OK
Vise bi mi odgovaralo to sto je jednostavnije i pouzdano znaci nesto sto si mozda probao i radi bez problema.
Imam 2 x 42V ispravljac.
Ja bih te zamolio ako mozes da postiras danas.
Hvala Unaprijed...
I am sorry...We are just guys from same country and sometimes is easier to communicate on our language.I asked him to post schematic today for amp I requested.
Vise bi mi odgovaralo to sto je jednostavnije i pouzdano znaci nesto sto si mozda probao i radi bez problema.
Imam 2 x 42V ispravljac.
Ja bih te zamolio ako mozes da postiras danas.
Hvala Unaprijed...
I am sorry...We are just guys from same country and sometimes is easier to communicate on our language.I asked him to post schematic today for amp I requested.
Here is a simple amp I've built many times...
Notes:
Q6, 7, 8 will work best with some small heatsinks. I used a double sided PCB with ground plane on top and just bent the transistors to lie on it, with a drop of heat transfer grease between the transistor body and the copper. It will work with no heatsink but the transistor temperatures will be about 60C. Q6, 7, 8, 10 may be changed for something in a case more suitable for cooling but I would not recomend it because most transistors in cases like that have lower gain. There is a better way to solve this, see below under improvements.
Q1, 2, and 3 are mounted on appropriately large heatsink. Mount Q1 between Q2 and Q3 if Q2 and Q3 are close, else mount Q1 close to either Q2 or Q3 if Q2 and Q3 are further apart.
R1 = 10k trimmer has '40%' beside it because that was left from the simulator. It sets bias current, wiper towards Q8 increases bias current. About 25mA will minimise distortion into 8ohms resistive at 1kHz but will have unreasonably high HF distortion. Depending on application I use from 40-150mA.
Although not obvious, R7, R6, D3 form a current limiter which is quite symetrical for both N and P output MOSFET. Changeing of the power devices will require changes in their values, so know what you are doing before you change this!
Some things not shown on the schematic:
No Zobel/Boucherot network has been included in the schematic. When configured as a standalone amp, i normally include it, typical examples are easily found.
Input cap - if DC coupling is used from a low impedance source, change C3, R4, R8 to 470u, 3.3k 2W, 100ohm respectively. If input cap is used, leave as is. This reduces output offset voltage.
Power rails to output tansistors need fuses. 4A fast blow max for your power rails. you can push it higher, IRFPs are quite robust, but be careful. Output current limit is set to about 8-9A. If you trust your fuses, you can remove the zener...
Possible improvements:
Small non-inductive foil caps should be placed in parallel with C1, C2, C3, C4 and input cap if used (if it's not a foil type already).
Remove diodes D1 and D2 and feed the driver stages with regulated 40-45V for better output power.
Use higher gain transistors for Q5 and Q6 (BC546B work well but R23 and R24 may need increasing so that Vcemax for BC546 is not compromised under any circumstances). The distortion will lower somewhat.
D4 and D5 may be replaced by a LED (R20 may have to be changed to keep the current through Q9 more or less the same).
Q4 and Q5 will be slightly unbalanced in this circuit due to the simple current mirror used. This increases distortion a bit and introduces 20-30mV of output offset. If offset is a problem, R22 can be slightly lowered by paralleling with a large (10k or so) resistor, to correct this. A wilson current source is the real answer but then it makes it a different amp
Increasing driver stage current will reduce distortion but since the LTP tail current is the reference, this can get complicated. Also, Q6, 7, 8 heat can become a serious problem. There are two ways to solve this:
Add a 'shadow' pair to R13, Q7, Q8, R21
Add cascode transistors to Q6, Q7, Q8
I've done both but then this becomes a different amp, so that's a nother story
Distortion will be lower if R2 and R5 are lowered, or removed entirely. If you remove them replace R6 by 100ohms, a zener diode must be added between gate of Q2 and output, both zeners should be about 5.1-5.6V. Thermal coupling between Q1, 2, 3 must be done carefully (use good insulators and grease, also mount transistors mechanically close). You also get increased output power. You may want to increase R18 to avoid accidentally being able to set an extremely high bias current. In any case, meddling with R2 and R5 requires change of the zener diode to get appropriate current limiting.
