Ft15k mosfet paralleling

[Flavio88]

how i can paralleling another output pair on this amp to let the fet's work with less current, and also obtain more power????

[Twaksak]

To add another mosfet pair to this amp I would connect another base stopper resistor the same value as R12, R15 to the node between R11,R12 and from there to the gate of the second mosfet that you want to add. The drain and source connecting to V+ and output as per the original mosfet T6. The negative side is similar with base stopper resistor between node R14,C6.

Practical though the resistor must be close to the gate of the mosfet and be carefull of long connections that cause oscillation problems.

Chris

[Flavio88]

ok
then 100 ohms resistors to the gates.
In the original circuit there isn't the classical source resistors(small value high power), i think that now with 2 pairs of fets these resistors are needed for current sharing...

what do you think?

[AndrewT]

Hi,
change the angle to double the length and hard wire the extra resistors between the legs (leads).

Add the source resistors to help with current sharing.

Test all your new and old FETs and select the closest pair matches for each channel.

You will not get extra power from this modification.
You need to change either the load impedance or the supply voltage or both to change the power.

If you change the supply voltage it will change the current flows in the resistors and upset the balance in the design parameters.
This amp will suffer badly due to rail modulation.

The extra gain in the doubled pairs may require different stabilisation components.

[Twaksak]

Hi Flavio88

Technically mosfet transistors are NTC devices and do not need source resistors to force current sharing. Are you re designing the amplifier circuit with higher voltage rails to supply more power?

Chris

[AndrewT]

Hi,

mosfet transistors are NTC devices and do not need source resistors to force current sharing

I don't agree, even lateral FETs need matching AND source resistors to try to balance the currents between devices.

I don't want one device of a group running hot. It is then likely to fail if stressed and take out it's lazy partners.

There is at least one manufacturer using common source in a vertical FET that does not use source resistors, but he goes for what I would call super matching and is careful to specify domestic duty for his power amps.

[Twaksak]

Hi AndrewT

I also use source resistors to help current sharing. The word technically says in theorie the harder working mosfet would warm up more first, then conduct less due to negative temperature cooficient properties thus forcing his partner to do more work. In this way sourse resistors are not needed.

I have seen Mosfet amplifier output stage schematics that do not incorporate source resistors. How your design principle is will be your own choice is what I think

Chris

[AndrewT]

Hi,

then conduct less due to negative temperature cooficient properties thus forcing his partner to do more work.

this is the bit we diverge on.

Yes the hotter device tries to get it's partners to take some of the share, but they NEVER take an equal share of the quiescent current. They do move a little closer. The problem, let's say with three pair output stage, is that one could be low on Vgs and the other five closely matched. That one WILL run hot and the five wil run cooler.

Even with source resistors the quiescent sharing only moves a very little closer. I think the source resistors have a much more effective function in sharing output current.

But I come back to the device that runs hotter. On average the hot one will fail first when the amp is stressed.

[unclejed613]

choose mosfets with super low RDSon. that increases the available current from each device. this also makes the device run cooler.

[AndrewT]

Hi Uncle,
I think Rds on only affects voltage loss and power dissipation when at or near saturation (if that is the correct term for a FET).

At quiescent operating conditions, Rds on is not relevant.
The dissipation in each device is Pd=[device Iq*Vds]
A low Vgs will demand higher Id compared to it's partners and this is the same as Iq at zero input signal. So, there is an inverse relationship between Vgs and Pd + Tj that higher Tj can never compensate for.