| PanzerLord |
Hello everyone, I`ve gathered a lot of useful hints and tips on this forum, so I though I`d share my project (me proud!).
I`m currently etching the PCB`s and waiting for heatsinks and output transistors to arrive, so it will probably take a couple of weeks before I can realize it in the real world. It would have been fun if someone else also has buildt this, in this way I could get some feedback on possible improvement and so on...
Okidoki, here are the facts
It`s a 50W class A balanced power amplifier.
-Differential input with cascode loading and current mirror
-All current sources are modified wilsons, the rolls royce of current sources
-2-pole compensation in the VA-stage(local feedback)
somewhat close to the ones I got in multisim.
-3 sets of push-pull output transistors.
-A heatsink with thermal resistance less than 0.3C/W per channel should do.
The reason I achieved these S/N, and THD levels, where the use of current sources, cascode at input , degeneration resistors, tail current tweaking (with respect to the CC), two-pole compensation, and of course the current mirror in the input stage.
The current sources and current mirror alone improves the THD+N by over 5000 times, the cascode then improves the THD by 40-50% (increased linearity).
In multisim, the simulation results where far beyond the spesifications I set as a minimum before I began designing the amplifier. (However, the results where even better when I drove the input unbalanced. The reason to this is probably differential input inbalance, I`ll look in to that in the next days).
(interpolation level was set to 12, and sampling frequenzy to 128kHz for more accurate results):
THD>0.0002% (3rd. harm: 0.4uV@1kHz, 7.5uV@20kHz
2nd. harm: 1uV@1kHz, 21uV@20kHz)
All harmonic distortion simulations with 8ohm load and full voltage swing (degeneration resistors where changed to 10ohm)!!
Due to the balanced input, noise analyzis isn`t possible (S/N>115db in unbalanced mode)
Pout 8ohm=52W RMS (with 1V loss in powersupply)
DC operating point=-6mV
I`ll post my pre-amp project in a couple weeks..
Give me some feedback on this one...
:) |
|
|
| PRR |
> Give me some feedback on this one...
You asked....
Looks like current in Q19 Q20 is undefined. I don't see a bias diode.
I assume Q25 Q26 are "protection". If that is your "bias diode", I never saw that scheme and will have to think about it. For a few months.
If Q25 Q26 are bias: you have no short-protection at all. For a home-brew with $1 transistors, that's fine, but don't loan it to your clumsy friends.
If you short the output, and feed signal, Q10 will die instantly (before the output smoke).
Some preamps may not like the low input impedances.
Input impedances are not balanced for both common-mode and differential inputs (many "balanced" input designs have this flaw).
Base bias for Q15 Q16 is referenced to the supply rail. This is good for the lower mirror, bad for the input diff-pair. If I had to choose, I'd try to keep the input diff-pair at constant collector voltage, and let the mirror suffer.
Triple-zero distortion is not going to happen outside the fantasy world of the simulator. Layout coupling will introduce more distortion than the simulator can simulate. If you commit no grave grounding sins, this amp will probably be double-zero THD through 1KHz.
The signal current variation in Q10 is quite large due to high load current and only 2 stage buffer. Many amps work fine this way, but if you are playing the numbers game you might consider either higher standing current in Q10 or a triple buffer.
As a practical matter: do you really want driver stage power supplies higher than output stage supply? It adds another power supply problem and cost. Generally cheaper to run both on the same supply rails and accept a little waste. Also it may be possible to forward-bias the C-B junctions of Q19 Q20 and blow them. |
|
|
| sreten |
I can't see what sets the standing current through the output
stage - at all - am I confused ?
:) /sreten. |
|
|
| tschrama |
Hi,
I like it very much, very spectacular :cool: , lots of features and proffesional looking... I only have some doubts about the CCS.. I'm not used to this arangement.. Seems like the Iq is refferenced to the Vsupply.. what if there's some 100Hz humm.. Seems like that humm will 'only' be attenuated by about 40dB .. but I could be wrong.. .. have been before;)
Succes!!,
Thijs |
|
|
| peranders |
R11, C6 D1 connected wrong? Constant voltage ACROSS emitter-collector of Q3 and Q4.
R11 to -35 V and C6, D1 to ground. You could also do a little overkill in adding a NPN instead of ground and then connect base between the emitters of Q3, Q4 (via two resitors) and C6 and D1 to it's emitter. Then you really will create constant voltage across Q3 and Q4. The collector of this NPN is connected to +35 V or something lower.