Expect a good clean 60W under all circumstances from this amp, with plenty headroom.
Notes:
Q6, 7, 8 will work best with some small heatsinks. I used a double sided PCB with ground plane on top and just bent the transistors to lie on it, with a drop of heat transfer grease between the transistor body and the copper. It will work with no heatsink but the transistor temperatures will be about 60C. Q6, 7, 8, 10 may be changed for something in a case more suitable for cooling but I would not recomend it because most transistors in cases like that have lower gain. There is a better way to solve this, see below under improvements.
Q1, 2, and 3 are mounted on appropriately large heatsink. Mount Q1 between Q2 and Q3 if Q2 and Q3 are close, else mount Q1 close to either Q2 or Q3 if Q2 and Q3 are further apart.
R1 = 10k trimmer has '40%' beside it because that was left from the simulator. It sets bias current, wiper towards Q8 increases bias current. About 25mA will minimise distortion into 8ohms resistive at 1kHz but will have unreasonably high HF distortion. Depending on application I use from 40-150mA.
Although not obvious, R7, R6, D3 form a current limiter which is quite symetrical for both N and P output MOSFET. Changeing of the power devices will require changes in their values, so know what you are doing before you change this!
Some things not shown on the schematic:
No Zobel/Boucherot network has been included in the schematic. When configured as a standalone amp, i normally include it, typical examples are easily found.
Input cap - if DC coupling is used from a low impedance source, change C3, R4, R8 to 470u, 3.3k 2W, 100ohm respectively. If input cap is used, leave as is. This reduces output offset voltage.
Power rails to output tansistors need fuses. 4A fast blow max for your power rails. you can push it higher, IRFPs are quite robust, but be careful. Output current limit is set to about 8-9A. If you trust your fuses, you can remove the zener...
Possible improvements:
Small non-inductive foil caps should be placed in parallel with C1, C2, C3, C4 and input cap if used (if it's not a foil type already).
Remove diodes D1 and D2 and feed the driver stages with regulated 40-45V for better output power.
Use higher gain transistors for Q5 and Q6 (BC546B work well but R23 and R24 may need increasing so that Vcemax for BC546 is not compromised under any circumstances). The distortion will lower somewhat.
D4 and D5 may be replaced by a LED (R20 may have to be changed to keep the current through Q9 more or less the same).
Q4 and Q5 will be slightly unbalanced in this circuit due to the simple current mirror used. This increases distortion a bit and introduces 20-30mV of output offset. If offset is a problem, R22 can be slightly lowered by paralleling with a large (10k or so) resistor, to correct this. A wilson current source is the real answer but then it makes it a different amp
Increasing driver stage current will reduce distortion but since the LTP tail current is the reference, this can get complicated. Also, Q6, 7, 8 heat can become a serious problem. There are two ways to solve this:
Add a 'shadow' pair to R13, Q7, Q8, R21
Add cascode transistors to Q6, Q7, Q8
I've done both but then this becomes a different amp, so that's a nother story
Distortion will be lower if R2 and R5 are lowered, or removed entirely. If you remove them replace R6 by 100ohms, a zener diode must be added between gate of Q2 and output, both zeners should be about 5.1-5.6V. Thermal coupling between Q1, 2, 3 must be done carefully (use good insulators and grease, also mount transistors mechanically close). You also get increased output power. You may want to increase R18 to avoid accidentally being able to set an extremely high bias current. In any case, meddling with R2 and R5 requires change of the zener diode to get appropriate current limiting.
Expect a good clean 60W under all circumstances from this amp, with plenty headroom.
