Not entirely true for the same load condition - laterals have extremely low Cgd. The dominant part of current goes to Cgs. The capacitances will increase by 50% - but the gate voltage swing required will also reduce by approximately 1/3. The only part that will change much is the Cgd contribution but this is really very small.
Re: Poles in the amp
There's not a simple answer to this question as there are many variables in play. All in all it should work, but the increase in cumulative Gate capacitance will increase distortion by some relatively small fraction.
Assuming that you are running three output MOSFET pairs against a power supply that was originally supplying four (total for both channels), then the rail voltage will be somewhat stiffer, which is all to the good. Honestly, unless you are just shooting for a power increase into low Z loads, I'd recommend leaving the channel as-is, but putting one pair of MOSFETs on one heatsink, leaving the other pair on the original heatsink and doubling the bias. That will give you a far better return in sound quality without placing undue demands on the power supply, as it's a rather modest design.
Nearly all amplifiers have something that could be termed a driver stage. The mere existence of a driver stage, however, doesn't mean you can drive any arbitrary number of output devices.
To a first approximation, the amount of current available in the driver stage places a real world limit on how many output devices can be driven. This is designed in--there's no pot to adjust it. You can change parts values as fab suggests, but you have to keep an eye on other parameters, not least of which being the driver devices themselves. How much voltage, current, and heat can they take? Don't assume that because the driver's heatsink is a reasonable temperature while you have the amp on a table with the case open that everything will be all right. As soon as you close the case, the ambient temperature inside the chassis will increase dramatically and the driver's heatsink will have a harder time getting rid of excess heat.
The output power will increase only into lower impedances. Output into an 8 Ohm load will remain unchanged. Power into low impedances becomes a function of how much drive current is available. Since most amps are rated into an 8 Ohm load, you can generally assume that they're able to supply that much current. It's when you see less than twice the power into a 4 Ohm load that you begin to suspect current limiting. Merely adding a third set of output devices will not increase the power at 4 Ohms if, for instance, the rail voltage has begun to collapse. I think you'll find this to be the case for the Hafler.
The output stage of the Hafler is a follower. To get, say, 10Vrms out, you have to put that much in, as follower output stages are (to a first approximation) unity gain stages regarding voltage. So, no, adding a third pair of output devices will not decrease the input voltage swing requirements.
Grey
Dick West said:
There's not a simple answer to this question as there are many variables in play. All in all it should work, but the increase in cumulative Gate capacitance will increase distortion by some relatively small fraction.
Assuming that you are running three output MOSFET pairs against a power supply that was originally supplying four (total for both channels), then the rail voltage will be somewhat stiffer, which is all to the good. Honestly, unless you are just shooting for a power increase into low Z loads, I'd recommend leaving the channel as-is, but putting one pair of MOSFETs on one heatsink, leaving the other pair on the original heatsink and doubling the bias. That will give you a far better return in sound quality without placing undue demands on the power supply, as it's a rather modest design.
megajocke said:
Nearly all amplifiers have something that could be termed a driver stage. The mere existence of a driver stage, however, doesn't mean you can drive any arbitrary number of output devices.
To a first approximation, the amount of current available in the driver stage places a real world limit on how many output devices can be driven. This is designed in--there's no pot to adjust it. You can change parts values as fab suggests, but you have to keep an eye on other parameters, not least of which being the driver devices themselves. How much voltage, current, and heat can they take? Don't assume that because the driver's heatsink is a reasonable temperature while you have the amp on a table with the case open that everything will be all right. As soon as you close the case, the ambient temperature inside the chassis will increase dramatically and the driver's heatsink will have a harder time getting rid of excess heat.
The output power will increase only into lower impedances. Output into an 8 Ohm load will remain unchanged. Power into low impedances becomes a function of how much drive current is available. Since most amps are rated into an 8 Ohm load, you can generally assume that they're able to supply that much current. It's when you see less than twice the power into a 4 Ohm load that you begin to suspect current limiting. Merely adding a third set of output devices will not increase the power at 4 Ohms if, for instance, the rail voltage has begun to collapse. I think you'll find this to be the case for the Hafler.
megajocke said:
The output stage of the Hafler is a follower. To get, say, 10Vrms out, you have to put that much in, as follower output stages are (to a first approximation) unity gain stages regarding voltage. So, no, adding a third pair of output devices will not decrease the input voltage swing requirements.
