Preventing the MOSFET diodes to dissipate power

[ali_asadzadeh]

Hi,
Guys do you have a cleaver way of bypassing the MOSFET diodes, so they would not dissipate power! even by adding more MOSFETS! here is a classic SSR circuit,do you have any Idea?

[langwadt]

the diodes doesn't dissipate any power, when the mosfets are on they conduct the same in both directions

[ali_asadzadeh]

In the SSR, when it's ON, One of MOSFET's internal Diode would conduct, base on the Voltage sign applied to it! just check the picture, and note the AC voltage across it

[2N3055]

Both mosfets will be on in the on state...

[Ian.M]

For SSRs, when on, the reversed MOSFET's channel shunts its body diode so no significant current flows through the diode.  When off, the normally-biassed MOSFET blocks current through the body diode of the reversed one.     

However, in high power MOSFET H-bridges that have a Hi-Z state (where both the upper and the lower MOSFETs on one or both sides of it turn off simultaneously), it can be a problem.    This is usually associated with letting a motor load free-wheel (coast).   If the mechanical load can cause the motor to over-run faster than its normal maximum speed, that can cause uncontrolled conduction through the body diodes.  If the body diodes don't have sufficient rating and you can't modify the system to eliminate the Hi-Z state, shunt them with suitably rated Schottky diodes, which will divert current from the body diodes.

[Siwastaja]

Proper driver obviously drives the MOSFET on whenever the diode would otherwise conduct. If this is a problem, use proper driver logic. Basically, in a bridge of two MOSFETs, always drive one of them on, problem solved.

During the deadtime, the diode may conduct very shortly. In this case, the reverse recovery loss of the diode is of a bigger concern than the Vf drop of the diode.

[langwadt]

In the SSR, when it's ON, One of MOSFET's internal Diode would conduct, base on the Voltage sign applied to it! just check the picture, and note the AC voltage across it

the picture is wrong, when the mosfets are on they short out the diodes

[ali_asadzadeh]

Thanks for responses, I think I did not make it clear that I want to be able to bypass the diodes so they would not make any dissipation in the ON state, you can check the TI app notes

Self-Powered AC Solid State Relay With MOSFETs or Solid State Relay 24-V AC Switch With Galvanic Isolation

Just google those names and you should be able to get the SSR appnotes.

The appnote's tell's how the SSR can be implemented and shows the calculations,the dissipated power can be calculated , most of the power would be dissipated on the Diode.

[wraper]

Thanks for responses, I think I did not make it clear that I want to be able to bypass the diodes so they would not make any dissipation in the ON state, you can check the TI app notes

Several people already said that diode is already bypassed by MOSFET in open state. Is it that hard to understand? I say it again, in this circuit both MOSFETs are in open or closed state simultaneously. In this circuit no significant current ever flows through the diodes.

[JackJones]

https://www.ti.com/lit/ug/tidubr5/tidubr5.pdf

The first picture in this thread makes it look like the gate drive is isolated in which case both of the MOSFETs would be on, bypassing the diodes.

But looking at the appnote it seems that the gate drive shares a common ground which would make current go through the diodes.

I'm sure if there was an easy way to prevent that it would be mentioned in the appnote.

Edit: I had a better look at the schematic, I think I misread it a bit at first glance. I don't think there should be any current through the body diodes.

[wraper]

https://www.ti.com/lit/ug/tidubr5/tidubr5.pdf

The first picture in this thread makes it look like the gate drive is isolated in which case both of the MOSFETs would be on, bypassing the diodes.

But looking at the appnote it seems that the gate drive shares a common ground which would make current go through the diodes.

I'm sure if there was an easy way to prevent that it would be mentioned in the appnote.

