I was simulating yesterday a few power MOSFETs, academically, just to see how they work, and whether they provide an advantage over a complementary transistor push-pull AB stage, or using them instead of a pass darlington transistor pair, and I have discovered that depending on the MOSFET used, they typically drop from 3V to 6V on the Gate to Source.
That means that it will be impossible to get close to the rail(s) ans that would preclude their use for whichever application requires output voltages near the rails.
Have I got that right ?
That means that it will be impossible to get close to the rail(s) ans that would preclude their use for whichever application requires output voltages near the rails.
Have I got that right ?
Yes.
But look at CFP, where the Vgs is not lost from the rail voltage. It is Vds sat that is lost and Vds sat can be very low.
The BIG disadvantage of running any output stage into saturation (applies to EF/SF and CFP) is how slowly the saturated device comes out of saturation. It is not where any audio amplifier should be.
An amplifier generally responds more quickly from overloading by allowing the drivers stages to clip. You will find quite a few designers inserting Baker clamps and similar to make sure the driver stages clip softly and thereby preventing saturation in the output stage and the driver stages.
Using raised supply rails to the driver stage is a certain way to get output stage saturation if you overload your amplifier.
But look at CFP, where the Vgs is not lost from the rail voltage. It is Vds sat that is lost and Vds sat can be very low.
The BIG disadvantage of running any output stage into saturation (applies to EF/SF and CFP) is how slowly the saturated device comes out of saturation. It is not where any audio amplifier should be.
An amplifier generally responds more quickly from overloading by allowing the drivers stages to clip. You will find quite a few designers inserting Baker clamps and similar to make sure the driver stages clip softly and thereby preventing saturation in the output stage and the driver stages.
Using raised supply rails to the driver stage is a certain way to get output stage saturation if you overload your amplifier.
No.
The FET gate-to-source current is ideally zero, and in that case there would be no voltage drop. The gate is electrically insulated from the source, so I'm unsure of your simulation setup which led to your hypothesis. But a high side FET driver can definitely put a load at very near the supply voltage.
The FET gate-to-source current is ideally zero, and in that case there would be no voltage drop. The gate is electrically insulated from the source, so I'm unsure of your simulation setup which led to your hypothesis. But a high side FET driver can definitely put a load at very near the supply voltage.
Possibly, depending on your type of output configuration. If you use an N channel on the negative portion (inverting) and a P channel on the positive portion (also inverting), there is no problem as the sources stay around their respective rails allowing the gates to drive at full voltage, typically protected with a Zener diode. Otherwise a conventional complimentary pair just require a bootstrap capacitor to move the gate voltage supply higher than the rails.
But the high side gate driver is required precisely because the gate must be brought several volts higher than the source in order to get that saturation you write about.
akis, your simulation is correct.
Here are some pictures of what I mean: there are 4 simulations, the first and fourth show no current passing through, even though we have a hefty voltage at the gate. The MOSFET starts to open at about 3.9V. Very similar, or worse results can be obtained with other MOSFETs I have tried. A comparable transistor would only drop around 650mV.
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you really don't want to run power Mosfet at low Vds anyway - for linear stability reasons
there is a large nonlinear jump in the device parasitic C when you get closer than ~2*Vth to the rails
if you are at all pushing the loop gain compensation you can run into a zoo of problems when you get close to saturation
I have simmed "gain scheduled" compensation to help but don't know if it works real world since spice models may be lacking in detail
there is a large nonlinear jump in the device parasitic C when you get closer than ~2*Vth to the rails
if you are at all pushing the loop gain compensation you can run into a zoo of problems when you get close to saturation
I have simmed "gain scheduled" compensation to help but don't know if it works real world since spice models may be lacking in detail
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Thanks. I am trying to see how close to the rails I can go because I am working with batteries and I cannot afford to throw away 3-4 volts. One application is very similar to an audio amplifier output stage class AB (except it is at 200KHz fixed), another application is a linear power supply type voltage controller, where I have to feed the gate, say, 10V to get 7V at the output. In both cases I lose less volts with a Darlington or even less with a complementary pair.
I am not sure I have done it right, is that what you mean? If yes, we still lose 3V and the MOSFET is still not conducting.
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The MOSFET needs a certain Vgs to conduct, so if you use a follower conficguration, the highest you can go on the outout is then the gate is at the same potential as the drain, i.e power supply. How much voltage drop there will be depends on MOSFET treshold and transcondutance, i.e. the current you need (more current = more Vgs is needed, but unless the treshold is crossed practically no current passes at all).
For regular vertical or HEXFETs, it takes about 6V Vgs to get the to conduct full current at that same voltage across Vds. The only way around that is using an extra power supply of some sort (this includes bootstrapping) to get Vg 'above' the rail the drain ios tied to, in effect making Vg more than Vd, so that you can produce the requied Vgs even when Vs is almost equal to Vd i(which is itself tied to the power rail).
The alternative is to use a common source conficuration where the source stays tied to the supply rail and Vgs is always between the supply rails up to the required voltage for full conduction. Alternatively you can use the inverted common source where the sources are at ground potential, and the drains are tied to a floating power supply where the midpoint of the power supply is the amps output.
In all cases driving the MOSFETs to Vds ~< 2Vth gets you into problems with nonlinearly and rapidly increasing input capacitances. The common source configuration also increases apparent input capacitance by multiplying the reverse transfer capacitance (Cgd) through the Miller effect.
Incidentally, this is why switching circuits rely on low charge storage MOSFETs for efficient operation as they do indeed try to drive the MOSFET to the lowest possible Vds as this is directly responsible for heat loss. This also applies to class D amplifiers, their output is a switching circuit, and one of the reasons for their high efficiency.
For regular vertical or HEXFETs, it takes about 6V Vgs to get the to conduct full current at that same voltage across Vds. The only way around that is using an extra power supply of some sort (this includes bootstrapping) to get Vg 'above' the rail the drain ios tied to, in effect making Vg more than Vd, so that you can produce the requied Vgs even when Vs is almost equal to Vd i(which is itself tied to the power rail).
The alternative is to use a common source conficuration where the source stays tied to the supply rail and Vgs is always between the supply rails up to the required voltage for full conduction. Alternatively you can use the inverted common source where the sources are at ground potential, and the drains are tied to a floating power supply where the midpoint of the power supply is the amps output.
In all cases driving the MOSFETs to Vds ~< 2Vth gets you into problems with nonlinearly and rapidly increasing input capacitances. The common source configuration also increases apparent input capacitance by multiplying the reverse transfer capacitance (Cgd) through the Miller effect.
Incidentally, this is why switching circuits rely on low charge storage MOSFETs for efficient operation as they do indeed try to drive the MOSFET to the lowest possible Vds as this is directly responsible for heat loss. This also applies to class D amplifiers, their output is a switching circuit, and one of the reasons for their high efficiency.
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akis,
MOSFETs can be used in the output stage of an amplifier. Just use a bootstrap type circuit in the driver stage.
MOSFETs can be used in the output stage of an amplifier. Just use a bootstrap type circuit in the driver stage.
For the class-ab output look at the upper P-MOS and lower N-MOS as Jon pointed out.
For the low drop out regulator you can use a P-MOS as the pass device and an op-amp. I've got a schematic I drew somewhere, but I can't locate it at the moment.
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