Help with amplifier design (high quality)

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Hello !

I´m not new to electronics, but new to amplifier design.
This would be my first DIY Amplifier.

I studied lots of schematics (many of them found in this forum...excellent forum!) and tried to make my first design.

The aim is to develop an amplifier with high accuracy, output power is less important (about 50W shall be enough).

Before I start to calculate exact values and operating points I have some questions to you.



What you see is:

Input Stage: differential amplifier with reference current souce, output formed by current mirror.

"Voltage gain" stage: bootstap configuration for constant input impedance to the input stage, a bit feedback provided by R18

"current gain" stage: cascoded emitter follower to get a linear output characteristic curve.

output stage: A pair of complementary mosfets



Questions:

Is effective bootstapping possible over TWO stages (current gain and output stage) ?!?

Do the cascoded transistors (current gain stage) work with only a zener and a resistor at the basis? Zeners should be powerfull!

I want to drive the output mosfets without emitter resistors to get a high damping factor. For this, is it better to match them? Do you see any offset problems ?!?

I have NO Ideas for the semiconductors, everyone uses other ones and the market is very large and complex. IR HEXFETS for the output stage shall be a good choice ?!?

Thanx for any comment !!!

greets
Peter

An externally hosted image should be here but it was not working when we last tested it.
 
CrazyChipMan said:


Is effective bootstapping possible over TWO stages (current gain and output stage) ?!?


Hi Peter

Yes, it is, as you have put in the schematic...over voltage gain stage and output stage.



Do the cascoded transistors (current gain stage) work with only a zener and a resistor at the basis? Zeners should be powerfull!

This cascoded transistor are in my view a very bad idea , because as the output mosfets don't have voltage gain , you loose in the cascoded stage , precious volts for full voltage swing.




I want to drive the output mosfets without emitter resistors to get a high damping factor. For this, is it better to match them?

Only if you use lateral mosfets, you can confidently dispense with source resistors , without fear of thermal runway .
In the case of vertical mosfets is not safe.:hot:

Use emitter degeneration resistors in the current mirror.

One last important point: why the inductor L1 at the input shorting the low frequencies to ground and inducing a large offset ?:eek:

Regards
 
Peter,
I can't comment a lot as I'm not an EE but shouldn't the feedback R be R6?
Also, many R's and transistors have no value which could make commenting more difficult.
Last, in many schematics I've seen from Mosfet specialists like Nelson the value of the gate R's is a bit higher (roughly 100-400 ohm).
Looks nice otherwise.

/Hugo :)
 
Hello!

Was a mistake with the L before the input stage, It should be a Resistor.
No values! because I have to choose the semiconductors first, waiting for your proposals....

@tube_dude
what are lateral mosfets ?

the voltage I loose at output voltage will be the voltage over the zeners, i suppose ?!?

greets
Peter
 
Hi Peter

The way you have drawn the cascoded driver stage would only allow a voltage swing that is limited to the value of D1 and D2 minus 0.68v.

Get rid of the cascode and try a CFP driver stage.
or just connect the collectors of each driver to the supply rails.
Move C7 to Q16 across the Base, Emitter, better still use a 2 pole compensation network.

Insert degeneration resistors in the first stage current mirror start with 100 ohm and emitter resistor in Q1 and Q2 values starting at 10 ohms to 100 ohms each depending on the current set in differential stage.

R1 I would start with 330 Ohms and R2 22k Ohms this will set the current at about 2ma

Get rid of R18 and link it to the supply rail.

move your supply decoupling to in front of R17 and the bootstrapping does not have very good PSRR.
In fact I would use a voltage regulator there

source resistors in the o/p fets would be good. start with 0.22 to 0.47 ohms. 0.22 would work well.

You are also going to need a zoble network on the output to ground. start with 10 Ohm 5 Watt in series with 100nf 250v AC x2 Cap.

This should get you started and I think you will end up with a nice little amplifier.
 
Anthony's comments are all on the money.......

However, another option to preserve your neat constant voltage driver.

Split the two Rs supplying the bases of the 'cascoding' transistors. Make them equal, and run two caps of 220uF from the midpoint back to the output rail.