Grey
1. No, amps using lateral mosfets often drive the gates directly from the VAS as they demand very low current. In an amp with bipolars it is very true that beefier drivers will be needed for higher output powers. And due to the extremely low Cgd three mosfets won't need much more current at all than two.
2. You are missing that the limiting factor is not the peak current (at least into high impedance loads) - it's the high on resistance of the lateral mosfets reducing available output swing. Rds(on) is about 1 ohms cold increasing to nearly 2 ohms when hot...
"The output stage of the Hafler is a follower. To get, say, 10Vrms out, you have to put that much in, as follower output stages are (to a first approximation) unity gain stages regarding voltage. So, no, adding a third pair of output devices will not decrease the input voltage swing requirements."
Irrelevant. What matters for output stage drive current consumption is the swing across Cgs capacitance. Cgd is very very small thus the current required by it even at full output is low compared to the current required by Cgs. The voltage swing (and thus dV/dt) over it will decrease approximately the same amount (for the same load resistance) as the capacitance increases when adding another pair as the transconductance is increased.
2. You are missing that the limiting factor is not the peak current (at least into high impedance loads) - it's the high on resistance of the lateral mosfets reducing available output swing. Rds(on) is about 1 ohms cold increasing to nearly 2 ohms when hot...
"The output stage of the Hafler is a follower. To get, say, 10Vrms out, you have to put that much in, as follower output stages are (to a first approximation) unity gain stages regarding voltage. So, no, adding a third pair of output devices will not decrease the input voltage swing requirements."
Irrelevant. What matters for output stage drive current consumption is the swing across Cgs capacitance. Cgd is very very small thus the current required by it even at full output is low compared to the current required by Cgs. The voltage swing (and thus dV/dt) over it will decrease approximately the same amount (for the same load resistance) as the capacitance increases when adding another pair as the transconductance is increased.
The P230 model uses three pairs of outputs per channel with the same circuit card (and values) as the DH220. They work very well in this configuration. Having the extra pair makes it easier on output stage to drive low impedances (still four ohm minimum load) but the output power is still limited by the transformer. The transformer in these amps is nice, but these amps were built to be cheap to build/buy. Its not oversized.
Now, just running a single channel on these transformers it would make sense to have the extra pair of outputs. With a single channel the power supply has current to spare.
Now, just running a single channel on these transformers it would make sense to have the extra pair of outputs. With a single channel the power supply has current to spare.
Re: Re: Re: Poles in the amp
R33 (DH220) = R31 (DH200).
As I said the resistor change is not required for this to work - when adding an extra pair. However, increasing the drive current for the mosfets is recommended for the sonics. If someone wants to try it, it is the easiest mod you can do since you do not have to desolder any part of the pcb but only solder a resistor to the junctions of the mosfet gate resistors on the heatsink connections. This resistor will appear in parallel with the one mounted on the pcb.
Stormrider said:
R33 (DH220) = R31 (DH200).
As I said the resistor change is not required for this to work - when adding an extra pair. However, increasing the drive current for the mosfets is recommended for the sonics. If someone wants to try it, it is the easiest mod you can do since you do not have to desolder any part of the pcb but only solder a resistor to the junctions of the mosfet gate resistors on the heatsink connections. This resistor will appear in parallel with the one mounted on the pcb.
megajocke said:
Methinks I detect someone who wants to argue semantics. If you're accustomed to thinking of bipolar circuits that use one or more pairs of followers to drive a set of output bipolars, then perhaps you're trying to argue that a voltage gain stage can't be a "driver" stage. Nonetheless, it can function as one. A driver stage need not be comprised of followers.
Of course MOSFETs have a high Zin--effectively infinite--but that doesn't negate the fact that adding output pairs increases the cumulative Gate capacitance, which increases distortion, particularly at high frequencies. More drive current will offset this tendency. As a real world example, an amplifier I recently built dropped a full 0.1% distortion simply due to increasing the current in the driver stage.
You're also assuming that the Hafler's power supply is a theoretically ideal power supply, which it isn't--not by a long shot.
You can't generate voltage swing out of thin air with a follower output. If you expect to see 10Vrms output, then you're going to have to put 10Vrms into the follower. Actually, there's always a small loss of voltage involved (which I alluded to above), not an increase.
