Fanuc said:
MOSFETs in a common source connection will tend to have a higher input capacitance for two reasons. First, you will see the full gate-source capacitance because the source is not moving with the gate signal (unless it is heavily degenerated). Secondly, you will likely get Miller multiplication of the gate-drain capacitance by the amount of the common-source gain of the stage. These two effects together can add up to quite a bit.
In my design, if I drove the main output rails with a series regulator I would arrange it so that I could still have at least 1000 uF of reservoir capacitance AFTER the regulator to supply transient current needs. The regulator then needs to be designed to be stable with such a load. A soft regulator or a capacitance-multiplier type regulator would tend to be stable with such a load capacitance, at the expense of the output being not quite as stiff; i.e., the output impedance might be a few tenths of an ohm, rather than milli-ohms. In my designs, I refer to this as a soft rail regulator, and its main component is a MOSFET running essentially as a source follower with an appropriate voltage fed to the gate. I do not close the loop from the output of the regulator back to the gate.
The JLH design is a fairly simple and straightforward design. Its not great in my opinion because he drives the output MOSFETs directly from the VAs without buffering. I don't know how much current he runs through the VAS. I also wonder how much high frequency nonlinearity he gets from the gate-drain capacitance of his MOSFET VAS transistor - I don't know how big that transistor is.
Bob
Hi all
just an observation on the last couple of posts - I think it is a bit of a misconception to assume that the output MOSFET gate capacitance is really bootstrapped. That is only true at low frequencies. If low distortion is required at high frequencies, the MOSFET gate capacitance has to be driven as though it were common source, not common drain.
I built an early MOSFET amp similar to JLH's as discussed with 2SJ48/2SK135 outputs. It sounded dull, comparable with old 2N3055 BJT's which was really disappointing at the time.
When I later compared the gate drive charge between MOSFETs and bipolars, I found that the charge needed was almost the same in both cases. The effective capacitance of a forward-biased bipolar junction is enormous - really, this is charge in the base - which turned out to be very similar to MOSFET gate Qg.
The only differences are that the MOSFET has a lower capacitance but needs a higher voltage swing (typically 5V) compared with the bipolar which operates at lower voltage (less than 1V swing) but needing much higher currents to drive.
At that time I realised that MOSFET amps better have those drivers!
cheers
John
just an observation on the last couple of posts - I think it is a bit of a misconception to assume that the output MOSFET gate capacitance is really bootstrapped. That is only true at low frequencies. If low distortion is required at high frequencies, the MOSFET gate capacitance has to be driven as though it were common source, not common drain.
I built an early MOSFET amp similar to JLH's as discussed with 2SJ48/2SK135 outputs. It sounded dull, comparable with old 2N3055 BJT's which was really disappointing at the time.
When I later compared the gate drive charge between MOSFETs and bipolars, I found that the charge needed was almost the same in both cases. The effective capacitance of a forward-biased bipolar junction is enormous - really, this is charge in the base - which turned out to be very similar to MOSFET gate Qg.
The only differences are that the MOSFET has a lower capacitance but needs a higher voltage swing (typically 5V) compared with the bipolar which operates at lower voltage (less than 1V swing) but needing much higher currents to drive.
At that time I realised that MOSFET amps better have those drivers!
cheers
John
In bipolars we can calculate the input and output impedance. For example, if we have NPN bipolar with Hfe=100. Used as emitor follower.
If the base impedance is 100ohm, then the emitor impedance is 1ohm.
The other way around, if the emitor impedance is 100ohm, the at the base we will se 10k impedance.
Is mosfets is totally free from this mechanism? No matter how high or how low the source impedance is, the impedance seen at the gate will not affected?
What is the impedance of mosfet's gate?
If the base impedance is 100ohm, then the emitor impedance is 1ohm.
The other way around, if the emitor impedance is 100ohm, the at the base we will se 10k impedance.
Is mosfets is totally free from this mechanism? No matter how high or how low the source impedance is, the impedance seen at the gate will not affected?
What is the impedance of mosfet's gate?
Hi,
The disagreement we often see is in trying to evaluate the effective capacitance/change of charge seen from the preceeding stage.
On top of that there is the Millar/Miller? effect where the capacitance is multiplied by the gain in the stage, now the variation in capacitance is also multiplied by the gain. Oh, what the heck, let's just swamp it with a big Miller compensation cap and it will bury the distortion, but not cure/eliminate it and ensure stability as well.
near enough it's a pure capacitor. The problem is the capacitance varies with the voltage applied to the gate i.e. non linear. A source resistance combined with this variable capacitance leads to distortion.
The disagreement we often see is in trying to evaluate the effective capacitance/change of charge seen from the preceeding stage.
On top of that there is the Millar/Miller? effect where the capacitance is multiplied by the gain in the stage, now the variation in capacitance is also multiplied by the gain. Oh, what the heck, let's just swamp it with a big Miller compensation cap and it will bury the distortion, but not cure/eliminate it and ensure stability as well.