I only see how body diodes are used for rectification a bit later, to form a full fridge rectifier together with 2 other diodes. And this works only in off state. As of circuit itself and it's description, looks like someone did not do their job properly and then someone approved this crap.

https://e2e.ti.com/blogs_/b/industrial_strength/archive/2016/07/26/a-modern-approach-to-solid-state-relay-design


Figure 3: Current flow through a MOSFET SSR during on-time (a); and off-time (b)

[langwadt]

https://www.ti.com/lit/ug/tidubr5/tidubr5.pdf

The first picture in this thread makes it look like the gate drive is isolated in which case both of the MOSFETs would be on, bypassing the diodes.

But looking at the appnote it seems that the gate drive shares a common ground which would make current go through the diodes.

I'm sure if there was an easy way to prevent that it would be mentioned in the appnote.

The common ground is the source of the FETs so they will both be either on or off bypassing the diodes

The figure 9 in that app note is simple wrong

[ali_asadzadeh]

Several people already said that diode is already bypassed by MOSFET in open state. Is it that hard to understand? I say it again, in this circuit both MOSFETs are in open or closed state simultaneously. In this circuit no significant current ever flows through the diodes.

Fair enough, But If this is the case and diodes would not dissipate much, why most of the SSR's that are made are big, retaliative to the size of current MOSFETs, we can achieve easily 60-100V VDs and under 10 mOhm Rdson for MOSFETs, with Id in the order of more than 50A,in a 8-PowerTDFN package, like (BSC100N06LS3G), But we can not find any SSR with a current rating of say 25A even 5-10 times the size of these MOSFETs! where does the dissipation came from that the SSR sizes are so big!

[wraper]

Fair enough, But If this is the case and diodes would not dissipate much, why most of the SSR's that are made are big, retaliative to the size of current MOSFETs,

Because it's standard size of SSR and most of them are not meant to be tiny. Mostly made for industrial equipment, not consumer devices. Not to say you forget about insulation.

we can achieve easily 60-100V VDs and under 10 mOhm Rdson for MOSFETs, with Id in the order of more than 50A,in a 8-PowerTDFN package, like (BSC100N06LS3G), But we can not find any SSR with a current rating of say 25A even 5-10 times the size of these MOSFETs! where does the dissipation came from that the SSR sizes are so big!

And how are you going to transfer power from those? 10 mOhm x2 (2 mosfets in series) @ 25A = 12.5W of power to dissipate. Also SSR need to have a lot of margin because of their intended use.

[David Hess]

In the SSR, when it's ON, One of MOSFET's internal Diode would conduct, base on the Voltage sign applied to it! just check the picture, and note the AC voltage across it

When on, both MOSFETs are enhanced.  The one with the diode which would conduct in the forward direction instead has the MOSFET carry the current instead of the diode as the drain-to-source voltage is below the diode's forward voltage drop which will commonly be the case.

Two diodes can be added externally, one in series and one in parallel to both, to each MOSFET to prevent the body diode from conducting in applications where this matters but the common solid state relay does not require this because the forward biased diode is shorted out by the MOSFET conducting backwards.

[fastbike]

I've just stumbled upon this thread when working through how to reduce MOSFET losses in an anti-series (source to source) configuration.

As the TI app note, and the OP, make clear - this is a common and valid configuration for using MOSFETs to switch AC loads.

The loss across the diodes dominates the total loss at low frequencies (i.e 50 or 60 Hz mains) because the diode loss is Vfwd_sd x I_load.

The MOSFET conduction loss is dominated by I_load: formula is I_load² x Rdson

At high frequencies the switching losses come into play as there is frequency component in the formula (the V_ds and I_load are constant)

So IMHO the OP is correct in looking for a solution to reduce the diode losses. Did they ever get a satisfactory answer ?

[Zero999]

I've just stumbled upon this thread when working through how to reduce MOSFET losses in an anti-series (source to source) configuration.

As the TI app note, and the OP, make clear - this is a common and valid configuration for using MOSFETs to switch AC loads.

The loss across the diodes dominates the total loss at low frequencies (i.e 50 or 60 Hz mains) because the diode loss is Vfwd_sd x I_load.

