Class-A Mosfet Source Follower

Hi,

I found a class-A MOSFET current boosting circuit on a headphone-oriented web site, thought it may be used successfully for some line amp circuits & have redrawn it ...

http://www.twin-x.com/groupdiy/displayimage.php?pid=305&fullsize=1

The circuit has two stages as shown: a voltage gain opamp stage followed by a separate MOSFET current booster. The idle current for the booster is set by the 150 ohm source resistor - in this case (-15v) & it provides 100 mA of current to the MOSFET source. Maybe a current source (BJT or LM337 & 12 ohm resistor) would work better. What do you think? Additionally, the 5k trimpot allows you to null out DC on the circuit's output.

I have a few questions about this circuit's use. As drawn the current booster is shown AFTER the opamp voltage gain stage. Would it be possible or advisable to include the current booster MOSFET within the opamp's feedback loop? I've seen that done in other types of source followers & assume it would be possible here. If so, what are the advantages / disadvantages of placing the current booster within the opamp's feedback loop? Any & all comments, etc. are welcome.

Skip Beach

What is the purpose of that global feedback loop you want?

Hi Wavebourn,

The answer to your question is what I'm asking, in a round about way ...

Would it be possible or advisable to include the current booster MOSFET within the opamp's feedback loop? ... If so, what are the advantages / disadvantages of placing the current booster within the opamp's feedback loop?

Click to expand...

Any thoughts, recommendations or knowledge to impart ? Thanks,

Skip Beach

Yes, you could include the output in the global feedback. You may have to modify the compensation slightly.

Also, as shown in the original schematic, I would delete the 118K, the 5K pot, the 121K and the zener diode. The resistors aren't going to do much in this circuit since this is a conventional voltage feedback opamp. They will just be fighting the low output impedance of the op amp. This opamp is also unity gain compensated so the 20pf feedback capacitor is probably not necessary either.

If you want to maintain the Class A operation and not add a second symetrical MOSFET to the negative supply, you might consider changing the 150Ohm resistor to a current source. This will provide better drive to the inductive load.

[quote author="sbeach"]Hi Wavebourn,


Any thoughts, recommendations or knowledge to impart ? Thanks,

Skip Beach[/quote]

Here you go: http://www.groupdiy.com/index.php?topic=19419

Hi burdij,

You wrote ...

Also, as shown in the original schematic, I would delete the 118K, the 5K pot, the 121K and the zener diode. The resistors aren't going to do much in this circuit since this is a conventional voltage feedback opamp. They will just be fighting the low output impedance of the op amp.

Click to expand...

Don't you need to bias the MOSFET gate by applying the appropriate voltage for the device? That's the reason for the trimpot & associated resistors, I think. If I'm wrong about that, please help me understand this. Also the zener is included there as insurance to protect the MOSFET's gate from voltage transients.
Skip Beach

Yes, I agree on the zener, but its actual effect is to clamp the source of the MOSFET a fixed voltage above or below the output of the opamp. It can't rise higher than .7V above the output or lower than the zener voltage below the opamp output. But on the bias network, you have to imagine that the output of the opamp is a resistor to ground of a small value, let's say 10 Ohms in series with a voltage source of zero Ohms impedance. The effect on the gate of the MOSFET is controlled principally by this resistor, not by the current flowing from the positive or negative supplies through high value resistors. At least, that is my understanding.

The resistor network and trimpot as shown anyway is superfluous---the closed loop output Z of the opamp is completely dominant. Perhaps there is a missing coupling capacitor between opamp output and the rest of the circuitry to the right.

(I posted this before seeing its predecessor of one minute).

a few things

First a mosfet gate often needs 3 to 4 volts gate to source to turn on for current to flow.

The zener is to limit the gate to source to under 15V at 20V, some mosfets can have a gate oxide breakdown.


The resistor string is to adjust the 3VDC or more voltage gate to source to turn the mosfet on enought and have better drive both ways to + and -(that is what it looks like to me).