Re: Question about upgrade
Naslo bi se, ali polako - nismo u onom vicu 'nakapajte mi rezervoar benzina jer kap ne kosta nista'
Double power into 4 ohms but you are asking for trouble because unless you have a huge heatsink and/or you intend to drive simple loads (not very reactive) you will be operating the output transistors close to maximum.
In theory, IF you supply the driver stage with at least 10V more than the output stage, you can get the output to go within fractions of a V of the rail voltage, but distortion will rise a lot for those last few V. So, with 40V rails into 8 ohms, assuming rail voltage does not droop, you get 100W into 8 ohms. THEORETICALLY 200 into 4 but you will need a HUGE heatsink and special mounting hardware to do this semi-reliably. I would not recomend more than 100W-120W continious out of one pair of IRFPs into any load. You could use multiple pairs but you need a driver stage for this. Alternatively, you can build two and bridge them, but I would recomend a lower output rail voltage (36V or so).
You cannot replace the IRF520 with a BJT, not without modifying nearby components. You could however use IRF530, 540, 640 instead, almost any MOSFET rated for at least 10A (the reason is higher transconductance in higher current types). If you want to use a BJT you will have to do a lot of experimenting with a resistor in the emitter and R2 and R5.
You cannot get more power with any kind of driver stage as power is limited by the output rail voltages and voltage/current/power ratings of Q2 and Q3.
The power you get out of the amp is defined by the voltage on voltage at C1 or C2 (max power rail - 0.5V) minus Vgs for the MOSFET at maximum current (about 5-7V depending on load) or the power rail voltage in that instant minus same 5-7V, whichever is lower. The power will be that voltage squared divided by 2xRload. Power rails should not be higher than 50V for the output stage AT ANY TIME for the given Q2 and Q3.
A driver stage may help if you want parallel MOSFET pairs, because the unmodified driver will not be able to drive more than one pair. In that case you need to connect the driver transistor collectors AFTER the diodes D1 and D2, and disconnect R6/R7 from the IRF520. Bases of drivers go to the point where you just disconnected R6 and R7. Emitters go to R6/R7 instead of the IRF520. Your R? in the schematic should be on the order of 150ohms. Drivers should be on heatsinks. C1 and C2 should be increased to 1000uF. Add 2 diodes in series with IRF520 drain, the diodes whould be in thermal contact with drivers (ideally). For this configuration a BJT instead of IRF520 could work better but still you need to change resistors around it and a resistor in the emitter of this BJT. Such a modified output stage can drive 2-3 pairs of MOSFETs.
From your questions and proposed mod I have to surmise you don't completely understand how this amp works, so I would advise starting small and with a proven design such as it originally is, before you start modding.
zox2003 said:da li imas jos koji?
Any more?
Naslo bi se, ali polako - nismo u onom vicu 'nakapajte mi rezervoar benzina jer kap ne kosta nista'
zox2003 said:Can I run this amp on 4 ohms?How much more power I will get? Can I replace IRF520 in your schematic with BJT transistor?
Thanks
Double power into 4 ohms but you are asking for trouble because unless you have a huge heatsink and/or you intend to drive simple loads (not very reactive) you will be operating the output transistors close to maximum.
In theory, IF you supply the driver stage with at least 10V more than the output stage, you can get the output to go within fractions of a V of the rail voltage, but distortion will rise a lot for those last few V. So, with 40V rails into 8 ohms, assuming rail voltage does not droop, you get 100W into 8 ohms. THEORETICALLY 200 into 4 but you will need a HUGE heatsink and special mounting hardware to do this semi-reliably. I would not recomend more than 100W-120W continious out of one pair of IRFPs into any load. You could use multiple pairs but you need a driver stage for this. Alternatively, you can build two and bridge them, but I would recomend a lower output rail voltage (36V or so).
