher than of a diode. In so far larger current spikes are expectable. What we see in the waveforms may be however simulation artefacts. There's far to little of information to determine what's going on (e.g. gate voltage rise time, generator impedance, kind of MOSFET models used in the simulation).

 

[BradtheRad]

Notice the polarity of current through the MOSFET. It has reversed.

I've had the same thing happen in simulations, even though it did not appear possible.

I believe it has to do with the timestep. Usually I need to shorten it.

When the timestep is long, short-lived events such as spikes are not tracked very well by the simulator algorithm.

The effect seems to show up particularly when there are inductors and capacitors and diodes interacting, but no resistors. I find it often helps to install high-ohm resistors across these components.

 

[mrinalmani]

I replaced the MOSFET with a 1nF capacitor, to ascertain whether the output capacitance could be a problem. Inserting the capacitor also gave spiky results, although the magnitude of surge current was reduced to approximately half. Therefore output capacitance is certainly a problem, but obviously this capacitance cannot be as large as 1nF. Perhaps there is something else also.
And as far as for the simulator, I simulated again with a resolution of 100 pico-seconds. The spike indeed is a spike with a rectangular waveform and not an sinusoidal oscillation.

 

[FvM]

You'll expect an current spike of 100 nAs when charging the 1 nF capacitor. What do you see?

In case of diode connected MOSFET, there will additonal reverse recovery charge of the slow substrate diode.

 

[mrinalmani]

Indeed I see 100 nAs!
I will now find a fast switching MOSFET and simulate...

 

[KlausST]

Hi,

I still hold on the idea that´s caused by the body diode of Q23, with it´s slow reverse recovery behavior.


Even the negative polarity.... isn´t it so that when the FET Q1 switches OFF the voltage gets reversed (on Q23) caused by the inductance?
Maybe an additional scope channel across d-s of Q23 may help to find a solution..

1nF gate to source is in a typical range, but is it for drain to source? I don´t know...


Klaus

 

[mrinalmani]

I simulated with a fast switching MOSFET "2SK3233" The transients have reduced from over 100A to less than 5A!
Since the capacitor alone wasn't producing enough surge, it seems reasonable to assume that the internal diode was not recovering. And why should it? Because the current had been freewheeling through it! Please correct me if I am wrong.
Thanks everyone for support. I am attaching the final results obtained form 2SK3233.

- - - Updated - - -

@ KlausST:
You are right. The diode does not recover due to the free-wheeling current during the OFF period. And the current's polarity is negative coz it is the free-wheeling current through the inductor and diode in series and not the current from the source.
And btw, I checked for the drain-source capacitance of IRF840 from Vishay. It is around 0.3nF to 0.4nF

 

[FvM]

Without showing Vds rise time, the waveforms don't have real relevance. According to datasheet, 2SK3233 has also a slow substrate diode. But nobody knows how far transistor models are representing the real device behaviour. You would need to verify each datasheet specification, e.g. reverse recovery behaviour, in a simulation test case to be sure that it's modelled correctly. In other words, this thread might turn out as a discussion about unrealistic expectations in simulation results.

I also wonder what's the purpose of using the MOSFET as diode. If you want synchronous rectification, you'll control the rectifier MOSFET in a way that substrate diode conduction is mostly avoided.

 

[mrinalmani]

The purpose of the MOSFET is not to act as a diode, but to extract charge from the output capacitor and dump it back into the source. I am trying to design a circuit to make the capacitor voltage follow a desired curve, hence extraction of charge is as important as injection.

 

[KlausST]

Hi,

The purpose of the MOSFET is not to act as a diode, but to extract charge from the output capacitor and dump it back into the source.

are you sure it makes a difference?

Or is the FET controlled in future?

Klaus

 

[mrinalmani]

When FET 1 is fired, the circuit should act as a simple buck converter and rise the output voltage by a small amount after each trigger. (thus increasing the output voltage in small steps)
When FET 2 is fired, the output capacitor's charge will be extracted and pumped back to the source by a small amount after each trigger. (thus decreasing the output voltage in small steps)

By effective controlling, curves of desired shape should be possible. (for eg, the curve of half sine-wave 0-180 degrees) Needless to say the frequency of the curve should be much smaller than the firing frequency. (firing frequency is variable)

 

[FvM]

O.K. the circuit is intended to act as a bidirectional converter rather than a simple buck converter. It would a good idea to use the "low side" MOSFET as synchronous rectifier in buck operation, avoiding forward bias of the substrate diode.

I also suggest an exact simulation of transistor gate control, including defined dV/dt and gate resistors.

 

[mrinalmani]

Thanks for the suggestion. Synchronous rectification of both the FETs seem possible. One for buck operation and the other for charge-extraction operation. But please note that FET 2 is a low current FET because the only current through it is that required to "discharge" the capacitor. However FET 1 needs to serve the load also, and thus is of a high rating. If FET 2 was used as a synchronous rectifier during buck operation, its current rating will have to be increased, perhaps by a factor of more than 5 to 6 times. (calculations will tell exact how many times). Do you not agree? On the other hand, an external fast-recovery diode will also be expensive. I think there is a lot of subjectivity related in deciding "what is the best".