Logic problem with wiring?

[Venusblue]

Hi, I'm real new at building pedals, but i'm always looking to learn more. I read through here all the time but rarely post, But this caught my attention. It seems that after Q1, the signal really just gets sent to the ground, Right?

The way i'm seeing it, Is the signal is coming through from the input, hitting C1, then Q1, then being bled into the ground. It would also seem that everything hitting R2 and R4 is just being attenuated to the ground and not serving any real purpose, but I know that this isnt the case, so how is it that they're still making a difference in the circuit if the output is just being bled to ground?

[G. Hoffman]

Hi, I'm real new at building pedals, but i'm always looking to learn more. I read through here all the time but rarely post, But this caught my attention. It seems that after Q1, the signal really just gets sent to the ground, Right?

The way i'm seeing it, Is the signal is coming through from the input, hitting C1, then Q1, then being bled into the ground. It would also seem that everything hitting R2 and R4 is just being attenuated to the ground and not serving any real purpose, but I know that this isnt the case, so how is it that they're still making a difference in the circuit if the output is just being bled to ground?

The output comes out through C2, through R5 to the output.  Your V_s (the 9VDC) gets shunted to ground, relative to the input signal on Q1's base.  R1 and R2 are biasing the input of Q1, holding it at the middle of it's operating range when there is no input signal, which allows about half the current through R3 to go to ground.  C2 blocks this DC voltage from the output.  When you put an AC signal (like a guitar signal) at the input, you change the voltage at the base, so when the guitar signal is high, more current flows through Q1, and when the signal is low, less current flows through Q1.  The voltage at the collector of Q1 wiggles inversely to the input signal, which looks like AC to the cap (because it is), so it passes right through.  This gives you a polarity inverted output signal.  And the normal human ear can't tell if a signal has been inverted, unless there is something to compare it too.  (I've got some record engineer friends who actually CAN tell that something is weird, but most people can't.)

That's the rough theory.  There is a bit more going on there, though not much.  I'm sure there are people here who can explain it better, but I hope that helps.


Gabriel

[GibsonGM]

Hi Guys,

The LPB-1 is a great example of a simple transistor boost circuit, and of volume controls
in general.  The signal inputs to Q1 thru the input capacitor.  That cap filters the signal (trims the bass, if you like) and assures that it is only AC that gets in.   The AC signal is boosted by the transistor, which uses DC to 'raise' the AC signal by a certain amount (the gain of the circuit).  So you now have an AC signal riding on a DC level...

The output capacitor C2 blocks that DC, and "resets" the AC signal, which is also larger now, back to ride around a zero level point.

Now picture R5 as TWO resistors, which make a voltage divider - which that pot actually is.  As you move the point of contact higher 'up', the ouput (wiper in our case, since it's really a pot) has less resistance in its way and therefore you get more output.  At the 'highest' setting of the pot, the signal is going directly to the output with just a resistance to ground in parallel (which affects the output impedance, but that's another topic...). 
If you move the wiper down, you are placing more resistance 'in the way' of the signal. 

These are simple resistive voltage dividers, that's all.   Same thing with R2 and R4...they are setting the bias voltages the transistor needs to operate at say "1/2 the battery supply" so that you can amplify the + AND - half of the AC signal coming in.    Think of R2 and R4 as 'holding the base and emitter of Q1 at a certain voltage ABOVE ground" at the places where they connect to the device, and if they weren't there, those points would be grounded (shorted).   If they were omitted, the transistor wouldn't operate well at all (it would be amplifying in a different class, and part of the signal would be 'ignored').   This is all in the article you sited.   

Google "voltage divider",  "biasing transistors" and check out "Geofex" link at top of page, and use the search function to read up on transistor biasing, it is interesting and important to designing your own stuff!  Voltage division & Ohm's Law is hugely important, too   )

[PRR]

> everything hitting R2 and R4 is just being attenuated to the ground

For audio purposes, the "9VDC" line is ALSO "ground". It does not budge when signal pushes it.

So by your theory, R1 R2 R3 R4 and R5 ALL leak signal to ground.

I didn't read Gabriel or Mike's essays, sorry if this says a same thing. In short, whatever signal limps through to Base (maybe <10% of guitar's available power) influences the large steady current flowing from battery through R3 Q1 R4. The signal voltage on R4 is the same (nearly) as the input. But the signal current is much higher-- that's part of the magic of a transistor. Any signal current flowing in R4 must come from R3 (well, 1% from the guitar, negligible). R3 has R4's signal current but is 25 times higher resistance value: it drops 25 times the signal voltage found at R4. And R4's signal voltage is roughly the base voltage and therefore the input voltage. Voltage gain is almost 25.

> everything ..... is ...being attenuated to the ground

True, except: Q1 has a potential power-gain of 50,000 between lo-Z source and hi-Z load, power gain 100 in identical cascade. The exact amount of gain is not known: transistors vary a lot. So we stick some resistors in to swamp-out the exact transistor parameters and make the overall gain mostly a function of resistor values. Also we need resistors to get and set DC power in and extract signal power out.

So yeah, there is a LOT of "attenuation" happening all over. But that little Q1 part (correctly used) more than makes up for it.

[Venusblue]

Hmmm I see. Thanks to all of you guys for posting, That does clear a lot up.

