my_amp design, ver.02

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you got me off that first design ILIMZN!
what dou you think about this one?
 

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Your first amp was actually a very interesting design, it could have proven well worth the effort of figuring ot the bias scheme to make it work.
The one you have here is more 'classical' and I don't see any obvious problems. I do have some pointers, though:
1) If you are using a single MOSFET pair, it will prove worthwhile to omit the source resistors (0.22 ohm) if you can find a well mached MOSFET pair. MOSFET gm is relatively low, and if you can avoid making it lower by including source resistors, do so - of course when youa re using multiple pairs, resistors are needed for current sharing, and besides, two MOSFETs in parallel = 2x gm.
2) There are benefits to replacing the two diodes used as a reference for your current sources by something with a higher voltage drop. Consider this: the gain of a single common emitter BJT stage is roughly Rc/Re for large values of beta. In your current source, Rc is very high, so making Re small, due to the small voltage drop required to get the required current (0.6/Re roughly), increases the 'gain' of the current source with respect to it's reference voltage. Fortunately, two forward biassed diodes do not make much noise, but the nosie they make is amplified by the current source. You can effectively reduce the noise by increasing Re. In order to keep the current the same, your reference voltage needs to increase (more voltage drop on Re). Try two LEDs in series, for instance. Also, increased Re gives you better thermal stability, though this is not overly important in this design.
3) Your bias arrangement will overcompensate the thermal gradient of the MOSFETs. It is also unnecesairly symetrical, although this can have benefits for some heatsink mounting arrangements. Be sure to bypass it with a quality capacitor, it needs to emulate a low impedance voltage source as much as possible. This is especially important with MOSFETs as their Cgd is nonlinear. However, because youa re driving two, in a 'differential mode', as one Vgd increases, the other decreases, so the nonlinearity is partially canceled - but only as long as the path between the gates of the MOSFETs is low impedance! Regarding temperature compensation, MOSFET Vgs at constant current rises about twice as fast as BJT Vbe. However, the treshold voltage is about 4x as high compared to BJT, so a Vbe multiplier has to multiply Vbe - and it's temperature gradient! - by 4 to bias a MOSFET. Because the temperature gradient multiplies by 4 also, it ends up 2x higher than needed to compensate the MOSFET. Even worse if you put source resistors on the MOSFET as these will reduce gm variation with temperature. For amps where you can avoid source resistors on the MOSFETs (i.e. ones that use a single pair of MOSFETs), I have found a Vgs multiplier to be the ebst option. A TO220 MOSFET like the IRF510, 520, 610 work great for this. Alternatively, and also for cases where source resistorsa re needed, you can use a Vbe multiplier but you need to add some emitter degeneration to reduce it's temperature coefficient. This is best done by including a small resistor in the emitter line of the Vbe multiplier - in your case, between the emitters of the BD139/140. For cases where driver current is well controled (i.e. current sources), you can also use a Vbe multiplier to get the required thermal gradient, then pad the remaining voltage drop to get to the proper threshold voltage for the MOSFETs using a regular resistor. The Vbe multiplier will need to multiply with about 2, so it will have about 1.5V voltage drop. The remaining 1.5 you get using a suitable series resistor - this is for your type of Vbe multiplier. For a single BJT Vbe multiplier, everything is doubled as it has to cater for both the N and P MOSFET thermal gradient, so it needs to multiply by about 4, dropping about 3V, then add resistor to get t o about 6V total across gates for proper idle current.
 
You might want to increase values of VR1 R18 and R19.
Maybe two times higher values.

For example VR1 = 1k ( or 470 Ohm + 500 Ohm trimmer )
and R18/R19 1k5.

This will run a bit more of the current through Q9/Q10
and a bit less in R18-VR1-R19.


R9 and R13 (2 k) can be removed.
They do nothing useful.
( Except if you want to measure current balance in input stage. )


I would also, as mentioned, reduce 0R220 emitter resistors of MOSFETs.
At 5 Ampere peak current, there will be a drop of 1.1 Volt across those resistors.
I would say make them 0R1 or maybe even two 0.1 ohm in parallel (= 0.050 Ohm )

How much idle bias are you planning?
 
ok.

-I made some changes, lowered source resistors and decidet to use Vgs multiplier and changed diodes with zenner(is 2.7Volts enough?

-What should be the values of C6 and C7 ?


-Q7 and Q8 emmiter current is 15.5mA & Output Mosfet current is 100mA. What`s the best relation between these stages idle currents?