The MOSFET conduction loss is dominated by I_load: formula is I_load² x Rdson

At high frequencies the switching losses come into play as there is frequency component in the formula (the V_ds and I_load are constant)

So IMHO the OP is correct in looking for a solution to reduce the diode losses. Did they ever get a satisfactory answer ?

Read all of the replies to the thread.

To summarise: the diode never conducts, because it's bypassed by the MOSFET, when it's turned on. MOSFETs work in both directions and would be symetrical, if it wasn't for the fact that one of the body diodes is connected to the source. In fact, there are four terminal MOSFETs, with an additonal substrate connection, which are symetrical, but they're not that common. Checkout the MIC94050.
https://www.microchip.com/wwwproducts/en/MIC94050

[wraper]

I've just stumbled upon this thread when working through how to reduce MOSFET losses in an anti-series (source to source) configuration.

As the TI app note, and the OP, make clear - this is a common and valid configuration for using MOSFETs to switch AC loads.

The loss across the diodes dominates the total loss at low frequencies (i.e 50 or 60 Hz mains) because the diode loss is Vfwd_sd x I_load.

The MOSFET conduction loss is dominated by I_load: formula is I_load² x Rdson

At high frequencies the switching losses come into play as there is frequency component in the formula (the V_ds and I_load are constant)

So IMHO the OP is correct in looking for a solution to reduce the diode losses. Did they ever get a satisfactory answer ?

It's an error from multiple crappy appnotes from TI where they had mistakenly drawn the current flow.
https://www.ti.com/lit/ug/tiduc87a/tiduc87a.pdf?ts=1613458846974&ref_url=https%253A%252F%252Fwww.google.com%252F (page 5)
https://www.ti.com/lit/ug/tiduc91/tiduc91.pdf?ts=1613453059857&ref_url=https%253A%252F%252Fwww.google.com%252F (page 6)
https://www.ti.com/lit/ug/tidubr5/tidubr5.pdf?ts=1613477702154&ref_url=https%253A%252F%252Fwww.ti.com%252Ftool%252FTIDA-00377 (page 12)

But on TI forum there was a discussion about it and TI employee provided correct diagram. It's lame they still did not fix those crappy appnotes. 
https://e2e.ti.com/blogs_/b/industrial_strength/archive/2016/07/26/a-modern-approach-to-solid-state-relay-design

[fastbike]

To summarise: the diode never conducts, because it's bypassed by the MOSFET, when it's turned on. MOSFETs work in both directions and would be symetrical, if it wasn't for the fact that one of the body diodes is connected to the source. In fact, there are four terminal MOSFETs, with an additonal substrate connection, which are symetrical, but they're not that common.

Now I'm a little wiser (I wasn't aware they were bidirectional) so the source connected pair makes sense to block an AC load when off.  That also improves my power dissipation calcs as I was wondering what the V_fwd at 0.9v would do to power losses with any significant current !

Checkout the MIC94050.
https://www.microchip.com/wwwproducts/en/MIC94050

Interesting although that part is designed for low voltage applications so not suitable for my design.

Thanks for your time in explaining the concept.

[fastbike]

It's an error from multiple crappy appnotes from TI where they had mistakenly drawn the current flow.

But on TI forum there was a discussion about it and TI employee provided correct diagram. It's lame they still did not fix those crappy appnotes. 
https://e2e.ti.com/blogs_/b/industrial_strength/archive/2016/07/26/a-modern-approach-to-solid-state-relay-design

That makes a whole lot more sense now
Thanks for the clarification.

[Benta]

What you have to know is, that at many semiconductor manufacturers the application notes are often written by interns in the product marketing department if they have nothing else to do.
Those TI notes seem to be that kind.

[Zero999]

Even the SPICE model knows MOSFETs work backwards. Look at how the voltage falls to well below the normal diode voltage drop, when the MOSFET turns on.

[Benta]

Yeah, it's actually done routinely in buck converters with active rectifier (meaning the normal Schottky is replaced with an N-ch MOSFET).