You cannot replace the IRF520 with a BJT, not without modifying nearby components. You could however use IRF530, 540, 640 instead, almost any MOSFET rated for at least 10A (the reason is higher transconductance in higher current types). If you want to use a BJT you will have to do a lot of experimenting with a resistor in the emitter and R2 and R5.
zox2003 said:
You cannot get more power with any kind of driver stage as power is limited by the output rail voltages and voltage/current/power ratings of Q2 and Q3.
The power you get out of the amp is defined by the voltage on voltage at C1 or C2 (max power rail - 0.5V) minus Vgs for the MOSFET at maximum current (about 5-7V depending on load) or the power rail voltage in that instant minus same 5-7V, whichever is lower. The power will be that voltage squared divided by 2xRload. Power rails should not be higher than 50V for the output stage AT ANY TIME for the given Q2 and Q3.
A driver stage may help if you want parallel MOSFET pairs, because the unmodified driver will not be able to drive more than one pair. In that case you need to connect the driver transistor collectors AFTER the diodes D1 and D2, and disconnect R6/R7 from the IRF520. Bases of drivers go to the point where you just disconnected R6 and R7. Emitters go to R6/R7 instead of the IRF520. Your R? in the schematic should be on the order of 150ohms. Drivers should be on heatsinks. C1 and C2 should be increased to 1000uF. Add 2 diodes in series with IRF520 drain, the diodes whould be in thermal contact with drivers (ideally). For this configuration a BJT instead of IRF520 could work better but still you need to change resistors around it and a resistor in the emitter of this BJT. Such a modified output stage can drive 2-3 pairs of MOSFETs.
From your questions and proposed mod I have to surmise you don't completely understand how this amp works, so I would advise starting small and with a proven design such as it originally is, before you start modding.
Thanks...
Thanks you on your answers.You are helpfull.
Let's sumarize:
With same circuit as your original can I get 100w/4ohms on +/- 42V rails without any modifications (as is)?
Do I have to possible reduce something to have this amp running that power?
Can I exclude R2 and R5 for just a pair of IRFP240/9240?
Can I use IRFP9240 instead 9140?
Thanks a lot.
Thanks you on your answers.You are helpfull.
Let's sumarize:
With same circuit as your original can I get 100w/4ohms on +/- 42V rails without any modifications (as is)?
Do I have to possible reduce something to have this amp running that power?
Can I exclude R2 and R5 for just a pair of IRFP240/9240?
Can I use IRFP9240 instead 9140?
Thanks a lot.
Re: Thanks...
Yes - a quick sim shows you will get 100W into 4 ohms if the rails do not drop below 36V at full load. Just use a big heatsink!!!
No, you have to increase - the heatsink
Yes, but be careful with thermal coupling. IF your heatsink is too small or thermal interface between transistors and heatsink is not good, you will have thermal runaway, and quickly!!!
On the other hand, excluding R2 and R5 alow for a bit more power, or more rail droop. You can get 100W into 4 ohms if rails do not drop below 33V or if, as above you alow for drop to 36V, you get 120W.
Also, you need to set both gate damper resistors to 100 ohms and put a zener in parralel with G-S of both MOSFETs, 5.1-5.6V.
The single zener limiter works by using the bias voltage and drop on R2/R5 as part of the limiting circuit, without R2/R5 it needs to be different obviously.
You can use anything you want but there is no guarantee it will work OK, just kidding...
IRFP240 in this application robs you of about 1.5V of headroom, but just on the negative half-period which means you get less power and more heat, and asymetric clipping. You also get up to 50% more distortion in any case. Also, at 100W into 4 ohms you will be operating the 9240 right at the edge. Also, you need to modify the current limiting zeners because 9240 needs more voltage at input for same current as 240. 9140 does not.
That 9140 was not chosen for nothing
zox2003 said:
Yes - a quick sim shows you will get 100W into 4 ohms if the rails do not drop below 36V at full load. Just use a big heatsink!!!
