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Old 25th October 2007, 02:21 PM   #1
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Default How much current to drive output MOSFETs?

How much current is required from the driver to achieve a given speed (dv/dt) from the output stage of a typical power amp? Is it something that can easily be calculated based on the published parameters of the output devices used, or is some empirical rule of thumb the way to go?

This is not intended as a theoretical question but more a case of where to aim when designing a particular amplifier, prior to building and measuring in practice. From my perspective, it would also be nice to understand a little of what influences this aspect of amplifier performance.

Iím sure it is a topic that must have been covered before but I canít find any matches using the search facility. So, Iíve done some thinking that I would like to share in the hope that it will stimulate discussion. Feel free to shoot me down as I may be way off track.

By way of an example intended to make things more specific, I will consider a class A complementary source follower output stage driven by a class A complementary common source configuration. Many of you will recognise this as bearing more than a slight resemblance to more recent NP X type designs

What are the key parameters that influence how fast the output stage can be driven? As I understand it, this comes down to how fast the current from the driver can charge (and discharge) the capacitance presented by the output devices. Simplistically the relationship is given by I = C dv/dt. The difficulty is that C is not constant but instead is a function of DC and AC conditions.

Using the IRFP240 (its complement the IRFP9240 is similar but not identical) as a specific example of an output MOSFET, the various capacitances are given by the manufacturer for a limited set of conditions. In the case of the International Rectifier datasheet, various capacitances are plotted for Vgs = 0, the axes representing capacitance against Vds. Three lines are shown, the most interesting from a driving point of view being Cgs and Cgd. Note that all three have the same general characteristic of being largest when Vds = 0 and decreasing with larger values of Vds, although the graph only goes up to around 25V. Perhaps that is because the lines are approaching horizontal at this point, i.e. it appears there is a minimum value for larger Vds.

So what value of capacitance is appropriate when calculating the required drive current? Vds for each device of the complementary pair in our output stage will vary from almost zero when the device is fully on to close to that of the power supply when it is turned off. In addition, Vgs will be around 4V or more (rather than zero) Ė might this have a significant effect on C?

An observation is that when driving a complementary class A pair, one device of the pair is at its worst case (highest C) when the other is best case. One way of looking at the total C seen by the driver is as the sum of the two which is fortunately better than 2 x worst case. However, Iím unconvinced about this being the correct value to use as it seems to give an unrealistically high current requirement even when other considerations, discussed in the next paragraph, are taken into account.

In a source follower output configuration, Vgs changes relatively little with device current (Ids) due to the effect of feedback from the source resistance and this in turn reduces the effective Cgs seen by the driver by almost an order of magnitude. Unfortunately, the same cannot be said of Cgd and so this tends to predominate.

Using the IRFP240 as an example, the worst case Cgs (Vds = 0, Vgs = 0) is around 1300pF, whilst Cgd is about 1250pF. For our source follower pair the effective worst case Cw is therefore 1380pF (130pF + 1250pF). Best case (Vds = 25V, Vgs = 0), Cgs is around 1250pF whilst Cgd is about 40pF and so total effective Cb is 165pF as a source follower. This gives a combined C for the pair of 1545pF. The drive current required for a relatively modest 50V/uS is therefore 1545pF * 50V/us, i.e. 77mA. For our complementary class A driver pair, this translates to a bias of 38.5mA per device.

A bias of 38.5mA for the driver for a single pair of output devices seems excessive and is near the limit for what a 2-stage X-style front end can easily provide. Given that an amplifier of decent output power will require many such pairs, say up to 10; clearly we have a problem. Or more likely there are flaws in my reasoning.

It occurs to me that the situation is eased slightly in the case of bridge amplifiers like the X-series, since each half of the amplifier need only provide half the overall speed. Another way of looking at this is that each half contributes half the overall voltage swing to the speaker. Even so, 19.25mA bias for each driver device is still rather high and will only allow us to drive two pairs of output devices per half, i.e. a total of 8 output devices in all.

Iím guessing that other factors need to be taken into account to arrive at a more appropriate conclusion. Perhaps the flaw is in using capacitance values based on Vgs = 0 since this is clearly not the value it will be in practice? Maybe the fact that the worst case capacitance occurs only as the amplifier approaches clipping is significant?

Iíd appreciate your thoughts.

Ian.
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Old 25th October 2007, 07:33 PM   #2
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I'll give you a rough ballpark. A reasonable example is
output follower pairs of IRF240/9240.

noting that 1 amp / 1 Farad = 1 Volt / 1 second:

I recall an example where a 80 mA drive gave about 40 V/uS
slew into 12 pairs, which works out to be about a 2 nF load,
which gives about 200 pF per pair.

You can apply similar logic to other devices using the Cgd as
your most important number.

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Old 25th October 2007, 08:37 PM   #3
gl is offline gl  United States
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Nelson,

Was that 80 ma of bias in the driver circuit or 80 ma peak?

Graeme
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Old 25th October 2007, 08:46 PM   #4
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From discussions with Nelson back when I was fiddling with the tube front end for the Monster Amp, I believe he tends to think in terms of peak in cases like this.

Grey
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Old 25th October 2007, 09:22 PM   #5
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Quote:
Originally posted by gl
Was that 80 ma of bias in the driver circuit or 80 ma peak?
For the duration of that slew, which generally quite short

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Old 25th October 2007, 10:36 PM   #6
gl is offline gl  United States
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Thank you.
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Old 26th October 2007, 08:32 AM   #7
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Quote:
Originally posted by Nelson Pass
I'll give you a rough ballpark. A reasonable example is
output follower pairs of IRF240/9240.

noting that 1 amp / 1 Farad = 1 Volt / 1 second:

I recall an example where a 80 mA drive gave about 40 V/uS
slew into 12 pairs, which works out to be about a 2 nF load,
which gives about 200 pF per pair.

You can apply similar logic to other devices using the Cgd as
your most important number.

Thank you Nelson.

I'm not sure how to relate the 200 pF per pair to the datasheet graph but perhaps I don't need to. I guess that I can use your ballpark as a basis for scaling based on Cgd relative to these devices.

Ian.
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Old 26th October 2007, 03:05 PM   #8
MarkAMC is offline MarkAMC  Hong Kong
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> I'm not sure how to relate the 200 pF per pair to the datasheet graph

In a complementary follower pair, one can assume that the load impedance is substantially higher than the output impedance, and Cgs effects can thus be ignored.

The capacitance that the driver stage will see is then dominated by Cgd. Since the gain of the follower is close to but less than 1, one can also ignore Miller effect as n approximation.

In the datasheets, both IRFP240 & 9240 have Cgd values of about 100~150p with a Vds of around 20V. So the pair together gives a value of 200p~300p, Vds dependent.

Please felll free to correct if I am wrong.


Patrick
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Old 26th October 2007, 03:10 PM   #9
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Patrick,

I agree with you but the puzzle for me is that Vds is not going to be constant - it will vary between almost zero and close to the supply rails as the device follows the signal (assuming the amp is driven to full output). What is the effective C under these conditions? Is it reasonable to simply take the average value?

Ian.
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Old 26th October 2007, 03:17 PM   #10
EUVL is offline EUVL  Europe
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Ian,

How often do you clip your power amp?

I think 90% of the time, you are running not more than a few Vs output.

The effect of Cgd variation with Vds becomes significant when you are within 8-10V of rail voltage. Of course one can always take worst case and use the Cgd value at Vds=5V or so.

I don't think many manufacturers quote their slew rate at max. output, do you ?


Patrick
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