Why don't you build your VAS stage with small signal transistors and then add an extra emitter follower? |
|
|
| PanzerLord |
| quote: | | R11, C6 D1 connected wrong? Constant voltage ACROSS emitter-collector of Q3 and Q4. |
Cascode load.
| quote: | | I like it very much, very spectacular , lots of features and proffesional looking... I only have some doubts about the CCS.. I'm not used to this arangement.. Seems like the Iq is refferenced to the Vsupply.. what if there's some 100Hz humm.. Seems like that humm will 'only' be attenuated by about 40dB .. but I could be wrong.. .. have been before |
The power supply for the input stage is seaparated and will consist of a bootstrapped regulator, and, the psrr with the wilson CCS is far better that with the standard diode, just try it out in your school lab!
| quote: | | Looks like current in Q19 Q20 is undefined. I don't see a bias diode. |
the emitter resistors sets the value of Vbias (the value of these must be changed to 0.91ohm->0.7volts/ca0.8 amps), if you study the curcuit loong enough, you`ll understand that this will work. This type of biasing is used in the class A amp in R. Slone "High-Power Audio Construction Manual.
| quote: | | If Q25 Q26 are bias: you have no short-protection at all. For a home-brew with $1 transistors, that's fine, but don't loan it to your clumsy friends. |
study the schematic a bit longer, more on protection to come
| quote: | | s a practical matter: do you really want driver stage power supplies higher than output stage supply? It adds another power supply problem and cost. Generally cheaper to run both on the same supply rails and accept a little waste. Also it may be possible to forward-bias the C-B junctions of Q19 Q20 and blow them |
See above (the trick is to build a voltage doubler at the diode
bridge, then attach a ultra low noise regulator after the doubler, better than the one used in the A75 at www.passdiy.com)
More to come (sorry about the bad english), I was in a hurry writing this.... |
|
|
| Steven |
Hi Panzerlord,
At first glance (no time for details yet) I would say it looks good. The trick with Q25 and Q26 for biasing is also used in some Yamaha pre-amps. This works only for class A output stages and I wonder what happens if your speaker impedance drops and the amplifier needs to shift into class AB operation for higher output currents. It seems to me that at that moment the output stage will get into current limiting by the bias circuit.
Steven |
|
|
| sreten |
Q25 and Q26 looked like protection to me, I've never seen
that form of bias circuit before - so I was confused.
Probably still confused but what sets the Vas standing current ?
:) /sreten. |
|
|
| PanzerLord |
The current through R19 and R13 sets a voltage across Vbe on Q25 and Q26 (-> VbeQ26/R19 ), just stare at the coupling for five mins or so and you`ll get it. I`ve been lying all night thinking about.
However:| quote: | | This works only for class A output stages and I wonder what happens if your speaker impedance drops and the amplifier needs to shift into class AB operation for higher output currents. It seems to me that at that moment the output stage will get into current limiting by the bias circuit. |
Is a good point, thanks for bringing this up Steven, I have been so exalted about my project laltely that I`ve forgotten this.
I`ve also thought about what PRR said about input impedance, and I will look into that in the next couple of hours.
I`m thinking about replacing Q26 and Q25 with a modified Vbe multiplier....This is the first time I actually design and build an amplifier, before I have buildt from other peoples schematics, so you have to be a bit paicent. I want to put most of my effort in the design before I actually build it, in that way I`ll avoid falling into traps (I hope :cannotbe: ).
And I`m very grateful for your questions and for your criticism, it makes me think twice about the various soulutions regarding the topologies I`ve used, and hopefully make my design closer to perfection..(if it at works that is....)
;) ;) ;) ;) |
|
|
| PanzerLord |
| Yeah, what should the input impedance be in balanced mode, I was thinking about 10kohms, but is 5kohms to low? |
|
|
| PanzerLord |
OK, have increased the input impedance to 10kohms
I have also replaced the current mirror with a wilson current source active load (Q1, Q2 and Q29), the results where better linearity (I had some problems with instability around 50MHz, that`s why the values of the CC`s was so high), and far better bandwith, and of course a minute improvevement in THD.
But the main task of the wilson active load is to increase the bandwith and decrease the top I had around 50MHz.
I was then allowed to lowere the values of the Two-pole CC`s to around 50pF for C2, and 300-500pF for C1, the results here where even more improvement in reducing harmonic distortion...
I have taken what Steven mentioned about lower impedance loads into considerations, and replaced the "mystic" Q25 and Q26 with a Vbe multiplier. I realised (he was right:) ..hehe), that, obvious, Q25 and Q26 didn`t just determine just the bias current in the output transistors, but also, on the other hand, functioned as a current limiter.
Thus, the output current was set to a total of 3 amps no matter what impedance the load had.
Once again, sorry about that bad english, guess I have to work even more on that...;) |
|
|
| PanzerLord |
| here is the new input and VA stage |
|
|
| PanzerLord |
Here is the new output, I`ve resized the emitter resistors, Q25 and Q27 functions as ashort circuit protection, it allows the output to deliver ca 150-200W max to the load (around 360W total, 2A max per output pair).
Have also inserted a zobel network.
The 8ohm resistor is for simulation purposes only |
|
|
| PanzerLord |
| quote: | | just try it out in your school lab! quote: |
or at work or in your own minilab at home, if you got one.