The mosfet will most likey be able to turn on the the + rail better(RDSon) than the source resistor can pull it to the - rail(need to look at the mosfet curves):
however you are running out of voltage drive (going +) you might want the drive voltage to be higher than the + rail (hint look at some old school bootstrap quasi constant current source power amp drivers Pass labs HK citation 12mosfet mod IIRC)

Look at the mosfet curves

This circuit might not clip nice. It might be fine small signal. close to clipping(power supply rail) who knows?

What about the tori?(.6VDC) nothing wrong with BJTs is this just to use a mosfet?




EDIT I should look TWICE at the circuit before a post.

IT is not a good design(as drawn) IMO the opamp low output Z will overpower the attempt to bias the opamp via the resistor string connected to the base. So the string does next to nothing if I worked this out right. It would need a cap from the opamp to gate(as posted by bcarso)

One could offset the input at pin 3 of the opamp to adjust the gate voltage of the mosfet

So one might ask what if you use a resistor from the opamp(no coupling cap) to the gate to "decouple" the opamps low Z output from the bias string? There are a few problems one power mosfets have lots of gate cap (work out the math, sometimes a few 1000Pf).
Then you need to work out the interaction with the bias string.

I think this falls into the class of "self biasing" devices. The source will naturally float to a voltage that allows sufficient field strength within the gate to turn the device on. When you close the feedback loop, then the opamp output would naturally rise (or fall depending on whether the device is P or N channel) in voltage to turn on the MOSFET enough for the voltage at the source to balance through negative feedback, the input to the opamp.

[quote author="burdij"]I think this falls into the class of "self biasing" devices. The source will naturally float to a voltage that allows sufficient field strength within the gate to turn the device on. When you close the feedback loop, then the opamp output would naturally rise (or fall depending on whether the device is P or N channel) in voltage to turn on the MOSFET enough for the voltage at the source to balance through negative feedback, the input to the opamp.[/quote]

A first, an additional pole will cause the amp to oscillate. Second, as were mentioned already, such bias will not work. Third, as soon as opamp's input is biased on zero volts you will see the same voltage on output, it means very low swing of output voltage to the negative rail in respect with positive one when working on the real load before horrible distortions of high order due to deep feedback. Far the more, voltage drop on 100 Ohm resistors will be huge. Also, if opamp have own output ptotection it will saturate when Zener opens, causing zero feedback and high gain, it means horrible distortions... Especially, if capacitance on output presents...

The whole design resembles a dish made from good salad, good gamburger, blueberry jam, cookies, sugar and pepper, all taken from good cookbook, but mixed together. Anyway, a brave attempt to develop something own and learn on own mistakes is a good idea!

It has crossed my mind a few times to use a class A Mosfet circuit to some how drive a gapped output transformer (neve style), just to see how it sounds as opposed to the bipolar method. But heck, I can’t even seem to find time to build things I know are going to work.

Hello again,

Well I guess it is just a crappy design. As I mentioned, the schematic was on a diy headphone amp site & I thought it looked simple enough for even me to build successfully. Even as a failed choice, the Q&A about this circuit has helped me learn something. Thanks everyone,

Skip

You have at least couple of optinos: to build my amp, or try this one to make to sound good. The first one is the easiest way, the second one promices more skills and understanding.

Ok, let's start from the source follower.
What output voltage do you need on which load resistance?
It is the basic.
Current to the positive rail is limited by the source follower, it may be huge potentially.
Current to the negative rail is limited by a resistor that loads the follower. Assume it and your headphones in series. Use Ohm's law to calculate it's value. Now, for the highest symmetrical output level on the given load you have to figure out what DC must be on the output of the source follower, right? It have to be positive. Now, let's see what voltage must be on the FET's gate, assuming that for the FET to start conducting it needs 2-5V depending of the type.
The result will be some relatively high bias with relatively low maximal output voltage swing. Also, transfer function's value of your source follower will depend on output voltage, because its current bias will depend on it.
Try and calculate what resistor in source do you need for your load, and what output bias voltage do you need.
It is for a starter.

Um, not sure what you mean by:

The whole design resembles a dish made from good salad, good gamburger, blueberry jam, cookies, sugar and pepper, all taken from good cookbook, but mixed together.

Click to expand...

