output drive of preamp (fetboy)

Hey,

sorry if this is a dumb question, but i am trying to fully undertand the concept of output drive in a preamp. i am looking on the FetBoy article (attached). Hampton says that if output resistance is too high (of one stage), the stage wont have enough drive and will be clipped.

now this confuses me,
first, lets say i have a 100k load.  the less current we provide, the smaller signal we will get on the load - but why would it clip? i dont get it. i would just expect a smaller amplitude.
second, we're in fact not interested in output current, right?  voltage is what matters, right?

Thanks!

He did not say of the output resistance of one stage is too high. The 47K resistor he refers to sets the current in the output stage which in turn determines its drive capability. More standing current equals greater drive capability. if you increase the 47K resistor the standing current is reduced and hence so is the drive capability. The 57K is not an output resistance.

Cheers

Ian

aviel said:

now this confuses me,
first, lets say i have a 100k load.  the less current we provide, the smaller signal we will get on the load - but why would it clip? i dont get it. i would just expect a smaller amplitude.
second, we're in fact not interested in output current, right?  voltage is what matters, right?

Click to expand...

Because if you think of a transistor as a sort of variable resistor (which is a greatly simplified view) then if it's resistance doesn't go that low, it will not be able to sink that much current and thus the voltage will not go lower than a certain point. In other words, it will be the lower half of a voltage divider where the upper half is define by the other transistor (which should be very much "off" when the lower one is "on") and the load impedance. More specifically the output impedance of the transistor is going to be vaguely the impedance on the base (47k) divided by the beta of the transistor which we'll say is 100 so that's 470 ohms.

hay...  I don't care for that example... start with more basic discrete device topologies. The emitter follower is AKA common collector topology (as compared to common emitter or common base)..

Ability to source current in one direction (into base for npn) is a function of the available current driving the base, the current gain of beta of the transistor, and in the limit PS current from + supply into collector. In the other direction current is limited to the pull down (or up for a pnp) emitter load. The drive current availability is completely different from output impedance.

This is a simplification but surely there are lots of tutorials on the WWW...

keep it easy and start with the basics, like one transistor circuits... then you can cobble a few one transistor circuits together. .

JR

> More specifically the output impedance of the transistor is going to be vaguely the impedance on the base (47k) divided by the beta of the transistor which we'll say is 100 so that's 470 ohms.

No.

If Rb and C are connected together you are on path to figuring the DC current. But here they aren't.
-------------------------
> we're in fact not interested in output current, right?  voltage is what matters, right?

We ALWAYS need some current in the load. (There are no truly infinite loads.)

If you are talking 100K with normal signal voltages, the current may be "small" (small to a flea?  small to a power company?). Maybe 0.1mA. For "good" (how good is good?) drive we would normally want the DC current to be much larger than the peak signal current. So you may be fine with 1mA. Another guy may have a lot of 10K loads. Or a 300 foot cable (about 1K at the top of the audio band). Or a 600 Ohm transformer, as noted in the figure.

Note that if he needs 39mA to drive 600 Ohms, "obviously" 100K load can use 39mA*(600/100K) or a quarter of a mA. HOWever that plan is reputed to be extensively ear-tweaked, so you are 'on your own' if you change it.

PRR said:

> More specifically the output impedance of the transistor is going to be vaguely the impedance on the base (47k) divided by the beta of the transistor which we'll say is 100 so that's 470 ohms.

No.

If Rb and C are connected together you are on path to figuring the DC current. But here they aren't.

Click to expand...

Not sure what you mean. I'm just saying that when the output swings high the impedance of the lower transistor is going to have limit that is vaguely 47k/100. I just simulated this [.asc attached as .txt to pass attachment filter] and although the JFET is pinching off, the output goes up to ~17V and the lower transistor + 100R is sinking ~40mA which is 17/0.04 = 420 ohms.

But I don't know why I'm getting into this because this circuit isn't that great in any respect.

Thanks for you answers!

PRR said:

If you are talking 100K with normal signal voltages, the current may be "small" (small to a flea?  small to a power company?). Maybe 0.1mA. For "good" (how good is good?) drive we would normally want the DC current to be much larger than the peak signal current. So you may be fine with 1mA. Another guy may have a lot of 10K loads. Or a 300 foot cable (about 1K at the top of the audio band). Or a 600 Ohm transformer, as noted in the figure.

Click to expand...

i am not sure again what you mean by drive? and why would we want the dc current to be higher? in theory, we would want just the AC signal (current).

squarewave said:

Because if you think of a transistor as a sort of variable resistor (which is a greatly simplified view) then if it's resistance doesn't go that low, it will not be able to sink that much current and thus the voltage will not go lower than a certain point. In other words, it will be the lower half of a voltage divider where the upper half is define by the other transistor (which should be very much "off" when the lower one is "on") and the load impedance. More specifically the output impedance of the transistor is going to be vaguely the impedance on the base (47k) divided by the beta of the transistor which we'll say is 100 so that's 470 ohms.