I wasn`t assuming you where a student (sorry about that), I just forgot to extend the sentence, my fault, sorry!! |
|
|
| Steven |
Maybe you are aware of this article, but for others it might be interesting: a nice tutorial about balanced inputs (and outputs). Especially input impedances are not always what you think they are, depending on how you drive the input.
http://www.dself.dsl.pipex.com/ampi...ed/balanced.htm
Anyway, it was a good thing that you increased your input resistor at the negative input. That one was far too low in the first diagram. Even now your input is probably not 10k, as you think it is, but less. See the article above; it is described somewhere in the middle.
A nice mind-game: how can we extend the original Q25/26 construction to have its good properties for class A, while still being able to deliver more than twice the idle current if required? I have not thought about it yet, the question just pops up.
Steven |
|
|
| PanzerLord |
| quote: | | Some of these impedances are not exactly what you would expect. In Case 3, where the input is driven as from a transformer with its centre-tap grounded, the unequal input impedances are often claimed to "unbalance the line". However, since it is common-mode interference we are trying to reject, the CM impedance is what counts, and this is the same for both inputs. The vital point is that the line output amplifier will have output impedances of 100 Ohms or less, completely dominating the line impedance. These input impedance imbalances are therefore of little significance in practice; audio connections are not transmission lines (unless they are telephone circuits several miles long) so the input impedances do not have to provide a matched and balanced termination. The low impedance of 6.7k on the cold input sounds impossible as the first thing the signal encounters is a 10k series resistor, but the crucial point is that the hot input is driven simultaneously, so the inverting op-amp input is moving in the opposite direction to the cold input, due to negative feedback, a sort of anti-bootstrapping that reduces the effective value of the 10k resistor to 6.7k. The input impedances in this mode can be made equal by manipulating resistor values, but this makes the CM impedances (to ground) unequal, which seems more undesirable. |
The way I interpretered it,it is better to have an unequal differential intput inbalance, than have an unequal CM impedance, what do you think? Is there any way I could get both equal?
I dont`t think I have enough mathematical skills to solve this problem, but thinking ALOT about ways to solve this problem.
Thanks for letting me know...
I`m stilll thinking about using the first solution using Q25 and Q26 as bias transistors...:smash: |
|
|
| PanzerLord |
| quote: | | differential intput inbalance |
unequal input impedance! |
|
|
| PRR |
> Is there any way I could get both equal?
Not with just one op-amp. |
|
|
| PanzerLord |
| This is for Johanlo;) |
|
|
| Christer |
| quote: | Originally posted by PanzerLord
This is for Johanlo;) |
Haven't really followed the thread, but two ímmediate comments
on the graphs.
1) What is the unit of magnitude in the first? dB I guess.
2) What is the level of the signal for the distorsion graphs? They
tell us nothing withoug that info. |
|
|
| PanzerLord |
Ok, so, as I understand it:
The only way to achieve symmetry is to actually use two amplifiers, is that something that`s common when relating to power amplifiers with balanced inputs?
I`m pretty sure I know what`s going on inside an unbalanced system, when it comes to balancing, I have absolutely no experience (I don`t know why I set so low values on the input resistors in the beginning, it`s pretty obvious they where to low).
Should I just drop the whole balancing issue, and go for unbalanced? I`m getting far better results in balanced mode after the last modifications. But it seems that the input impedance on the - input is pretty lower than the positive input...which of course makes sense.
Again, thanks for your contributions! |
|
|
| PanzerLord |
The unit of mahnitude is dB.
The signal level for the distortion graphs is 28V peak with a 8ohm load (50W). |
|
|
| PanzerLord |
| I`m just about finished with the power sypply schematics for the input and VA stage, I`ve attached a preliminary schematic over the voltage regulator for one rail... |
|
|
| Steven |
| quote: | Originally posted by Steven
A nice mind-game: how can we extend the original Q25/26 construction to have its good properties for class A, while still being able to deliver more than twice the idle current if required? I have not thought about it yet, the question just pops up. |
| quote: | Originally posted by PanzerLord
I`m stilll thinking about using the first solution using Q25 and Q26 as bias transistors...:smash: |
Hi PanzerLord,
I think I've found a way to keep your Q25 and Q26 for the biasing of the output stage. As long as the output currents are easily within the class A region, you get the advantage of this bias method, i.e. a current increase in the upper half causes an (almost) equal current decrease in lower half, vice versa. For fixed voltage bias systems you don't get this linear behavior, but a reciprocal behavior instead: upper half current doubling gives lower half current halving, vice versa. Well, that is only the case for emitter resistors of 0 Ohm and BJT's. In practice it is between these extremes.
The circuit I have in mind acts as the original one, but if the currents are getting too large (for normal class A) it will keep the almost cut-off transistor conducting and allows the other one to deliver more current. The circuit will shift into class AB, but none of the output transistors will be completeley cut-off ever, so the amplifier is essentially non-switching, although operating in class AB for large output signals.
