Amp basics: loading

[xtian]

I have successfully remained ignorant of amplification stage loading. Not sure where to look. I have Merlin's preamp book.

For example, Merlin talks about the design of FX loops, and how the component choices can load the previous (or is it next?) stage. I also hear about this subject regarding tone stacks. I barely know enough to ask the right questions!

[ampgeek]

My generic understanding is that loading is, simply put, impedance to ground.

The lower the impedance, the more the signal is "loaded down" or attenuated.

Dave O.

[matt h]

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[pdf64]

Are you familiar with the concepts behind Thevenin and Norton equivalent circuits?
http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
http://en.wikipedia.org/wiki/Norton%27s_theorem

eg a voltage / current source combined with its equivalent series / parallel source impedance.
Without a grasp on those source models, it may be tricky to make sense of the loading on them.
Pete

[vibratoking]

I wont' attempt to explain because everything I say might be wrong or different in some other context. For most guitar amps, matt is on the right track. Zout and Zin form a divider because most signals are voltage signals. There is nothing magic about a 10x ratio. That's just some rule of thumb developed for those that don't want to understand. If we were talking power transfer, then we would want the Zout = Zin. Things change if the signal is current or power.

Grid current causes the input impedance to decrease and load the driver. This creates non-linear behavior which is more difficult mathematically.

If you really want to understand all the details, then I would suggest Richard Kuehnel's books. He goes through this like an engineer would. He does the full mathematical treatment and then he does it with simplifying approximations. He then compares the difference to prove the effectiveness of the simplifications. It is a good read and a good topic. You might even learn some linear algebra if you are persistent. You really can't effectively design without understanding this.

[matt h]

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[vibratoking]

I would, however, also offer the point that plenty of EE's who do understand this have designed *terrible* amps... and some of the best sounding guitar amps were clearly built by someone who'd never heard of a Thevnorton equivadonkey. So don't feel like unwillingness to learn linear algebra is actually placing you out of designing great guitar amps.

This is true to some extent. Anyone can assemble blocks and with some intuition or luck end up with a great sounding amp. And do it without really understanding how it works. But that isn't designing in my book. BTW, this is true of engineers as well. I can't tell you how many products I have seen that worked, but not as designed. There is a lot that can go wrong. I thought xtian was saying that he truly wanted to understand how loading worked. For that, Kuehnel is a good resource that almost completely illustrates the process...or I could direct you to a plumbing manual and we can start from there.

I have heard the terrible amp argument many times. I an not aware of any EEs worth their salt that have designed terrible guitar amps. It's pretty hard to design a bad one if you know what you are doing.

[matt h]

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[Phil_S]

Well, this is a great topic, but it seems not so easy to make comprehensible to the ordinary amateur. I like the voltage divider explanation. I can wrap my mind around that. However, I still don't fully comprehend the difference between resistance and impedance and I surely am fairly clueless about figuring Zin and Zout! I am hoping this topic will unfold a little more and not in an overly technical way.

[ampgeek]

Resistance is DC voltage drop at a given current flow. Impedance is the AC voltage drop at a given current and will vary with frequency.

Laymans terms for molecular biologists!

Dave O.

[Phil_S]

Egad, I know how lame this sounds...I'm not afraid to bare my soul on this. So, a resistor is an object that is a fixed value.

We calculate V=I*R where V and R are known, and solve for I. In this case, V is DCV. So that stated value of the resistor is, in fact, "resistance?"

Over the same exact resistor, if VAC is also flowing, the drop across that resistor allows us to find the impedance if we also know Frequency and Current? Did we just find current, above, V=I*R? (Look in RDH4 for a formula? Believe it or not I have a copy, it's just not very "accessible" to me.)

Aside to others...sorry if I am dragging this thread down. I see this as a rare opportunity to move up several rungs in one fell swoop.

[xtian]

No prob, Phil! I've wondered all these things too.

I'll take a shot: One does not consider a resistor's impedance, because its resistance to AC does not change with frequency. That's why a big old resistor is a FIXED load when you use it in place of a speaker.

A speaker (coil of wire and magnet) is a REACTIVE load, and we measure its impedance (i.e., 8ohms).

[V2]

(sorry I was too tired to read all the preceding replies).

Consider a typical hot-rodded JCM800 gain stage, where gain is increased by using a larger plate resistor (220k-330k instead of 100k). Then consider the interstage attenuator (voltage divider or pot) that comes next. The resistance in that attenuator will be in parallel with the preceding plate resistor, and resistors that are in parallel combine to make a resistance that is LOWER than the lowest resistance. So if you have a 1M pot after a 220k plate resistor, you have 180k resistance. That's a bit lower than your plate resistor, and that lowers the gain of your previous gain stage (look at Merlin's AC load lines). Now consider having a 100k pot after the gain stage; in that case, the parallel resistance is ~69k, which is FAR less. So your gain will be reduced when by the smaller resistance that follows; it 'loads down' the prior gain stage.

The same thing happens for tone stacks. The slope resistor is in parallel with the prior gain stage. If the output of the prior gain stage is taken from the plate (usually 100k), then a 33k slope resistor that is in parallel 'loads down' that gain stage. If the output is taken from the cathode instead, the impedance is much higher, so a 33k slope resistor doesn't load it down as much.

[jazbo8]

Here is a good visualization to help with understanding impedance. Simplily put, the impedance of an inductor is 2xPixFxL, and the impedance of a capacitor is 1/(2xPixFxC), where F is the frequency of interest, L is the inductance in Henry and C is the capacitance in Farads. Once you know the impedance of a L or C, you can then treat it like a resistor for your calaculations BUT only at one frequency.

Also check out the other apps on Falstad, there are many good lessons for basic (and some quite advanced) electronics.

[passfan]

Not to simplify the subject but I'll attempt to anyway ; google impedance matching and you'll find a wealth of info on the subject. To me matching stage impedance for proper loading is the same as matching components in a stereo system chain back in the good ole days. It's also why we run active effects loops which serve as simple current buffers to keep from loading down preceding stages. Do I completely understand it from an EE viewpoint ,no? But I get the basic jist of the , why do we do it. Had to study it's effects from a component standpoint when I was in audio engineering school.