Can I Parallel 2 Mosfets in VVR?

[bcmatt]

Just for knowledge

At the begin I didn't understand well why you want to use a pair of Mosfet in parallel for VVR

for the same reason ....

I can't imagine a tube amp needing parallel MOSFETs for VVR.

however reading your post I remembered a circuit a friend sent to me some years ago

Double Mosfet VVR.jpg

in this circuit you have 2 mosfet that aren't in parallel, they are in series

the first Mosfet (on the left) act as Voltage Regulator the second one (on the right) works like a short

if the first Mosfet blows and go short the second Mosfet replace the first as Voltage regulator

Why this is wanted ? Because if you use the output as B+ or use it to supply a G2 (especially in a tube that don't like High Voltage as B+ on G2 and require a less voltage to feed it) if the Mosfet burns you easily burn also the tube, instead, this way, if one mosfet burns you have the other that act as safety component

Franco

Interesting! But would the user know when the first one blew and they were operating on their last Mosfet?

[Kagliostro]

Unfortunately no till the second one don't blast

Franco

[Ten Over]

Those MOSFET's have wild variation in their Gate Threshold Voltage. If a parallel pair were not well matched, I could see one conducting most of the current while the other is barely turning on. To make matters worse, the Gate Threshold Voltage decreases as the temperature rises making the conducting MOSFET an even bigger hog as it warms up more than the other one.

You might want to consider a series arrangement as Franco suggested.

VVR Basic Two MOSFET.png

[Kagliostro]

One other way, less simple but I think working, will be to build a pair of complete VVR and to connect it in seires

then adjust the first pot as the first VVR lower the voltage by 1/2 of what you require and the second one to reduce the remaining excess of voltage

this way you are sure the power dissipation will share between the two mosfet

Franco

[bcmatt]

One other way, less simple but I think working, will be to build a pair of complete VVR and to connect it in seires

then adjust the first pot as the first VVR lower the voltage by 1/2 of what you require and the second one to reduce the remaining excess of voltage

this way you are sure the power dissipation will share between the two mosfet

Franco

I like the thought process here.
Although, I wonder if it would actually help much.
I had heard (and I don't fully understand why) that the Mosfets actually produce the most heat around the middle of the reduction area. They had said it was actually easier for them closer to full voltage as well as the more extreme bottom 1/3 voltage setting. Can someone verify or explain this?

[Kagliostro]

They had said it was actually easier for them closer to full voltage as well as the more extreme bottom 1/3 voltage setting. Can someone verify or explain this?

As far as I can know the mosfet are used in this application as Linear redoucers (something like a rheostat)

I can agree that if you reduce the voltage of only a little fraction dissipation will be small

but I'm not able to figure why at 1/3 of the voltage as output the dissipation will not be high

---

The way to have a low dissipation on the component will be to use a PWM circuit (more difficult to be implemented and that induces other problems depending on the frequency of work), not a viable path for our simple purposes

Franco

[martin manning]

but I'm not able to figure why at 1/3 of the voltage as output the dissipation will not ne high

At very low voltage the current draw is reduced. Power is I^2R, and the reduction in I^2 becomes dominant.

[Ten Over]

I had heard (and I don't fully understand why) that the Mosfets actually produce the most heat around the middle of the reduction area. They had said it was actually easier for them closer to full voltage as well as the more extreme bottom 1/3 voltage setting. Can someone verify or explain this?

When the voltage drop across the MOSFET is larger, the voltage on the tube plate is smaller. When the voltage on the tube plate is smaller, the tube draws less current. So when the drain-to-source voltage increases, the drain-to-source current decreases. Let's graph that:

MOSFET Dissipation 1.JPG

The more power that a MOSFET is handling, the hotter it gets. The maximum power is dead center on the graph and the power tapers off toward each extreme.

With the voltage split equally between two MOSFET's in series, the voltages on the X-axis will be one-half and the power at each step will be one-half. The power will still peak at the mid-point, but it will be half the single MOSFET power.

[Kagliostro]

With the voltage split equally between two MOSFET's in series, the voltages on the X-axis will be one-half and the power at each step will be one-half. The power will still peak at the mid-point, but it will be half the single MOSFET power.

English isn't my language, please can you explain this thing with other words

Thanks

Franco

p.s.: Do you mean that also power is divided in two parts so if one Mosfet is subject to 10W dissipation having two mosfet each one is subject to 5W dissipation ? ( 5W + 5W = 10W)

[Ten Over]

p.s.: Do you mean that also power is divided in two parts so if one Mosfet is subject to 10W dissipation having two mosfet each one is subject to 5W dissipation ? ( 5W + 5W = 10W)

Yes.

[martin manning]

It's the same as placing two same-value resistors in series to reduce voltage instead of one at the same total resistance. In the case of two resistors, half of the voltage is dropped in each one, current is the same, so half the power (V*I) is dissipated in each.

[Kagliostro]

Thanks Ten Over & martin manning

Previously I understand exactly this but was not sure you were saying that

With my post of in series mosfet I was trying to say the same thing

Franco

[martin manning]

When the voltage drop across the MOSFET is larger, the voltage on the tube plate is smaller. When the voltage on the tube plate is smaller, the tube draws less current. So when the drain-to-source voltage increases, the drain-to-source current decreases. Let's graph that:
MOSFET Dissipation 1.JPG
The more power that a MOSFET is handling, the hotter it gets. The maximum power is dead center on the graph and the power tapers off toward each extreme.

The maximum power dissipation will be at 50% voltage reduction if the load is purely resistive. Since vacuum tube current is proportional to voltage to the 3/2 power, it drops off faster. With that assumption, maximum dissipation is lower, and it occurs at around 40% reduction in voltage.

[Ten Over]

When the voltage drop across the MOSFET is larger, the voltage on the tube plate is smaller. When the voltage on the tube plate is smaller, the tube draws less current. So when the drain-to-source voltage increases, the drain-to-source current decreases. Let's graph that:
MOSFET Dissipation 1.JPG
The more power that a MOSFET is handling, the hotter it gets. The maximum power is dead center on the graph and the power tapers off toward each extreme.

The maximum power dissipation will be at 50% voltage reduction if the load is purely resistive. Since vacuum tube current is proportional to voltage to the 3/2 power, it drops off faster. With that assumption, maximum dissipation is lower, and it occurs at around 40% reduction in voltage.

So? I neither explicitly stated nor did I mean to imply that the actual plate-to-cathode voltage had a linear relationship with the actual plate-to-cathode current. I merely wanted to show the direction each one went relative to the other in order to illuminate a basic thermodynamic phenomenon for the OP. The straight line on the graph was chosen for simplicity and clarity. A curve would only have served to complicate the demonstration and to add a degree of confusion.

[martin manning]

No slight intended, I just thought it made an interesting plot.