I keep you guessing for the moment, leaving you :confused: or :dodgy: , but I would like to do some more simulations before posting. Hint: my circuit adds only two diodes and two resistors to your original circuit. :cool:
Steven |
|
|
| PanzerLord |
| Hi steven, thanks for your interest, I`m a bit :confused: , but I think I know where you are going. I`m curious about your solution! I can hardly wait....!:) |
|
|
| PanzerLord |
| C2 and C1 are reduced to respectevely 39pF and 390pF, reducing THD with 20%, no influence on 2nd and 3rd harm, but higher ones were reduced. In the real world, this would probaly hav little or no effect, was just trying to tweak the circuit... |
|
|
| PanzerLord |
After som soing some serious mind work around the balanced input issue, I`ve come to the conclusion that the only way to achieve 100% balancing is going for bridged mode.
I just doubled the cost!!:bawling:
In theory, I quadrupled the power output, well I actually did it in reality also. But my main goal was to construct a 50W class A power amplifier, correct me if I`m wrong but:
-If I let the bias current be the way it is right now (total 2A per amp), the balanced (and bridged) amp will deliver 100W class A, and when the voltage swing is +14V RMS for the left one, and -14V for the right one, which gives 28V->100W in 8ohm`s; it will leave the class A mode an switch to AB when delivering 100W-200W (ok ca180W).
Look at the schematic and correct me if I`m wrong (Vcc is still the same, +/-38 for the input/VA-stage, and +/-32V for the output stage.
However in the original design I increased the input impedance so the balance was 20k/7k, I don`think I`ll construct the "bridged" design, because I honestly think there isn`t any sonic benefits when the input impedance is so much greater than the output imp of the preamp. On the other hand, if I construct my new design, my power output will be kick-!"#¤"!#!!:devilr: |
|
|
| PRR |
> only way to achieve 100% balancing is going for bridged mode.
No. In fact it is normally best to separate the differential-input function from the power amp function.
In your plan, if you ever must feed it an unbalanced input, you can only get 50 watts of the 200 watts you paid for.
There are two main ways to take a balanced differential floating input:
The left one, without buffers, is commonly done and often better than unbalanced, but flawed. With buffers it works very well, at the cost of three separate amplifiers (though the input buffers may be unity-gain which may offer a simple solution).
The right one needs two amplifiers and has a fixed medium input impedance. It uses only inverters, which may allow a simple design or makes CMRR of an op-amp unimportant. You can't use high value resistors or noise becomes an issue. You probably don't want to try to use the power amp as the lower amp, because if you need gain (power amps usually do) then the feedback resistor value gets awful high for good stability.
There is another 2-amp diff-input which gives infinite input impedance but is confusing to gain-structure. |
|
|
| PanzerLord |
Hmm, I think I`ll go for the original idea anyway (without buffers,as the left schematic indicated), allthough the input impedance will be unequal, I don`t think it will become a problem...adding buffer stages will add more active components in the signal path, I really don`t want to do that. I`d rather live with the impedance imbalance....
The circuit above, however, will deliver 200W (theoretically) into 8 ohm, the question is at what powerlevel...assuming the -signal is 180 degrees phasshifted related to the +signal (referred to gnd).
I not going to building an amp with option for unbalanced input, since the preamp has only balanced out.
I`ve buildt a power amp for my sub some years ago using this kind of design, and it delivered around 150W into 8ohms. I designed a phaseshifter with a inverter and a buffer...it worked nicely untill a capacitor made a hole in the ceiling (to low voltage spec.... :xeye: ). |
|
|
| Steven |
Another way to get a balanced input and single ended output is to use a circuit with a long tail pair input, apply input signal on both bases and feedback to the emitters. Since it has a single ended output only feedback to one emitter is possible, but a resistor of equal value as the feedback resistor should be connected from the other emitter to the ground to balance the input. In this way the output will be referenced to ground. An example is the already quite old SSM2015 preamp from Solid State for Music (SSM) from the eighties. Later SSM was acquired by Analog Devices.
You can find a datasheet here.
A picture is shown below.
Don't get confused by the special biasing of the input stage by means of an additional amplifier that sets the bias curent of the input differential. This improves common mode rejection. Probably you can do without and just use current sources for the tail current.
Steven |
|
|
| Solid Snake |
| Looks like a great design. By the way, where do you get your heatsinks from? |
|
|
| PanzerLord |
You see, I have a 220V to 2x23V AC transformer and a 0,3C/W heatsink laying around.
At first I wanted to design a 150-200W class AB amp, but for that I needed a 220V to 2x38V transformer. I want to put my old one in use before I do anything else.
A 0,3C/W heatsink is a bit large for a 50W class AB amp, so I thought, why not put it in real use and design a Class A instead?
The backside is, I have to buy another heatsink (ca. $80 in US currency I think).