This is a very simple circuit. It is not a "new" design. You have two components, a voltage amplifier (the op-amp) and a source follower with a resistive load. Could have been an emitter follower or a cathode follower, choose your flavor. Placing the source follower within the feedback loop corrects for offsets that may occur. It also reduces harmonic distortion due to the negative feedback. This is the essential idea that "improves" on the original circuit. Its is hardly a weakness. This op-amp is a very good device. Read the datasheet. The changes stated by various posters, such as replacing the load resistor with a current source which will reduce the disipation and increase the drive capability are just good engineering practice.

Granted, there are probably much better circuits, but for a simple, getting started in DIY type circuit, this is a good choice, IMHO.

Source follower already have 100% feedback. What to reduce by additional feedbsck? Distortions generated by improper implementation?

[quote author="Wavebourn"]Source follower already have 100% feedback. What to reduce by additional feedbsck? Distortions generated by improper implementation?[/quote]

IMO that's a bit of a shoot-from-the-hip (or shoot-from-the-chip) response. I'm not quite sure what you imply.

Global feedback around a cathode/source/emitter follower may have disadvantages in other ways, but it does reduce the effects of the steady-state residual nonlinearity of the follower. And it will remove most of the offset voltage associated with the given part. Indeed, it's part and parcel of virtually all solid-state audio power amplifiers, although most of them have a complementary output configuration running far below class A bias.

A source follower has local feedback, yes; its effectiveness depends on the ratio of the load impedance (resistor(s), current sources, etc.) to the reciprocal of the device transconductance. The device drain impedance enters in, although less strongly (but should be accounted for when using triodes in cathode followers).

But all devices have finite transconductance and it varies with current and temperature and "drain-gate" voltage and frequency. Hence no simple single-stage "source" follower is distortionless. With more complicated designs the distortion can be reduced without using global feedback per se, but that's another vast subject.

[quote author="bcarso"][quote author="Wavebourn"]Source follower already have 100% feedback. What to reduce by additional feedbsck? Distortions generated by improper implementation?[/quote]

IMO that's a bit of a shoot-from-the-hip (or shoot-from-the-chip) response. I'm not quite sure what you imply.

[/quote]

I imply exactly what the voltage follower means: an amplification stage with 100% feedback by voltage applied in series with input.

However, it's dynamic output resistance is not linear. Open loop gain is not infinite. Transconductancy depends on source current and voltage between source and drain. That's why absolutely linear resistor in source will cause distortions, that's why absolutely linear capacitive load will cause more distortions, especially when it was not taken in calculations when bias voltage and resistor in source were calculated.

Yes, traditional operational amplifiers (and power amplifiers designed this way) use symmetrical emitter followers. Yes, as the result they generate crossover distortions and distortions caused by beta droop with currents. Yes, global NFB is usually applied to make this distortions better measurable.

Drawbacks of such approach are: NFB increases order of transfer function generating less "natural" distortions, it's deepth falls down with a frequency, it means higher increase of THD on higher end of audible spectrum, that again is "unnatural".

When lot of people repeat the same mistakes it does not mean mistakes do not exist. It just means that people repeat mistakes. Period.

I still suggest to calculate bias voltage and load resistor value for given swing on given load, to better understand what and why you are doing, instead of memorizing and repeating what others do and write. Electronics is not the Bible, it is an art of choosing compromises for given results, using well defined laws of physics. And, sometimes, experience with human perceptions, if you speak about audio.

In this topic I show how to use source and emitter followers without need to load them by a resistor or/and complementary follower to heroically overcome resulting distortions using GNFB loop:

http://www.groupdiy.com/index.php?topic=19419

Well Anatoliy, I could respond in a number of ways. But I'll extend the Prodigy warm-welcome benefit-of-the-doubt for just a bit longer, lest I be admonished for over-reaction.

If I may inquire, what are you running as idle current in your headphone amplifier?

[quote author="bcarso"]

If I may inquire, what are you running as idle current in your headphone amplifier?[/quote]

Half of maximal expected peak current on the load, including cable capacitance impact on 20 KHz. I.e. it dissipates much less than the follower with resistive load, and half of what dissipates CCS - loaded follower, and introduces much-much less distortions than each of them, also such distortions if heard sound more "natural". Similar stereo amplifier with maximal sinusoidal output 2x300W dissipates about 1300W.