A thought occurred to me as I read a line in the X1000 Bio...80 Mosfets per channel...How do you drive 80 mosfets with all the gate capacitance and have any high end left?
Does the nature of the X-series output stage negate that? or what am i missing?
Zc😎
Does the nature of the X-series output stage negate that? or what am i missing?
Zc😎
One way to do it is to partition the set of 80 MOSFETs into groups. For example: 60 output stage devices, 16 driver devices, 4 predriver devices. Now each MOSFET has a "fanout" of 4. That's not a challenging load at all.
Another way to do it is to bootstrap both Cgs and Cgd. Namely, arrange the circuit design so both the drain waveform and the source waveform, exactly follow the gate waveform. Thus Vgs and Vgd are constant, they never change. Since dQ = C*dV , if dV equals zero then dQ also equals zero. The effective capacitance at the gate terminal is zero. That's not a challenging load at all.
Whether the amplifier you are studying uses these, or other, approaches: I have no idea.
Another way to do it is to bootstrap both Cgs and Cgd. Namely, arrange the circuit design so both the drain waveform and the source waveform, exactly follow the gate waveform. Thus Vgs and Vgd are constant, they never change. Since dQ = C*dV , if dV equals zero then dQ also equals zero. The effective capacitance at the gate terminal is zero. That's not a challenging load at all.
Whether the amplifier you are studying uses these, or other, approaches: I have no idea.
Well, it's really simple. Think about current slew rate.
Let's say the amp needs to raise its output voltage to follow the signal. In order to do that, more current must be supplied to the load. Therefore, the Vgs of the MOSFETs must be increased so that they pass the required current.
Your MOSFET has a transconductance Gm. If the load needs an extra current dI, then Vgs needs to rise by dI/Gm volts.
Now, if you got N MOSFETs in parallel, but each MOSFET has the same Gm than in the previous case, then the current in each MOSFET only needs to rise by deltaI divided by N. Therefore, the Vgs of each MOSFET will need to move N times less also.
The gate drive current (which is proportional to the MOSFET's capacitances multiplied by dVgs and N) will therefore not be that different whether you got 1 FET or 10... Gm will probably be less with many FETs unless the idle current is increased a lot, so the drive current should increase, but not by N times.
Now, if you increase the required current by increasing the amplifier rated power, then drive current will increase, but it is not proportional to the number of MOSFETs.
In other words, no driving problems.
Also, if you don't want your MOSFET to oscillate, you need to add a gate resistor. This creates a pole. Many small MOSFETs, each with its gate resistance, will have a higher pole than a huge MOSFET with the same resistance, even if the total capacitances in both cases are the same. This means several smaller FETs in parallel will have higher open loop bandwidth that a single large one.
FYI, let's consider a FET with:
Input cap = 1nF
Gm = 1 S
And we play a 20kHz sine at 2 amps peak amplitude.
If you do the math, you'll notice the drive current is below 500 µA...
You'll only need high drive if you let the amp clip and squeeze the FETs against the rails. Then yeah, to come out of clipping you need to pump the gate charge out.
Let's say the amp needs to raise its output voltage to follow the signal. In order to do that, more current must be supplied to the load. Therefore, the Vgs of the MOSFETs must be increased so that they pass the required current.
Your MOSFET has a transconductance Gm. If the load needs an extra current dI, then Vgs needs to rise by dI/Gm volts.
Now, if you got N MOSFETs in parallel, but each MOSFET has the same Gm than in the previous case, then the current in each MOSFET only needs to rise by deltaI divided by N. Therefore, the Vgs of each MOSFET will need to move N times less also.
The gate drive current (which is proportional to the MOSFET's capacitances multiplied by dVgs and N) will therefore not be that different whether you got 1 FET or 10... Gm will probably be less with many FETs unless the idle current is increased a lot, so the drive current should increase, but not by N times.
Now, if you increase the required current by increasing the amplifier rated power, then drive current will increase, but it is not proportional to the number of MOSFETs.
In other words, no driving problems.
Also, if you don't want your MOSFET to oscillate, you need to add a gate resistor. This creates a pole. Many small MOSFETs, each with its gate resistance, will have a higher pole than a huge MOSFET with the same resistance, even if the total capacitances in both cases are the same. This means several smaller FETs in parallel will have higher open loop bandwidth that a single large one.
FYI, let's consider a FET with:
Input cap = 1nF
Gm = 1 S
And we play a 20kHz sine at 2 amps peak amplitude.
If you do the math, you'll notice the drive current is below 500 µA...
You'll only need high drive if you let the amp clip and squeeze the FETs against the rails. Then yeah, to come out of clipping you need to pump the gate charge out.
The amplifier you mentioned has balanced outputs, so the 80 gets divided
into 20 pairs to be driven, and because it's balanced, the slewrate is
multiplied by 2, so it's as if it was 10 pair.
The earlier X amplifiers had slew rates somewhere above 50 v/us. When you
parallel outputs, the Ciss and Crss go up proportionally, but the apparent
Ciss is dependent on the transconductance, which has also gone up, so
the current required to charge Ciss increases less than you would think.
into 20 pairs to be driven, and because it's balanced, the slewrate is
multiplied by 2, so it's as if it was 10 pair.
The earlier X amplifiers had slew rates somewhere above 50 v/us. When you
parallel outputs, the Ciss and Crss go up proportionally, but the apparent
Ciss is dependent on the transconductance, which has also gone up, so
the current required to charge Ciss increases less than you would think.
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