I`m still having problems with the balancing issue, the schematic steven posted seemed logical, but that would ruin my original design (which is my first serious attempt in the upper class Hi-Fi-amps, have buildt some mid-range amps before, hence the heatsink and the tramsformer). The reason to this is, it was very important, from day one, that the design should consist of as few active components as possible in the signal path (I don`t consider the cascode as an active component in the signal path).
Adding op-amps or more gain-stages is from my point of view, undesirable, I`m actually considering going unbalanced, but we`ll see...
I Have a schematic of a matching unbalanced preamp lying around also, of course; i could drop the global NFB, but in my opinion, that is far more undesirable. This will increase THD dramtically!
I`m already working on the PCB`s, but it will take some time. This is 50% of the design process, electromagnetic influence and ground loops must be avoided at all costs |
|
|
| gustav |
When Q10 saturates hard, when the amp goes into heavy clipping, Q9 will try to clamp the supply voltage.
Guess what happens ! Q9 will burn (yes I have seen it)
A series resistor will help Q9 survive.
What about HF stability ? Large signal stability requires more phase margin than one might think.
Good luck with testing !! |
|
|
| Steven |
| quote: | Originally posted by PanzerLord
I`m still having problems with the balancing issue, the schematic steven posted seemed logical, but that would ruin my original design (which is my first serious attempt in the upper class Hi-Fi-amps, have buildt some mid-range amps before, hence the heatsink and the tramsformer). The reason to this is, it was very important, from day one, that the design should consist of as few active components as possible in the signal path (I don`t consider the cascode as an active component in the signal path).
Adding op-amps or more gain-stages is from my point of view, undesirable, I`m actually considering going unbalanced, but we`ll see... |
I think the concept of the SSM2015 can be incorporated in your design without adding any active component and actually leaving almost everything intact. Only the connection of the feedback should be changed and one balancing resistor added. Of course some resistor values will change. Note that now at least the input pair should be able to deal with the full input signal, because that will be applied between the input transistors, so Ic(Q5)*(R1+R2) should be big enough. This might affect the distortion in a negative way (not simulated), but that will not make the sound worse necessarily.
In the diagram I used the component labels of your first diagram in this thread.
Steven
PS I know I still have to post my Q25/26 solution. |
|
|
| PanzerLord |
The input voltage on q9 will never exceed more than a few tens of millivolts, q9 and q10 is (in simple terms) a I/V converter, the "q9-stage" has a gain=<1, it acts as a buffer for the very large output impedance from the gm stage (diff amp).
So Q9 will never saturate "hard". In fact it operates in a very small area of AC voltage. |
|
|
| PanzerLord |
| Hi Steven, thanks for the schematic (i gotta have it on a sliver plate :)), it made sense to me at once. I`ve done some simulations on it, the THD went from 0,0004% to 0.007%, nothing big, the 2nd harm went from 10uV to 400uV, and 3rd from 1uV to 100uV. These are pretty large jumps, plus, DC offset rose from -20mV to -0,7V....less CMRR? |
|
|
| Steven |
Distortion will go up for sure, especially the lower harmonics. You get far less loop gain, so also less feedback. This is because of the large values for R1 and R2 to make sure that the input pair can handle the complete differential input voltage. Because of less overall feedback non-linearities further on in the amplifier will also be surpressed less.
I'm a little surprised about the high offset though. Since you are simulating I would expect that you have perfect matching of the input transistors and also perfect no tolerance resistors. There seems to be a kind of unbalance around the input stage. The tail current of the differential pair is quite high compared to the base current demand of Q9, so that can hardly give an offset. Maybe you can find the cause by analysis of the simulation results. In practice the offset can be trimmed by connecting the balance resistor from Q3e not directly to ground but to a variable voltage close to ground, or even to the output of an integrator (servo) from the output, compared to ground.
Steven |
|
|
| gustav |
Frode,
Q10 (not Q9 !) will saturate and then draw a large base current !
The feedback loop will provide basecurrent to Q9 and the power in Q9 could reach about 1mA*100*40V=5W.
Q9 will then be destroyed !
I beleive a good amp should survive hard clipping. |
|
|
| PanzerLord |
Sorry, I mixed them, Gotta go to bed earlier...:)!
The max power capabilities for BD139 is 12,5W (ON semiconductor, not Philips which is only 8W), It can handle at 40mA at 70Vdc. I`ll of course attach it to a heatsink I think that`ll do...Im` considering placing a Re with 4-10 ohm`s for temp-stability though....and maybe...hmm..I`ll get back in a moment!:xeye: |
|
|
| PanzerLord |
Ok, here`s a possible protection circuit:
Q32 and R29 limits the power dissipation in Q8 to ca. 3W. |
|
|
| gustav |
Hello again,
the problem is not BD137. It is 2sc2240 which only can handle something like 0.3W.
All you need to do (in the first scm) is put a resistor (maybee 2kOhms) in the collector of the 2240 ! |
|
|
| gustav |
| Sorry , BD137 should have been BD139 |
|
|
| Lelak53 |
Dare I disturb?