Had you said current instead of voltage, we might have a somewhat better basis for discussion, although even that wouldn't be correct as the current requirement is primarily for driving the Gate capacitance, and would still increase assuming that you wanted the distortion characteristics to remain more or less constant compared to two pairs of outputs.
Stormrider,
If I recall correctly, the bias for the DH-200 was something like .25A per channel. I don't remember whether the DH-220 increased the stock bias and I've never looked at the P230 in detail. My recommendation for increasing the bias while staying with two pairs of output was based primarily on power supply limitations and keeping the heat dissipation equal on both heatsinks. If the same front end can do an effective job of driving three pairs of outputs, and if you can get away with increasing the bias beyond the simple proportioning of adding a third to the Iq (something like .36A or so), perhaps as far as .5A...all without having the rail collapse too badly, then that would be better still. My recollection is that the rail voltage was about 60V at idle, but started falling rather quickly as demand increased.
You'll still need to look at keeping the heat dissipation equal, though. I'd suggest running all Ns on one side and all Ps on the other, rather than have one N/P pair on each heatsink and one split between them.
Grey
GRollins said:
Beware of the heatsink temperature if you use 0.5A of bias. You will double the power dissipation in the big heatsink. Ensure that the temperatuer of heatsink does not go higher than 55 deg C. I have already experimented in the past with increasing bias but the amp heatsink size tends to reduce it anyway after a while (lateral mosfets temperature coefficient)....I suppose that the Vbe multiplier would need to be modified to sustain such a stable high bias...
I'm setting my monoblocks up like DickW.'s. One heatsink per chassis, per power supply. I won't be increasing the bias much if at all. I have them set for 375ma. It would be a good idea to use two heatsinks as you could increase the bias current some but I'm not because i want to have the monoblocks and a stereo amp as i explained above.
"Methinks I detect someone who wants to argue semantics. If you're accustomed to thinking of bipolar circuits that use one or more pairs of followers to drive a set of output bipolars, then perhaps you're trying to argue that a voltage gain stage can't be a "driver" stage. Nonetheless, it can function as one. A driver stage need not be comprised of followers."
No I don't care about the semantics. You are right of course that if the mosfets are driven directly from a voltage gain stage that stage will be the driver stage. What I meant is that this amp has a follower stage between - lessening the load on the VAS substantially.
"Of course MOSFETs have a high Zin--effectively infinite--but that doesn't negate the fact that adding output pairs increases the cumulative Gate capacitance, which increases distortion, particularly at high frequencies. More drive current will offset this tendency. As a real world example, an amplifier I recently built dropped a full 0.1% distortion simply due to increasing the current in the driver stage."
But the increased capacitance won't increase the current requirements in this case because of the increased transconductance. Did your amp use vertical fets or laterals?
"You're also assuming that the Hafler's power supply is a theoretically ideal power supply, which it isn't--not by a long shot.
You can't generate voltage swing out of thin air with a follower output. If you expect to see 10Vrms output, then you're going to have to put 10Vrms into the follower. Actually, there's always a small loss of voltage involved (which I alluded to above), not an increase."
What is that supposed to mean? The output power increase comes from the decreased on resistance. 0.7 ohms or 1 ohm in series with the speaker will make a difference. Also, the peak current before current limit per FET will be 4 amps or so when hot and this will limit power into 4 ohms a lot!
"Had you said current instead of voltage, we might have a somewhat better basis for discussion, although even that wouldn't be correct as the current requirement is primarily for driving the Gate capacitance, and would still increase assuming that you wanted the distortion characteristics to remain more or less constant compared to two pairs of outputs."
But the more capacitance won't require more current. And even if it did, it's not a problem as the follower will make sure the VAS doesn't get loaded down much. The follower has much lower output impedance than what is presented by the gate resistors.
No I don't care about the semantics. You are right of course that if the mosfets are driven directly from a voltage gain stage that stage will be the driver stage. What I meant is that this amp has a follower stage between - lessening the load on the VAS substantially.
"Of course MOSFETs have a high Zin--effectively infinite--but that doesn't negate the fact that adding output pairs increases the cumulative Gate capacitance, which increases distortion, particularly at high frequencies. More drive current will offset this tendency. As a real world example, an amplifier I recently built dropped a full 0.1% distortion simply due to increasing the current in the driver stage."