This double bootstrap will enable you to swing the cascode bases well beyond the rail, giving very high rail efficiency despite using mosfets.

Cheers,

Hugh
 
CrazyChipMan said:
what are lateral mosfets ?
The current flows horisontally in the chip therefore a bit higher Rdson but on the other hand rather low capacitance's. Hitachi/Renesas make those. Lateral mosfets are very unsual these days. Now (sinces one decade back) you'll have vertical mosfets which means that the current flow vertically in the chip => much lower Rdson, much higher capactiance's also.

Rdson is an unimportant parameter as long as it's under 0.5 ohms or so.
 
Hi Peranders,

....."Rdson is an unimportant parameter as long as it's under 0.5 ohms or so."

In a common source output amplifier, like my SKA, it's important as it determines the voltage gain in the output stage, very significant for preceding stage operational signal levels. Such amplifiers are coming to the fore as they make better, more efficient use of supply wattage, and circumvent the high Vgs losses of modern vertical mosfets.

Cheers,
Greg:D
 
The Saint said:
Hi Peter
Insert degeneration resistors in the first stage current mirror start with 100 ohm and emitter resistor in Q1 and Q2 values starting at 10 ohms to 100 ohms each depending on the current set in differential stage.

You can use 0.050-0.100 Volt as a target value,
for voltage aross these Q1, Q2 emitter resistors.
1 mA in each input transistor would make it 47-100 Ohm.

Same way you can put emitter resistors in Q5, Q6,
to improve precision in this current mirror.
Making voltage across these also 0.100 Volt or so.


I would also put supply rail filter ( R3 C1 ) at the other side of VAS. ( Between R10 R18 )
I think, but I'm not sure, same goes for R4 C4.

=================

What we all want to know, is of course, what is voltage supply of this amplifier.
What is your power output target value?
Will output run in Class AB, Class A?
Also, what output power transistors you will use?

This will effect how the rest of circuit should be.

HEXFET could be IRF540/9540 or IRF640/9640
 
amplifierguru said:
Hi Peranders,

....."Rdson is an unimportant parameter as long as it's under 0.5 ohms or so."

In a common source output amplifier, like my SKA, it's important as it determines the voltage gain in the output stage,..
Do we talk about the same parameter? Rdson is only valid when the transistor is saturated = not very important in a class A or AB amp. You may loose some power but in dB's its nothing.
 
Hello,

thanks for your very helpful comments! :)

I´ve made some changes of topology, I like Hugh´s Idea. Now it´s a really bootstapping amplifier... :xeye:

But I see the problems in the cascoded driver stage "The Saint" mentioned. Zeners must have half the rail voltage and limit output swing to half rail voltage, too ?!? :confused:

Sorry for my questions, I´m just a beginner... :rolleyes:

greets
Peter

An externally hosted image should be here but it was not working when we last tested it.
 
You started with quite a nice circuit, for a beginner.

You moved supply rail filters, C1 R3 C4 R4, one step too far.
Should be to the right of Voltage Amp Stage, VAS.
Transistor Q16 is doing the voltage amplifiation.

R10 and R19 belongs to output stage drivers.
 
for the mosfet amp you have (hopefully) Rout about 1/gm open loop, reduced for closed loop.

the major downside to all of this is the output swing

not sure what Q10 or Q11 do.

Mos devices may not have a low Vgs-sat, especially at high currents. if Vth is 2V, Vgs for 4A might be 4V, depending on the device construction. this might limit output swing. at some point you wonder if it wouldn't make more sense just to move to BJTs in class A if you want low efficiency. of course class A idles a lot hotter.
 
This cascoded transistor are in my view a very bad idea , because as the output mosfets don't have voltage gain , you loose in the cascoded stage , precious volts for full voltage swing.

Well ..does'nt the entire prestage need a higher voltage anyway, so what you are saying is.. That the prestages will end up too high in voltage to get the correct voltage swing?
And C9,R11,R17 how is that working compared to a real current source?
And remember, a classAB amp will play in classA on low powers :-D, otherwise it would be called classB, hehe..
And dont worry about the damping, it will be enough..
You might want to try a 2uF capacitor on the output and see how it will handle that(or not)..
 