Whatever is the purpose of the circuit, lowering the value of R30 would make more reliable design. Bypassing it with a capacitor would speed-up the reaction. |
|
|
| PanzerLord |
| Thanks Gustav, have inserted a 1.6k resistor in series with the collector. |
|
|
| Steven |
PanzerLord,
I think it is time to post my solution for the Q25/26 problem. :cool:
The circuit below shows it. I only added two resistors R5 and R8 and two diodes D6 and D1 (strange default labels BTW). The diodes are Infra Red LEDs of the GaAs type with a forward voltage of 1-1.2V, like CQY36. Most GaAlAs types have a higher Vf of about 1.3-1.5V and cannot be used, just as other LEDs with visible colors.
For small output currents the voltage across R3 and R4 is determined mainly by the Class A biasing and Q2 and Q4 keeps that voltage around 0.6V. The Vf of the LED is too high to have influence. If the output current is increased by a bigger input voltage or a heavier load the voltage across R3 increases for positive signals and the voltage across R4 decreases, for negative signals the other way around. Without the LEDs the maximum output current would be twice the bias current, since at that moment the voltage across R3 (or R4) becomes zero and the voltage across R4 (or R3) cannot increase beyond VbeQ2+VbeQ4. But with the LEDs the contribution of the voltage across R3 to the Q2/4 is gradually clipped, leaving some voltage left for R4. For negative signals the other way around.
This can be seen on the two graphs. The top one shows the output current through R3 and R4 with 10Vp input sine. The output stage is clearly in class A with a bias current of 70mA. The second graph shows the same but for an input signal of 30Vp. Now the stage is in class AB. But note that the current through R3 and R4 never reaches zero, just as the current through the output transistors will not go to zero. This is a "non-switching" class AB, i.e. the current is never switched off completely, nor reverse biased as you can get with normal fixed voltage biasing (Vbe multiplier).
Instead of using IR LEDs for this trick, you can also use a normal small signal Si diode and a Schottky diode in series, giving a Vf of about 1V. A third method is to use a Vbe multiplier with a multiplication factor of approx 1.6. The bigger the Vf, the smaller the minimum current will be in the output stage. If Vf becomes more than 2Vbe it will have no effect.
Of course, distortion will rise as soon the output current halves are "shaped", actually pre-distorted by the action of the LEDs. But this rise in distortion will not be so dramatic as the currents through R3 and R4 suggest. They will be subtracted from each other. The output voltage will still have a distortion less than 0.1% (open loop!). In this simulation most of the distortion comes from R1 and R9 due to varying base currents of Q1 and Q3. This will improve by using darlingtons for the ouput transistors, as in the diagram at the beginning of this thread.
The emitter resistor values are quite high in this example and only small signal transistors are used, but that is only because of my lack of models for power transistors in my free downloaded Circuitmaker program. The circuit can easily be used in power amplifiers. The value of R5 and R8 can be from 20 Ohm - 1kOhm. A small value will keep the output longer in class A without starting to shape the currents, a bigger value will increase the influence of the LEDs. This is simply because R5 and R8 are voltage dividers with the Rd of the LED as second resistor.
Steven
PS Just an idea: if you use optocouplers for the LEDs you can use the output current as an indicator whether the amplifier is in class A (no current) or class B (current). Of course these should be IR optocouplers, but I have not checked yet if these exist (or are these always IR?). |
|
|
| PanzerLord |
Oki, some new modifications, added a cascode in the VA stage and VA CS (ok, I cheated, I stole a bit from peufeu`s design), the cascode at the input stage has also been transformed in to a hawksford (thanks Andy C.). I`ve replaced the BD139/140 with
MJE340/350, it may become a 100W class A......:), -higher rail-voltage.
CC`s have been scrapped due to the excellent stability!! MAN, look at the bandwith and phase (without zobel-network)...
The distortion curves are almost flat throughout the audiospectrum!
I`m designing the PCB as I speak, I`m making an option for CC`s and VA without cascode, just in case the simulator`s fooling around with me). |
|
|
| PanzerLord |
AC analysis, red line is without zobel, blue with zobel. There will be an additional low-pass filter at the input.
Magnitude is dB. It`s flat all the way past 10MHz! |
|
|
| PanzerLord |
| The 2nd and 3rd harmonics (all simulations with 8 ohm load at full voltage swing +/-28volts peak-peak). |
|
|
| Konrad |
Hello
Here is one eksample of aktive loading to the innput stage.
Q1- Q2 is input transistors, Q3 Q4 is the aktive load. and Q5 Q6 the Vamp stage. Resistor values has non realistic values. |
|
|
| Workhorse |
| The Cascode loaded Diffrential improves the high frequency response of amplifiers, this is quite a good reason why u have used so many of them in clusters and current mirrors also |
|
|
| thanh |
hi!it's distortion is quite high above 10khz,
sorry to bad english |
|
|
| PanzerLord |
high above 10kHz? The distortion at 50kHz is less that
0.00000714% in both cases! I won`t consider that as high, I consider that extremely low...