Hi Lumanauw
At low frequencies your calculations could apply: a load of 1 ohm with a gm of 1A/V requires 2V input to achieve 1v out, but the input current is negligible. The input current remains negligible for almost any load, at low frequencies.
In general terms, the input impedance at low frequencies might be regarded as "very high".
But if you are driving a low impedance load, say a loudspeaker at 8 ohms, and with a relatively large current swing (5A or so) then the MOSFET gate capacitance has to be charged to get the MOSFET to provide the current, and this is when a substantial drive current might be necessary.
The input impedance for most frequencies will be capacitive and this is obviously frequency dependent. There may also be an equivalent series resistance for the gate, which will affect the response near "ft".
Further another reason that a driver transistor is needed for a MOSFET is that this input impedance (capacitance, mostly) is non-linear. Buffering will remove this dependency from the VAS somewhat.
The answer of input impedance therefore depends on frequency as I am sure you would expect, and load impedance, and gm ... so there is no single answer. Just points to take note of!
cheers
John
At low frequencies your calculations could apply: a load of 1 ohm with a gm of 1A/V requires 2V input to achieve 1v out, but the input current is negligible. The input current remains negligible for almost any load, at low frequencies.
In general terms, the input impedance at low frequencies might be regarded as "very high".
But if you are driving a low impedance load, say a loudspeaker at 8 ohms, and with a relatively large current swing (5A or so) then the MOSFET gate capacitance has to be charged to get the MOSFET to provide the current, and this is when a substantial drive current might be necessary.
The input impedance for most frequencies will be capacitive and this is obviously frequency dependent. There may also be an equivalent series resistance for the gate, which will affect the response near "ft".
Further another reason that a driver transistor is needed for a MOSFET is that this input impedance (capacitance, mostly) is non-linear. Buffering will remove this dependency from the VAS somewhat.
The answer of input impedance therefore depends on frequency as I am sure you would expect, and load impedance, and gm ... so there is no single answer. Just points to take note of!
cheers
John
john_ellis said:
John,
I totally agree that MOSFET output stages want to have decent drivers.
However, I disagree with you on the bootstrap issue. Even out to quite high frequencies, the source in a source-follower will follow the gate voltage to within about 10%. In actuality, it differs only to the extent that the source follower gain is slightly less than unity driving the load. This effecyively bootstraps the gate-source capacitance, reducing it by about 90%. HOWEVER, over the full current swing of the output circuit, one still must supply the amount of current to the gate that it takes to move the gate voltage from its value at low currents to its value at the peak currents. This will usually be a couple of volts. But the current that it took to do this was in reference to the full swing, so this is another way of saying that the effective input capacitance is lower. So if, for example, you were looking to find out where the effective pole was at the input as a result of driving the gate from some source resistance, the bootstrapped value of input capacitance is the proper way to look at it and calculate the pole frequency.
No matter what, one needs to look at the actual current requirements from the driver to swing the output stage over the voltage range needed at the rate of change needed.
Cheers,
Bob
Bob
You are of course correct in saying that the input impedance will be higher for a source follower than for a common source FET. I should have been more careful!
I hope I corrected this in my second response - this makes it clear that the load, gm etc are all to be considered.
However, regarding actual bootstrapping, one reason real bootstrap circuits might not have been so good is that they fail at the same rate as the frequency response of the output device. This is basically, I think, what spoils JLH's original 69 desgin using slow output devices, and why I now recommend the faster trannys.
cheers
John
You are of course correct in saying that the input impedance will be higher for a source follower than for a common source FET. I should have been more careful!
I hope I corrected this in my second response - this makes it clear that the load, gm etc are all to be considered.
However, regarding actual bootstrapping, one reason real bootstrap circuits might not have been so good is that they fail at the same rate as the frequency response of the output device. This is basically, I think, what spoils JLH's original 69 desgin using slow output devices, and why I now recommend the faster trannys.
cheers
John
HI Bob
but... just a further comment to clarify why I said at high frequencies the input impedance may seem as though it were common source.
There is a drain to source parasitic capacitance, and this will progressively by-pass the source loading, whatever the source load.
I'd agree that this depends on the frequency, but power MOSFETs tend to have quite high values of capacitance ...
cheers
John
but... just a further comment to clarify why I said at high frequencies the input impedance may seem as though it were common source.
There is a drain to source parasitic capacitance, and this will progressively by-pass the source loading, whatever the source load.
I'd agree that this depends on the frequency, but power MOSFETs tend to have quite high values of capacitance ...
cheers
John
john_ellis said:
Hi John,
While in principle the drain-source parasitic capacitance can have an effect in regard to the bootstrapping function versus frequency, when we plug in the numbers, we see it is pretty far out in frequency.
The drain-source capacitance will have its 3-dB "bypass" effect on the source load in a source follower configuration at a frequency where the impedance of the drain-source capacitance is approximately equal to 1/gm.
For an IRFP240, Cds varies from a high of about 1000 pf at 0V Vds to 550 pF at Vds = 5V, then lower beyond that. GM for the device ranges from 0.9S at Id = 300 mA up to 2.6S at 1.3A and upward beyond that.