AKSA said:
Anthony's comments are all on the money.......
However, another option to preserve your neat constant voltage driver.

The neat thing about the constant voltage driver is that it largely removes the driver transistors from thermal stability considerations, making dissipation on them more constant on account of constant voltage. They can be kept on the board and off heatsinks, which may prove important as follows:

In order to get the idle current stable without source resistors, you can use either a single pair of lateral MOSFETs on the output (unfortunately you get higher rail), call stis option (L) or a single pair of vertical mosfets with a Vgs multiplier to set the bias current, that would be option (V).

In both cases the bias setting string (currrently your BJT and resistors) needs to have two diodes in series with it, in thermal contact with the driver stage (just the current gain transistors in the cascode). This is where having them on the board and off the heatsink helps a lot.

For option (L) you only need a variable resistor instead of the Vbe multiplier, assuming you set Ibias to >=100mA. For option (V) you need a Vgs multiplier. Basically, replace your BJT in the bias generator with a small MOSFET, like IRF510 for instance. Resistors around it will need recalculating. This MOSFET needs to be located on the same heatsink as the output pairs. I have used this approach with verticals using no source resistors to keep gm high, and it works very well. Thermal coupling is excellent without the need for exotic mounting arrangements due to the IRF510 TO220 case with it's big metal tab.

Oh, and another thing: even though IRF9140 are getting difficult to find, you might want to use them as your Pch MOSFET and IRF240 as the Nch. This is about as good as you can get regarding vertical MOSFET complementarity, which is slightly more important in a design with only one pair of MOSFETs and without source resistors, than just using the 'traditional' complementaries, which really are not. They also have a signifficantly higher power dissipation rating than laterals, and cost a lot less, while providing about 2x the max current capacity compared to single die laterals, and limiting rail loss - again important if you use only one pair.
 
AKSA said:
Split the two Rs supplying the bases of the 'cascoding' transistors. Make them equal, and run two caps of 220uF from the midpoint back to the output rail.

This double bootstrap will enable you to swing the cascode bases well beyond the rail, giving very high rail efficiency despite using mosfets.


This is only true in the negative voltage swing , but in the positive voltage swing , it will be limited by the voltage of the positive rail ( the voltage of the VAS transistor and his CE saturation voltage more the Gate Source voltage of the mosfets ,as the output stage has no voltage gain).
 
as for the source resistors: actually you get a higher damping factor with them, cause of a slight current feedback loop that forms.. they also take care of thermal stability issues and identical load distribution to your mosfets..
why use irfp140 and not irfp240+irfp9240?..
 
CrazyChipMan said:
@tube_dude
what are lateral mosfets ?

Hi Crazychipman

Lateral Mosfets have a lateral construction of the Channel, but what made them diferent from the vertical are their electrical characteristics .

The following advantages are offered by the lateral construction:

- It has a very low zero-temperature coefficient drain current ( usually near 100 mA and the vertical 3 A )

-Lower threshold voltage than vertical.

-The Gate-to-Drain capacitance is low ant that offers the advantage for high frequency operation.


Now the disadvantages:

-Less tranconductance than the vertical.

-Price is higher and availabality is less than the vertical.

-Higher Rds(on) than the vertical.

Cheers
 
Yes, that was just what I was about to ask ;)
Seeing that the output impedance is 1/gm and the source resistors are in series with that...

Why use IRFP9140 instead of 9240? Very simple - check out the data sheet. Compared to 240, 9240 has half the current handling, about 60% transconductance. Hardly what I would call complementary, so I would ask why use the 9240 as complement to 240?
Then check the same for 9140 and you will find that they are a MUCH better mach, 9140 has half the breakdown voltage (not important if + to - rails is below it), but practically the same current handling, transconductance and gate capacitance, (keep in mind it gm is normally given for different operating points on different parts so you need to compare Id vs Vgs graphs, which unfortunately are on different scales).
 
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