I`ve also experimented with other bjt types, 2N3904/3906, 2SA872/2SC1775(very high beta -low noise). The design is very little dependent on bjt parameters, the only thing one has to do when replacing bjt is adjusting the current sources.
This shows that the amplifiers sound characteristic isn`t "that" dependent on the transistor type itself , but more on the actual design topology (which I belive isn`t the whole truth).
This is, however thruth when it comes to the overall distortion , no matter what bjt`s I replaced (the wilson must have equal beta bjt`s, and current sources must have high beta "reference" bjt`s), the 2nd and 3rd harmonics was below 500nV@20kHz/5uV@50kHz. That is low, (even for a pre-amp with 2V out).
In fact I measured distortion below 40mV@2MHz just now, that`s pretty wild...(this is without the zobel) |
|
|
| jcx |
I am an advocate of high feedback through high open loop gain but I worry that your results are “too good to be true” – I am having a hard time believing the ~ 160MHz gain-bandwidth from the 50pF * 20 Ohm (20 ~=1/input stage gm) compensation time constant working with the TIP output devices
With ~4 MHz ft , the TIP output devices should limit the usable loop gain-bandwidth too much for the reported distortion to be likely, are you sure you don’t have a “disable semiconductor capacitance” switch set in your simulator software? – some simulators have this option
And I feel that the compound Baxendall super-pair VAS is wasted with the relatively low impedance from the Darlington output stage, the load on the VAS from the Darlington is ~ 100 – 300K @DC and the load should start to increase > ~20KHz (from TIP35/6 output device Hfe roll off) – with such a low VAS load a simple cascode should be fine without the potential instability we are discussing in the Baxendall thread
The two pole compensation also loads the VAS; if the compensation network isn’t driven from a low impedance such as the mje emitter or the output, the two pole cap ratio should be resized to ~ 1:1 to minimize the load impedance |
|
|
| PanzerLord |
I don`t use the TIP devices for anymore, the simulation is done with MJ15001/15002 (hmm, ft=2MHz...?), and the 2-pole CC is gone...
By the way, I think you got a point regarding the baxandall, I replaced it with an ordinary cascode, and it didn`t show significant changes in distortion (or bandwith)... |
|
|
| thanh |
can you explain to me why the distortion become higher at high frequency above 10khz?
have you ever seen stochino's schematic,he connect 1 diode from C to B of mirror? |
|
|
| thanh |
| i think that how is the distortion if Q7 and Q8 is replaced by the CFB ,do you try to do like that? |
|
|
| maylar |
| Forgive my ignorance - and for comming into this thread late - but how does this design qualify as Class A? What is the criteria for that designation? |
|
|
| thanh |
panzerlord,where are you?
i'm sorry.i should say Q7 and Q8 are replaced by a pair of CFP,i saw peufeu's site |
|
|
| rbroer |
| quote: | Originally posted by PanzerLord
I`m currently etching the PCB`s
...
Give me some feedback on this one...
:) |
is how he started this thread :D
What a waste of boards since :(
I think he's busy etching PCB's ;) |
|
|
| PanzerLord |
lol.. Didn`t come out the way I meant it...
PCB`s went straight to *****. They had very poor quality...nuff said.
My computer crashed afterwards (got a really annoying message in XP startup, claiming system32 directory was lost or something.
Anyway, all my design and so on where lost. I`ve formatted and installed a new system, and I`ve redrawn all circuits, and a new pcb layoutprogram installed (ultiboard 2001). The PCB layouts are soon finished.
Also, the 50W project suddenly went 180W! Yey...new job=more money=bigger amp! |
|
|
| thanh |
| Panzerlord!Should i add 2 compensation capacitor into Q5 and Q6 from B to C? |
|
|
| thanh |
I'm sorry!
This is lavardin's patent.I have got it from peufeu's site.How is the distortion of it?This is only simplified schematic.Can you design a schematic from it? |
|
|
| thanh |
This is Aiace's circuit.I have just added R3 and R4.Should i do this?If i'm right ,how can i calculate the value of R3 R4?
thanks! |
|
|
| PanzerLord |
Hi Tahn, sorry for my late reply!
I don`t think q5/q6 need any CC`s, however, I`m a bit sceptic about stability due to the lack of CC between the collector on U9 an base on Q3. Things have been going slow lately, and I still haven`t buildt the amp. :whazzat: . But I`m still simulating and modifying the circuit.
I also tried swapping the diff bjt`s with jfet`s (out of curiousity of course), and the simulator showed very good numbers, not better than bjt`s...but better than I thought.
Regarding the lavardin schematic: I don`t know anything about the distortion preformance...I haven`t done any simulations...yet.