No, you have to increase - the heatsink
Yes, but be careful with thermal coupling. IF your heatsink is too small or thermal interface between transistors and heatsink is not good, you will have thermal runaway, and quickly!!!
On the other hand, excluding R2 and R5 alow for a bit more power, or more rail droop. You can get 100W into 4 ohms if rails do not drop below 33V or if, as above you alow for drop to 36V, you get 120W.
Also, you need to set both gate damper resistors to 100 ohms and put a zener in parralel with G-S of both MOSFETs, 5.1-5.6V.
The single zener limiter works by using the bias voltage and drop on R2/R5 as part of the limiting circuit, without R2/R5 it needs to be different obviously.
You can use anything you want but there is no guarantee it will work
OK, just kidding...IRFP240 in this application robs you of about 1.5V of headroom, but just on the negative half-period which means you get less power and more heat, and asymetric clipping. You also get up to 50% more distortion in any case. Also, at 100W into 4 ohms you will be operating the 9240 right at the edge. Also, you need to modify the current limiting zeners because 9240 needs more voltage at input for same current as 240. 9140 does not.
That 9140 was not chosen for nothing
640/9540 would be the right choice for proper symetry but these devices again have higher distortion, even higher than 240/9240 combinations. Eg, 240/9140 0.005%, 240/9240 0.008%, 640/9540 0.013%, all at 3kHz and 100W output. 640/9540 also produce a very undesirable distortion harmonic distribution, and there is no way except perhaps fans or liquid cooling, for cooling the single 640/9540 pair enough for 100W into 4 ohms. To be honest, I avoid IRF x30/9x30 and x40/9x40 for audio.
What is the problem with 9140? I can get it here from at least 4 different places (not always at reasonable prices, though). In the US I have yet to see it cost higher than about $2.5...
What is the problem with 9140? I can get it here from at least 4 different places (not always at reasonable prices, though). In the US I have yet to see it cost higher than about $2.5...
Depends heavily on bias current chosen. For 100mA, 100W into 4 ohms at 1kHz you typically get <0.005%, dominant 2nd harmonic. As usual, it rises with frequency and stays below 0.05 at 20kHz.
Higher bias currents usually do not do much for the midband distortion, but lower HF distortion.
Inclusion of R2 and R5 does not as much change the figures (they are very slightly higher) as much as the harmonic distribution becomes less favorable.
At lower power the situation is much more favorable. At 1W midband distortion drops to about half of the full power figure and the harmonic distribution becomes a textbook monotonously decaying series. A similar but slightly lower reduction happens at HF. Although the amp was designed for 8 ohm operation, the midband distortion figures stay almost the same into 4-8 ohms, but HF distortion at less than full power drops quite signifficantly for loads over 4 ohms. For example, at 8 ohms, 1W, you can expect 0.01% at 20kHz.
Of course, how much of harmonic distortion you can hear at 20kHz, the second harmonic already being at 40kHz, is debatable, but I did the measurements so there you have it
Frequency response is within 0.1dB inside 20Hz to 20kHz.
All of this of course assumes proper layout and wiring techniques.
Higher bias currents usually do not do much for the midband distortion, but lower HF distortion.
Inclusion of R2 and R5 does not as much change the figures (they are very slightly higher) as much as the harmonic distribution becomes less favorable.
At lower power the situation is much more favorable. At 1W midband distortion drops to about half of the full power figure and the harmonic distribution becomes a textbook monotonously decaying series. A similar but slightly lower reduction happens at HF. Although the amp was designed for 8 ohm operation, the midband distortion figures stay almost the same into 4-8 ohms, but HF distortion at less than full power drops quite signifficantly for loads over 4 ohms. For example, at 8 ohms, 1W, you can expect 0.01% at 20kHz.
Of course, how much of harmonic distortion you can hear at 20kHz, the second harmonic already being at 40kHz, is debatable, but I did the measurements so there you have it
Frequency response is within 0.1dB inside 20Hz to 20kHz.