But the increased capacitance won't increase the current requirements in this case because of the increased transconductance. Did your amp use vertical fets or laterals?
"You're also assuming that the Hafler's power supply is a theoretically ideal power supply, which it isn't--not by a long shot.
You can't generate voltage swing out of thin air with a follower output. If you expect to see 10Vrms output, then you're going to have to put 10Vrms into the follower. Actually, there's always a small loss of voltage involved (which I alluded to above), not an increase."
What is that supposed to mean? The output power increase comes from the decreased on resistance. 0.7 ohms or 1 ohm in series with the speaker will make a difference. Also, the peak current before current limit per FET will be 4 amps or so when hot and this will limit power into 4 ohms a lot!
"Had you said current instead of voltage, we might have a somewhat better basis for discussion, although even that wouldn't be correct as the current requirement is primarily for driving the Gate capacitance, and would still increase assuming that you wanted the distortion characteristics to remain more or less constant compared to two pairs of outputs."
But the more capacitance won't require more current. And even if it did, it's not a problem as the follower will make sure the VAS doesn't get loaded down much. The follower has much lower output impedance than what is presented by the gate resistors.
fab said:
Perhaps I didn't make myself clear. Let's try this again.
Assuming that I'm correct in remembering that each channel of a stock DH-200 was biased at .25A, then each heatsink would see .25A * 60V * 2 = 30W of power dissipation at idle. Total draw on the power supply being .5A.
If you were to rebuild a DH-200 to be a mono amplifier (N.B.: Two amps required) and put one pair of MOSFETs on each heatsink, then the stock .25A bias would now be spread over two heatsinks instead of one. In other words, each heatsink would now be dissipating 15W at idle.
You could then increase the bias for that channel (which now has one entire chassis [two heatsinks]) to itself to .5A. This returns you to the same power dissipation per heatsink (30W) and, for that matter, to the same current draw from the power supply (2 * .25A = 1 * .5A). The chassis as a whole will run cooler because you're actually shy one front end board...okay, so it's not all that much, but it will run a little cooler.
Three pairs of MOSFETs (six devices, total) biased at .5A total for the channel would not change the Pd calculations. In fact, I'd be tempted to start pushing the bias a bit harder still, depending on what the power supply could handle...
...and, yes, the heatsinks, too.
But I imagine the rail will collapse before the heatsinks get too hot.
megajocke,
Again, I'm getting the impression you're more familiar with bipolars than MOSFETs. Gate capacitance gets filed in the There Ain't No Such Thing As A Free Lunch bin.
Bipolars eat current directly. It literally flows in and out of the base. However, MOSFETs need current, too. The idea that the Gate capacitance can be ignored led to a lot of really evil sounding MOSFET designs back when people first started designing with them. Ignoring the charge/discharge requirements of MOSFETs leads to unpleasant consequences such as poor slew rate, distortion, and decreased bandwidth. Note that the capacitance problem is true for both lateral and vertical MOSFETs. Each has benefits. Each has drawbacks. (True for bipolars, for that matter.) But ignoring the increase in cumulative Gate capacitance due to adding a third (or fourth, or fifth...) pair of MOSFETs is Bad Juju.
If you want a trail of breadcrumbs, start by looking at the behavior of a capacitor charged by a current source. Then increase the frequency and watch what happens. This is easy to see on an oscilloscope. Or you can approach it mathematically. Or you can read textbooks. Or you can do it empirically with listening tests. Or you can read a bunch of threads here. Or you can just take my word for it.
Naaaaah, that'd take all the fun out of it!
However, I'm not interested in getting in a big fuss about it. My summer so far has ranged between "sucks" and "screw it, I'm outta here" and I'm just not interested in a ruckus here at DIY as I've already got a full dance card. So with that in mind, I'll leave you to believe or disbelieve what I'm saying, according to your nature and/or mood.
Grey
I'll try to explain it in another way, not that it matters much really - the Hafler is one of those amps that DID add an extra follower stage because they did think the current drawn was important. The other bigger models as others have pointed out use the same driver stage with the same standing current although driving more output devices in parallell.