-If I raise the bias current, to run the amp in class A, could it be loaded with 4 Ohm speaker?
-What will be max output power for this amp in class a (my power supply has +/-42V 4A)
- I`m gonna use IRFP240/IRFP9140.
 

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bogdan_borko said:
ok.
-I made some changes, lowered source resistors and decidet to use Vgs multiplier and changed diodes with zenner(is 2.7Volts enough?

The more the better! Since your input will not exceed about 6Vpp, in theory you could have up to (power supply) - 3V reference. of course, this is not a good idea, some safety factor is needed. Since your current source current is (Vzener-0.6)/Re, you can always adjust Re to fit. But you need to know two things: 1) zeners are noisy, so there has to be a cap across them to filter this out - forward biassed diodes are MUCH less noisy which is why LED diodes are often used, and 2) the most temperature stable zener voltage is around 6V, LEDs have a mixed temp coefficient (neg then pos) and are also overall quite stable.

-What should be the values of C6 and C7 ?

You will need to determine that with the simulator, and partly experimentally. That being said, this may not be the best place to use caps as frequency comp for this amp.

-Q7 and Q8 emmiter current is 15.5mA & Output Mosfet current is 100mA. What`s the best relation between these stages idle currents?

Eh, basics, basics!!!
There is nothing directing the RELATIONSHIP between them, these are not BJTs. You chose the current of the driver stage by calculating what is needed to charge and discharge gate capacitances of the MOSFETs. In your case, it is the Cgd that is most imoprtant, as Cgs is 'boottrapped' and largely invisible - actually, the more so the higher the idle current (because gm increases with current on MOSFETs and the gigher the gm the closer the gain of the output is to 1, at 1 the Cgs are completely invisible). The available current from the drivers sets up the maximum speed you can drive the output stage at. Also, it would be good if the actual bias current of that stage is signifficantly higher (like 10x) than the actual gate currents needed, so that gate charge nonlinearity is small compared to total current, as the current variation will modulate the BJT gm - in this case it is guite important as these are effectively the VAS. In your case, your limit is the power dissipation of the BD139/140 that youa re willing to accept, as well as their SOA. From this you have to calculate base current, and also from that, the diff amp currents - but i will get to this later.
As for output bias current, there is no real sweetpoint like with BJTs. However, higher bias current means more gm, and less distortion in that 'first watt'. It comes down to experiemnt, preference and how big your power supply and heatsinks are!

-If I raise the bias current, to run the amp in class A, could it be loaded with 4 Ohm speaker?
-What will be max output power for this amp in class a (my power supply has +/-42V 4A)

Your questions point to a lack of understanding on how power is derived in an amp. Assuming your transistors can take the heat and current, the ONLY thing that determines the output power is the power supply rails and the load. Secondairy tot his is how well you drive the output stage. In your case you will be losing some 7-8V off the power rails, not counting rail droop under load. So roughly your power output will be ((Vrail - 7)^2)/(2 x Rload), and that is a maximum.
Realistically, you CANNOT use this amp in class A with your power supply and 4 ohms anywhere even close to that maximum. First of all, because maximum current for 42V rails and 7V loss and 4 ohm load is (42-7)/4=8.75A, and for true class A you need a bias current of half that, i.e. 4.375A, already more than your power supply can handle. Also, at +-42V and 4.375A the power dissipated as heat is 367.5W - WAY over the limit for the transistors. All of it of course assuming perfectly 'stiff' supplies, which they will not be - at 4A current your supply lines will droop SIGNIFFICANTLY, perhaps as much as 10V. It's all very basic stuff. If you want class A, you will need to start your design with the heatsink, and derive everything from that, eventually ending at the power supply - which will have a signifficantly lower voltage than you have now, perhaps only half as much as now.
Even with an 8 ohm load, you are not much better off - half the bias current, half the dissipation - but it's still 183.75W. You woudl neeed a truly huge heatsink or a fan, and some VERY advanced mounting techniques for the MOSFETs to insure low thermal resistance. It's not impossible, just quite challenging. Assuming you do not cross the SOA of the transistors, of course you coudl use 4 ohm speakers - it's just that when the load current becomes more than the idle current, you get into class AB. That's how it always works on push-pull amps.
It is a whole different story if you are thinking deep class AB (high bias current) - for this, your limiting factor is the heatsink.
Also, consider that you need to design the amp for half the intended load anyway, to compensate for a reactive load. Because reactive loads change phase between current and voltage, the amp has to deliver a given current at much higher voltage drops across transistors than with the ideal resistive load. Fortunately, MOSFETs are much more robust here compared to BJTs.

- I`m gonna use IRFP240/IRFP9140.