All of this of course assumes proper layout and wiring techniques.
Re: Point???
Well, I know why I am using the pairs I am - no idea why other people are using the ones they do
The reasoning is really very simple. MOSFETs have lower Gm (a LOT lower) than BJTs. When pairing BJTs, as their gain rises, the actual perfect matching becomes less and less important. With MOSFETs you don't have that option as gain is low, so you have to work at finding the best P to N channel match. This is not at all easy.
P and N MOSFETs are intrinsically different, this is a matter of semiconductor physics. As a result, if you took an N ch VMOSFET and suddenly 'reversed' it so it was P, so all other things remain the same, you would at least get a little more than half current capability, as well as other subtle differences, one being different gain. With LMOSFETs (the usual 2Sj/2SK types) this is not as pronounced, they are better complements, but that design has other problems (reduced max current, higher Rdson, price). LMOSFETs are speciffically made for audio and their manufacturers take great pains to make them reasonably complementary. However, if one were to look at the silicon dies of the P and N part, they would notice that the parts are quite different, to make them appear electrically complementary.
It seems to me that people all too often take the manufacturer's markings for granted - when even a quick glance at the datasheet will prove that what may appear complementary is not. The solution is to look carefully at a number of parameters (current, Gm, capacitances, temperature dependancy would be the main ones) and find the best matches, not just take for granted that 240 and 9240 are complements.
I am sometimes amazed at the inertia of some designers. Even as limited as simulations are, they clearly show that the 'usual' complements are not really complementary. One of the funnyest things is when I see the gate-source zeners for current limiting being the same on the 240 and the 9240 - the current limit on the P MOSFET ends up only 50-70% of that of the N-MOSFET. This is VERY hard to miss and yet so many do... By the same token, I often listen to complaints about IRFP parts just not sounding good, and then they show me a schematic or an amp with the 'usual' 'complements' that are not.
zox2003 said:i do undestand what are you talking about MOSFET'S.
If that is a case why everybody is using different pairs then you recommend?
Example:
Everybody is using IRFP240 with IRFP9240 but you are only one that does not?
Same thing with IRF540 and 9540...
Why is that?
Well, I know why I am using the pairs I am - no idea why other people are using the ones they do
The reasoning is really very simple. MOSFETs have lower Gm (a LOT lower) than BJTs. When pairing BJTs, as their gain rises, the actual perfect matching becomes less and less important. With MOSFETs you don't have that option as gain is low, so you have to work at finding the best P to N channel match. This is not at all easy.
P and N MOSFETs are intrinsically different, this is a matter of semiconductor physics. As a result, if you took an N ch VMOSFET and suddenly 'reversed' it so it was P, so all other things remain the same, you would at least get a little more than half current capability, as well as other subtle differences, one being different gain. With LMOSFETs (the usual 2Sj/2SK types) this is not as pronounced, they are better complements, but that design has other problems (reduced max current, higher Rdson, price). LMOSFETs are speciffically made for audio and their manufacturers take great pains to make them reasonably complementary. However, if one were to look at the silicon dies of the P and N part, they would notice that the parts are quite different, to make them appear electrically complementary.
It seems to me that people all too often take the manufacturer's markings for granted - when even a quick glance at the datasheet will prove that what may appear complementary is not. The solution is to look carefully at a number of parameters (current, Gm, capacitances, temperature dependancy would be the main ones) and find the best matches, not just take for granted that 240 and 9240 are complements.
I am sometimes amazed at the inertia of some designers. Even as limited as simulations are, they clearly show that the 'usual' complements are not really complementary. One of the funnyest things is when I see the gate-source zeners for current limiting being the same on the 240 and the 9240 - the current limit on the P MOSFET ends up only 50-70% of that of the N-MOSFET. This is VERY hard to miss and yet so many do... By the same token, I often listen to complaints about IRFP parts just not sounding good, and then they show me a schematic or an amp with the 'usual' 'complements' that are not.
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