What I'm trying to say is that here is a significant difference between laterals and verticals:
Verticals have pretty high Cgd
Laterals have very, very low Cgd
This means:
Verticals need lots of current to just slew the output around. Extra current is needed to change output current too. Driving vertical mosfets rapidly gets out of hand - they do require a lot of current if you don't want slew rates of 5V/uS or so...
Laterals need most of their drive current to change output current - charging/discharging Cgs. This capacitance is lower too than in verticals.
The Cgs of about 600pF sees about +-10V swing at full load.
The Cgd of about 10pF sees about +-60V swing at full output.
At a specific frequency near maximum load Cgs will need ten times more current than Cgd.
Something lika an IRFP240 has about ten times the Cgd as a lateral - meaning that this is the significant draw for them as they have higher transconductance more than compensating for the higher Cgs. If one starts to parallell these without thinking the slew rate will start to degrade and current draw from driver stage will increase even with the same load used. In this case you are 100% right!
Now with 3 pairs instead of 2 the Cgs swing needed into a certain load will be about 1/3 less due to the increased transconductance.
The 50% higher Cgs will see 2/3 of the amplitude -> about the same current as the current needed for Cgd is pretty insignificant.
Conclusion is that for most verticals adding more pairs will need more current to not degrade output voltage slew rate. Not true for laterals though, here it is the output current slew rate that sets the current needed from the driver stage. Adding more laterals won't need significantly more drive current for the same output current slew rate.
I'm not suggesting that driving lots of laterals with a wimpy driver stage is OK though - might probably be OK for bass but as you say, the current needed should not be neglected. The Cgd will need quite a bit of current. It's just that it doesn't increase significantly when adding more devices IF you are driving the same load.
I'm sorry if you've had a bad summer. And I don't disbelieve you - you are definetly 100% right when it comes to the vertical fets or if you are decreasing load impedance the same amount as you add laterals.
edit:
It's possible many designers missed the current requirement by the laterals because they saw the low Cgd - thinking that very little current is enough and will still give good slew rate. 1mA is enough for 100V/uS... They probably missed that most current is drawn by Cgs and proportional to output *current* slew rate.
What I'm trying to say is that here is a significant difference between laterals and verticals:
Verticals have pretty high Cgd
Laterals have very, very low Cgd
This means:
Verticals need lots of current to just slew the output around. Extra current is needed to change output current too. Driving vertical mosfets rapidly gets out of hand - they do require a lot of current if you don't want slew rates of 5V/uS or so...
Laterals need most of their drive current to change output current - charging/discharging Cgs. This capacitance is lower too than in verticals.
The Cgs of about 600pF sees about +-10V swing at full load.
The Cgd of about 10pF sees about +-60V swing at full output.
At a specific frequency near maximum load Cgs will need ten times more current than Cgd.
Something lika an IRFP240 has about ten times the Cgd as a lateral - meaning that this is the significant draw for them as they have higher transconductance more than compensating for the higher Cgs. If one starts to parallell these without thinking the slew rate will start to degrade and current draw from driver stage will increase even with the same load used. In this case you are 100% right!
Now with 3 pairs instead of 2 the Cgs swing needed into a certain load will be about 1/3 less due to the increased transconductance.
The 50% higher Cgs will see 2/3 of the amplitude -> about the same current as the current needed for Cgd is pretty insignificant.Conclusion is that for most verticals adding more pairs will need more current to not degrade output voltage slew rate. Not true for laterals though, here it is the output current slew rate that sets the current needed from the driver stage. Adding more laterals won't need significantly more drive current for the same output current slew rate.
I'm not suggesting that driving lots of laterals with a wimpy driver stage is OK though - might probably be OK for bass but as you say, the current needed should not be neglected. The Cgd will need quite a bit of current. It's just that it doesn't increase significantly when adding more devices IF you are driving the same load.
I'm sorry if you've had a bad summer. And I don't disbelieve you
- you are definetly 100% right when it comes to the vertical fets or if you are decreasing load impedance the same amount as you add laterals.edit:
It's possible many designers missed the current requirement by the laterals because they saw the low Cgd - thinking that very little current is enough and will still give good slew rate. 1mA is enough for 100V/uS... They probably missed that most current is drawn by Cgs and proportional to output *current* slew rate.
Hey Dick, my P230 transformers have an unused pair of blue wires. Are these windings for the higher voltage supply the XL280 uses for its driver stage?