Click to expand...

i can understand that, but in that case i would expect the entire signal pattern to be shrinked. since the voltage is splitted acrros the voltage divider, in any point in time we would get (for example) 0.1*V(t) on lower, 0.9*V(t) on upper, so ptp signal will be shrinked, not clipped.

i did try to simulate this, and got even stranger results!
signal is not being affected if the current is low, BUT, if the current is too high, i get signal distortion (see attached picture).
i also did another test where i put a current source and didnt modify the base resistor, just to isolate the affect of curret.
results are are the same

aviel said:

Thanks for you answers!
i am not sure again what you mean by drive? and why would we want the dc current to be higher? in theory, we would want just the AC signal (current).

Click to expand...

The AC signal current available depends on the standing dc current. In a single ended output stage (like yours with a resistor in place of the bottom transistor) carrying a dc current I, the current can swing down from I to zero before clipping so the peak to peak current output goes from 0 to 2 times I. It does not matter what AC load you connect to the output, you cannot possible push more than this peak to peak current into it. This is what is meant by drive. The bigger I is the bigger the drive capability.

Cheers

Ian

Ians explantion seems easier to understand for me,

ruffrecords said:

The AC signal current available depends on the standing dc current. In a single ended output stage (like yours with a resistor in place of the bottom transistor) carrying a dc current I, the current can swing down from I to zero before clipping so the peak to peak current output goes from 0 to 2 times I. It does not matter what AC load you connect to the output, you cannot possible push more than this peak to peak current into it. This is what is meant by drive. The bigger I is the bigger the drive capability.

Cheers

Ian

Click to expand...

I had the impression that current can also change it direction..  I mean the ac current could shift to be negative..
Is this true?

aviel said:

I had the impression that current can also change it direction..  I mean the ac current could shift to be negative..
Is this true?

Click to expand...

If the output is connected to the load by a capacitor then the ac current in the load will flow positively and negatively. As the output voltage goes up, signal current flows from the top transistor via the capacitor into the load. Call this positive current flow direction.

As the output voltage goes down, the signal current flows in the opposite direction. But the current in the output stage itself just goes up and down.

Cheers

ian

aviel said:

i did try to simulate this, and got even stranger results!

Click to expand...

You're missing the 100R emitter resistor on the lower transistor.

ruffrecords said:

If the output is connected to the load by a capacitor then the ac current in the load will flow positively and negatively. As the output voltage goes up, signal current flows from the top transistor via the capacitor into the load. Call this positive current flow direction.

As the output voltage goes down, the signal current flows in the opposite direction. But the current in the output stage itself just goes up and down.

Cheers

ian

Click to expand...

Ok got this finally . Dc current always positive in this case, will flow from vcc to ground. The higher the dc is, dVac is also bigger

aviel said:

Ok got this finally . Dc current always positive in this case, will flow from vcc to ground. The higher the dc is, dVac is also bigger

Click to expand...

Nicely summarised.

Cheers

Ian

aviel said:

Hampton says that if output resistance is too high (of one stage), the stage wont have enough drive and will be clipped.

Click to expand...

And he's wrong! The output resistance of the stage is not governed by the 47k resistor and is largely irrelevant; it's the drive capability that is dependant on this 47k resistor, because it defines the current in the lower transistor. Current available to the load depends on what each transistor of the final pair can supply. Actually both transistors have their limits. For the lower transistor, it's simple and dependant essentially on the current injected in its base by the 47k resistor. For the upper transistor it's a little more complex, but not terribly (Ohm's law and 1st-degree equation).

> in theory, we would want just the AC signal (current).

We could use a passive microphone and SHOUT REAL LOUD, get only AC signal.

There is another trick: tubes/transistors. BUT these only conduct ONE WAY. We need BOTH ways. How can that be done? The usual simple technique is to "DC bias" the amplifier "half-way", swing both ways from there, then remove the steady bias. If you wave your arms in a small closet, you'll see that you can only get peak symmetrical swings *half* the size of the closet. The DC bias sets a limit on the AC signal possible.

> i did try to simulate this, and got even stranger results!

When I become king, you will need to work it out on a matchbook before you are allowed to use a computer simulation.

You have >0.9mA into bottom Base. By similarity, the top Base must be sucking 0.9mA. Where can this come from? Only from the 15.5K resistor. Which will drop about 15V. Which is "all" the drop we can afford, aiming for half-way on a 24V supply. This leaves NO current to the JFET. It's sitting dead.

There's more. Work on a matchbook and "dumb mistakes" tend to rise to the top. SPICE won't tell you that you put a 14.0K resistor under the JFET. So even without Q2 Base stealing all the current in the 15.5K, the 10X over-size JFET Source resistor holds the JFET essentially "off", not doing any good.

Hello dear comunity

i just built a pair of those preamps, gain is insane and clean, pretty beautifull distortion but the transistor and the 100r are getting super hot, burning hot

any advice to troubleshoot why? i checked all values of components as well as short circuits? tbanks a lot

pstcho said:

Hello dear comunity

i just built a pair of those preamps, gain is insane and clean, pretty beautifull distortion but the transistor and the 100r are getting super hot, burning hot

any advice to troubleshoot why? i checked all values of components as well as short circuits? tbanks a lot

Click to expand...

What is the voltage across it? You should have about 4V for the targetted 40 mA quiescent current. 4V at 40mA is 160mW, it's enough to make a 1/4W seriously hot.
What is the DC voltage at the output? Should be about 12V (a tad more should be better).