Regarding Aiace's circuit: I don`t see any reason why you should implement R3 and R4, I`ve tried values ranging from 100ohms up to 2kohm`s and the distortion increased with increasing values. No resistors at all showed best preformance..... |
|
|
| thanh |
| Panzerlord!What simulate sofware are you using?I have used ORCAD But I don't know how distortion is calculated |
|
|
| millwood |
| quote: | Originally posted by thanh
Panzerlord!What simulate sofware are you using?I have used ORCAD But I don't know how distortion is calculated |
you can do FFT and manually add it up, or you can find it under either "measurement" or "performance analysis". Just need to select a signal and apply the right measurement / analysis.
OrCad offers a video on how some of the functions work. pretty neat. |
|
|
| PanzerLord |
| I`m using Multisim 2001. Haven`t tried OrCAD yet, is it any good? |
|
|
| thanh |
This is Steven's current source.What is it called?How can i calculate the output resistance of it?
Thank a lot! |
|
|
| pjacobi |
| Voltage Gain of JFET * Early Voltage of Bipolar * (1/I + R/25mV) |
|
|
| thanh |
So how is voltage gain of JFET if i use 2SK30?
Is JFET noised by electric field?Can you understand me? |
|
|
| pjacobi |
| quote: | Originally posted by thanh
So how is voltage gain of JFET if i use 2SK30? |
You have to evaluate the diagram in its datasheet for this. Or look into its SPICE model parameter. I have neither of this for 2SK30.
But anyway, if you evaluate the formula just with typical values, you will see that the result is just 'very large'. So the total impedance of the node the current source is connected to, will be dominated by the other connections. For all practical purposes, the circuit drawn is a ideal current source.
Example BJT=BC337-40 JFET=J112 R=180 I=5mA
Simulated impedance is about 90 Meg
| quote: | Originally posted by thanh
Is JFET noised by electric field?Can you understand me? |
No, I don't understand the question.
Regards,
Peter Jacobi |
|
|
| thanh |
| SO how is the output resistance of Panzerlord's current source? |
|
|
| thanh |
| Electric field can effect to JFET.IT can be a kind of noise of JFET |
|
|
| PanzerLord |
Stevens current source has greater output resistance, but the jfet introduces more noise.
The simulator showed no significant improvement in distortion preformance (probably due to the greater VA and output stage distortion). I suppose the distortion figures in the diff-stage are somewhat better (marginally). |
|
|
| Steven |
| quote: | Originally posted by PanzerLord
Stevens current source has greater output resistance, but the jfet introduces more noise.
The simulator showed no significant improvement in distortion preformance (probably due to the greater VA and output stage distortion). I suppose the distortion figures in the diff-stage are somewhat better (marginally). |
I don't think the JFET will add noise in the case of a BJT current source with JFET cascode. Gate current is close to zero, so drain current is equal to source current, which is equal to BJT collector current. Ergo: the JFET does not add noise. All the noise comes from the BJT, its emitter resistor and what is in front of the BJT.
Steven |
|
|
| PanzerLord |
| hmm.. you`re probably right, I just took an "educated" guess, sorry....haven`t really experienced it (only simulation). :Popworm: |
|
|
| thanh |
| quote: | | jfet introduces more noise |
I think jfet doesn't introduces more noise.Its noise come from electric field strength |
|
|
| thanh |
| In orcad ,I did "transient",click button "performance analyst" but pspice say"more ... performance analyst".I want to see 2rd harmonic voltage to frequency and THD to frequency.So should i do? |
|
|
| PanzerLord |
| As an amplification device, JFETs are more noiser than bjts. |
|
|
| jam |
| As an amplification device either one could be more noisy if you you take the source impedence into account. |
|
|
| thanh |
io=2mA,how can i calculate the value of R1?
thanks a lot! |
|
|
| jam |
Divide 0.6 by 2ma. to get your answer. (300 ohms)
Vbe is the baes emitter drop which is 0.6 volts. You have to transpose I and R in the equation.
Regards,
Jam |
|
|
| thanh |
| Oh!Sam!thanks you!BUt i want to know the value of R1 not R2 |
|
|
| jam |
Thats Jam.....:D .
To calculate R1 you want about 3 to 4 mls running through the reference. First you need to find out how may volts there are across R1. Lets say it is 40 volts then if you need 4ma the resistor required is 40/4ml =10k ohms.
Regards,
Jam |
|
|
| jam |
R1 = 35/4ma = 8.75k ohms.
Regards,
Jam |
|
|
| thanh |
Hi Panzerlord!How are you?how are your amp?You haven't visitted this forum for a long time.
Can you reduce more distortion ?If your power transistors are biased in class B,how is the value of distortion?
You seem to not like symmetrical topo. |
|
|
| demogorgon |
it seems panzerlord har vannished..
he's dissapeared from www.overklokking.no too, a shame really.. i was looking forward to see the ressoults of this amp. |
|
|
| cunningham |
| Just a thought for Panzerlord...What about using a small audio transformer to couple (& stabaly bias) the input? This may make it easier to achive a balanced input. |
|
|
|