If i wanted to outfit my PC19c circuit cards for this higher voltage would it really be as simple as adding another bridge rectifier/filter caps and changing the wiring? I see that the XL280 has a diode across B+, B++ and another from B- to B--. Other than that i don't see where it differs from the power supply wiring of the PC19c.
If i wanted to outfit my PC19c circuit cards for this higher voltage would it really be as simple as adding another bridge rectifier/filter caps and changing the wiring? I see that the XL280 has a diode across B+, B++ and another from B- to B--. Other than that i don't see where it differs from the power supply wiring of the PC19c.
AJ,
I am inclined to say "go for it." The function of the diodes is not clear to me. Perhaps FAB will offer better advice.
Have you measured the voltage at this extra winding?
The idea in the XL-280 was to give the VAS and driver transistors about 5 VDC± more than the MOSFETs. This compensates for the extra voltage needed to turn on the MOSFETs and increases power a slight amount without adding very much to heat.
Why these two voltage sources need this diode separation is beyond my pay grade.
Note that fuse protection should remain for the MOSFETs which will continue to receive the lower voltages. Just unsolder the two wires at eyelet 3 and connect them together; ditto for the wires at eyelet 10. Then, the higher voltages from the extra little PS can go directly to the PCBs and fuse protection is probably not needed. But, in your mono blocks there is ample room for another pair of fuses so you can include them if desired.
My $0.02
I am inclined to say "go for it." The function of the diodes is not clear to me. Perhaps FAB will offer better advice.
Have you measured the voltage at this extra winding?
The idea in the XL-280 was to give the VAS and driver transistors about 5 VDC± more than the MOSFETs. This compensates for the extra voltage needed to turn on the MOSFETs and increases power a slight amount without adding very much to heat.
Why these two voltage sources need this diode separation is beyond my pay grade.
Note that fuse protection should remain for the MOSFETs which will continue to receive the lower voltages. Just unsolder the two wires at eyelet 3 and connect them together; ditto for the wires at eyelet 10. Then, the higher voltages from the extra little PS can go directly to the PCBs and fuse protection is probably not needed. But, in your mono blocks there is ample room for another pair of fuses so you can include them if desired.
My $0.02
Hafler Shut-Off issue
When I turn off the AC Power to my DH200; I can watch the cones of my woofers take an extrusion and there is an audible (60Hz?) buzz for about 2-3 seconds as the amp goes off (I assume large caps bleeding down?)
Is this something anyone has seen...Is it "normal" ? Should I put some sort of electronic disconnect (relay) to the speakers so when the AC to the amp is switched off, the speakers are instantly disconnected from the amp outputs?
When I turn off the AC Power to my DH200; I can watch the cones of my woofers take an extrusion and there is an audible (60Hz?) buzz for about 2-3 seconds as the amp goes off (I assume large caps bleeding down?)
Is this something anyone has seen...Is it "normal" ? Should I put some sort of electronic disconnect (relay) to the speakers so when the AC to the amp is switched off, the speakers are instantly disconnected from the amp outputs?
Does the amp have excessive dc off set when on?
When the amp is turned on and off there is a spike of dc offset and then it levels off. When the amp is on it should be between 0mV and 50mV.
You can put a 6800 to 10000ohm 2w resistor in parallel with the large filter caps so they will discharge faster.
When the amp is turned on and off there is a spike of dc offset and then it levels off. When the amp is on it should be between 0mV and 50mV.
You can put a 6800 to 10000ohm 2w resistor in parallel with the large filter caps so they will discharge faster.
Do both L&R cones move equally or does one move a lot more than the other? Is the slight buzzing sound equally loud in both channels or mostly in one?
When both channels are equally misbehaving one looks at the power supply's filter caps. If mostly one channel then one looks at the pair of 100 uF bypass electrolytic caps on that circuit card (among other things). Your electrolytic caps are 25+ years old and replacement of all of them would be a prudent thing to do.
And, for sure, you should check each channel for DC offset, a common malady of these older DH-200 amps.
Can you borrow a pair of caps suitable for the PS and try them? But be sure to observe all cautions about electrocution -- of course. Make sure these large caps are fully discharged before messing with them as they can pack a lethal charge.
Tell us more about the symptoms.