Well, i can only applaude the choice :)
Now, if you are using a single pair on the output, then you can completely forget the source resistors (0.1 ohm). They are not going to be doing anything useful, quite the oposite. But, you do need to add some output stage protection - fuses in the drain lines (with smaller bypass caps on the drain side), also zeners across GS of the MOSFETs to limit current. But when you do this, consider carefully what happens when your amp clips (and what happens if it clips as it is now, without gate protection).
Some pointers: current in the driver transistors, voltage across CE of input stage (the latter especially if an output fails short to rail, when C1 charges to full rail voltage...), charge removal from the driver transistors (watch for 'sticky' clipping).

Regarding the input stage, I would increase the tail current as much as possible while keeping input impedance reasonably high and nise reasonably low. Why? Well, in a dual diff input you have the advatage that input bias currents (Ib of Q1/Q2 and Q3/Q4) largely cancel each other out. Even though beta of the PNP and NPN transistors tends to be different, in this sort of configuration you can still get 2-3 times more current for equal input bias current compared on single diff stage. The higher tail current alowes you more linearisation, while still having higher gain, and most importantly, less critical selection of the current in the next stage, as well as a higher potential bandwidth of the next stage (just be careful with clipping!). Also, consider carefully what R12 and R14 are doing, there is a gain/lineraity/speed/stability compromise involved in their choice.
 
- I made some changes, lowered source resistors
and decided to use MOSFET Vgs multiplier
and changed diodes with zener in, CCS, constant current sources.
- Is 2.7 Volts enough?

-What should be the values of C6 and C7?

You can add an input filter for very high freqency.
A capacitor like 22pF-220pF parallel with R6 (44k).
This cap will short noise to ground, and it does not enter amplifier.
.......................

I never use zener diodes in amplifiers.
There are better ways which gives lower noise.

I would use RED LEDs. You can keep R15/16 at 10kOhm.
This will give like 4 mA current in LEDs and very low noise.
The voltage across a RED LED at 4 mA can be like
1.55 Volt (1.85 Volt for some types).
See this topic: Some noise measurements for LEDs and zener diodes

.......................

Current in Q7/Q8 is 15 mA.
I think this can be good.
If you use big MOSFETs with high input capacitance ( Ciss )
you can increase this to 20-30 mA.
........................

You can also reduce open loop gain, by add two resistors.
Value can be 4.7k - 47k. Lets call it R97
One resistor from collector of Q7 to ground.
The gain of Q7 will be: R97/ R12 (82 Ohm)

Likewise another resistor from collector of Q8 to Ground (0 Volt).
........................

Values of C6 and C7 across base-collector of VAS (volt amp stage) transistors Q7/Q8 is 39 pF.
This is okay.
You have to use oscilloscope to set these small caps to perfect value.
A simulation might give a hint.

Can be anything from 10-220 pF.
........................



Your amplifier is looking very good now, bogdan_borko!

It is a nice amplifier
:smash:
 
When looking at that topic, take notice of the noise of a 12V zener - in comparison with forward biassed diodes including LED and BE junction. It is a pity no measurement was done on a reverse biassed BE junction of small BJTs, this typically behaves as a somewhet good 8-10V zener...
 
Hi Bogdan,

If you want to calculate the maximum gate current to drive the mosfet then use this equation..

Ig=Qg X F...

Ig is gate current..

Qg is Total gate charge of the mosfet specified in the datasheet...

F is the highest frequncy of operation..which is about 30KHZ in audio amplifier with margin...

Hope this helps...

K a n w a r
 
Workhorse said:
If you want to calculate the maximum gate current to drive the mosfet then use this equation..

Ig=Qg X F...

Ig is gate current..

Qg is Total gate charge of the mosfet specified in the datasheet...

F is the highest frequncy of operation..which is about 30KHZ in audio amplifier with margin...

Hope this helps...

K a n w a r

Workhorse

Can you give a practical example from a datasheet from some MOSFETs we often use.

For example IRFP240, IRFP9140 orIRFP450.
How can we find out suitable gate current, Ig ?

you write: 'calculate the maximum gate current'
Shouldnt it be: minimum gate current?
 
Definitely minimum, and it's also worth extending that maximum frequency as well, I actually tend to use 200kHz. Sounds very high, but remember, some sources are 192k DACs, and I would prefer the amp not to have too high IMD in the aliasing region (don't want any leaks from the filters to fold down into the audio band). I am also not sure I would use Qg in the calculation, although that is definitely erring on the side of caution. Qg is not derived from a follower configuration - although, of course, the design may not be using a follower so...
 
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