If we take a fairly high value of Cds of 1000 pF and a fairly low value of GM of 0.9S, we see that the corner frequency of Cds against 1/gm is at 160 MHz - pretty far out.
Cheers,
Bob
Workhorse said:
The expense of a few driver transistors in a high power BJT amplifier in my opinion is pretty much trivial. MOSFET’s do require less drive current, but for the best performance (especially with multiple paralleled output devices) they require a low impedance drive source as well.
In a much more efficient way? I think you’re generalising a bit. BJT’s are cheap. Fabricating or having manufactured a copper heatsink interface is not.
Cheers,
Glen
Wrong. Copper bus bars are cheap on eBay and they're easy to cut up and drill holes for mounting. As a further improvement, I use beryllia wafers instead of mica or silpads, as they have even better heat conductivity than the aluminum of the heatsink. If you cut them under water with a handheld rotary tool and a cheap diamond wheel (50 cents on eBay), there's no danger from the dust.G.Kleinschmidt said:
Nixie said:
Ummm.....I was elaborating on a manufacturing cost comparison. Workhorse is talking about his commercial designs, not DIY projects made with bits sourced from eBay. The material cost of the copper is not the only consideration; there are manufacturing costs to consider as well……………
As I already wrote, but of course this is diyaudio so I have to repeat myself, the copper bus bars are very easy to work: just cut off a length that's large enough to it the power devices, and drill a hole for each one; drilling the copper is very easy and it can be done once it has been mounted on the heatsink, so that the number of holes drilled is the same as if the copper were omitted. As for your "bits sourced from eBay comment", I fail to see the cause for your derision towards this source--it was just an example, and obviously it would be even cheaper if one did a volume order for manufacturing. Copper is still a pretty cheap metal, and will remain so until a large portion of cars switch to electric.
G.Kleinschmidt said:
Glen, I tend to agree, and these comments largely apply to either MOSFETs or BJTs. For best performance, they both need good drive buffering, and driver transistors are generally a small part of the cost of an amplifier. Although the use of a copper bus bar is a nice idea, copper has gotten very expensive, and it is another piece part in addition to the heat sink itself. Finally, paralleling devices seems to be a necessary evil in larger power amps, and just from a thermal point of view it is hard to get around the basic limitation of the contact area of a given package like a TO247.
I like MOSFETs and they are easier to drive, but I think we all agree that that doesn't make all the problems go away. I will be the first to admit, for example, that designers using MOSFETs need to have a better understanding of high frequency issues and matters that can lead to high frequency parasitic oscillations.
Cheers,
Bob
G.Kleinschmidt said:
BJT's are much easier to implement in an amplifier for a given power output...A good complementary dual differential input with cascode loaded Vas and Triple Deep Darliington EF stage and a logically applied feedback network and you have something good thing to play reliable and sweet.....The Traditional way!
Have you ever Tried Vertical MOSFETS ? The Game is different here!
Bob Cordell said:
Mosfets do require good low impedance gate drive but not as low as BJT's need....
TO-264 [APT/IXYS]package is 1.7 times that of TO-247 [IRF HEXFETS]package...but it does require copper base to extract the heat from the package contact surface area
A simple example
In Class-TD [Tracking Rails not Class-H/G]
To obtain 2KW @ 2 ohms we need at least 4 pairs of 2SC5200/2SA1943 Bipolars
But only 2 pairs of IRFP260N Mosfets or one pair of APT20M22LVR all needed to obtain the same power ...this reduces PCB space and associated parts and boosts reliability
Cheers,
Kanwar
Nixie said:
Umm, I was not deriding eBay as a source of parts for DIY audio, or DIY anything, for that matter. I was simply pointing out the fact that your comments were tangential to the point I was making.
The topic I was discussing was the difficulty in effectively heatsinking a 520W MOSFET that comes in a TO-264 package VS the ease with which multiple BJT’s (or even lower power MOSFET’s, for that matter) parallel connected to give an equivalent SOA can be heatsunk.
If you wanted to get, say 400W out of a single MOSFET such as that APT part without cooking the buggery out of it, you’d need a ‘ell of a lot more than a few inches of copper bus bar.
The difficulty in heatsinking such a device arrises from the fact that the heatsink must have sufficient surface area to effectively transmit the heat from the dissipating device to the ambient air, but the efficiency of the heatsink degrades rapidly if a large proportion of the heatsink’s surface area is not in close proximity to the source of the heat.
It is not strictly the surface area of the actual TO-220, TO-247 or TO-264 package that is the issue here (Which some have assumed). It is the surface area of the heatsink with respect to the mechanical distribution of the dissipating devices connected to it.
For example, four individual transistors dissipating 100W each, evenly distributed over a 400mm long heatsink will run with a significantly lower junction temperature than would a single device dissipating 400W bung onto the middle of the same heatsink.
And yes, volume production of a part is cheaper than one-off tooling and manufacture, but BJT’s and 200W MOSFET’s bought in bulk are cheaper than when bought in one-off quantities too.
Cheers,
Glen