I still don't get something though. Say that the AC came in from the guitar to "B" on the diagram, meeting at with the DC "C", and coming out of Q1 as "E" to R4. R4 then goes to ground, and this amplified signal is attenuated to ground? I don't get how it gets from ground to output.

[PRR]

> R4 then goes to ground, ... I don't get how it gets from ground to output.

How does R4 get the larger signal current? It can only come -through- Q1 and R3. Which as you see, is the Output node (passing through C2 and a pot).

Two suggestions:

• Study basic transistor action. Yes, this is like banging your head against the wall. It took years to sink into my irish skull. (You CAN just "take it on faith": certain common transistor schemes will generally make signals bigger. That keeps you from going nuts while understanding develops.)
• You seem to think resistance==attenuation. This is not wrong, but not the only thing resistors do.

[PRR]

> I love the smell of baked tubes in the morning.

Do you understand tubes? Can you wave your hand at a Fender Champ first stage and show why signal comes out bigger?

The only fundamental difference, transistors or tubes, is that a transistor Base leaks a LOT more than a tube Grid. So much more that we design different. Also a tube grid is negative of cathode while an NPN base is positive of emitter, but that's minor (we add R1).

But the idea is the same. We take a small interesting signal, and a big boring battery. We use the interesting signal to control the boring battery power. The output has the interesting input with the battery power; is bigger.

[G. Hoffman]

Hmmm I see. Thanks to all of you guys for posting, That does clear a lot up.

I still don't get something though. Say that the AC came in from the guitar to "B" on the diagram, meeting at with the DC "C", and coming out of Q1 as "E" to R4. R4 then goes to ground, and this amplified signal is attenuated to ground? I don't get how it gets from ground to output.

The original signal all goes to ground.  But, as it goes through the transistor, the transistor gets turned more or less "on" (that sounds dirty!  ).  So, in theory, when you have no signal coming in the bias voltage from R1 and R2 keeps Q1 half way "on."  This keeps the voltage at the collector of Q1 ("C" on the schematic) at 4.5V (give or take - no need for the real math right now).  The other 4.5V is going to ground.  When the input signal is at its peaks (the top of the sine wave, if you will) opens Q1 completely (more or less), so the voltage at "C" is 0V (give or take).  When the input signal is at its lowest (the bottom of the sine wave) the voltage at C will be 9V.  The output cap (C2) filters the DC voltage, so that instead of swinging between 0V and 9V, you are swinging between -4.5V and +4.5V - more or less.  

I think the thing you are having a problem understanding is that the electrons which are output from the circuit never flow through the transistor.  I might be wrong, but that's what it feels like from the way you are wording you questions.  (That's actually not true, if we look at it from the point of view of real electron flow, but for some reason we persist in teaching "conventional" electron flow, which is backwards.  Don't worry about it.)

Of course, the actual values involved are WAY more complicated than this, but fortunately we don't have to worry about it.  As long as you understand that the voltage at "C" is inversely proportional to the voltage at "B", you'll be OK.  (Well, exponentially inversely proportional, but don't worry about that!)

(I know that PRR has already answered this, and he certainly understands it MUCH better than I, but I'm hoping that different wording will help with understanding, and perhaps my educational level being closer to your own will help me frame it in a way easier for you to understand.  But he knows what he's talking about MUCH better than I.)


Gabriel

[Venusblue]

> R4 then goes to ground, ... I don't get how it gets from ground to output.

How does R4 get the larger signal current? It can only come -through- Q1 and R3. Which as you see, is the Output node (passing through C2 and a pot).

Two suggestions:

• Study basic transistor action. Yes, this is like banging your head against the wall. It took years to sink into my irish skull. (You CAN just "take it on faith": certain common transistor schemes will generally make signals bigger. That keeps you from going nuts while understanding develops.)
• You seem to think resistance==attenuation. This is not wrong, but not the only thing resistors do.

I think resistors attenuate some signal, but not all of it. Looking at the schematics though, It looks like some frequencies are filtered by the resistors and then sent to ground, and  lost forever. The reason why i'm trying to figure this out is because I know (Read: think) this is wrong. Unless they're doing just a high pass filter to get rid of all those highs (Which makes sense for R2).

So now i read that the DC never actually goes through Q1. Um i dont understand. How does the signal get to the output, from the input, while going through the input and not all getting bled to the ground after coming out of point E? Sorry for having so many questions lol, but thank you guys so much. You've no idea how much i've learned in my short amount of postings.

I also have a very basic understanding of tubes. I mostly just know that I love it when the volumes turned up to 11 though. I like real harsh overdrive.

[G. Hoffman]

So now i read that the DC never actually goes through Q1. Um i dont understand. How does the signal get to the output, from the input, while going through the input and not all getting bled to the ground after coming out of point E?

It doesn't.  The input signal controls Q1, but after that it goes to ground.  Bye bye.  The actual electrons in the input signal are not at all a part of the input.

BUT, what Q1 does is it turns the 9V DC at "C" into a larger AC signal which is exactly the same as the signal at "B", except it is bigger and upside down (if you think of it pictures). 

The input signal is an AC signal, centered around 0V.  In a sine wave, half the time the signal is higher than 0V, and half the time it is lower than 0V.  Now, the biasing network (R1 and R2) hold B at (let's pretend) 4.5V.  C1 blocks that 4.5V from coming back through the cable to your guitar.  It also does some filtering, but that is not important to what you are trying to understand. 

So, you've got "B" sitting there at 4.5V.  The input signal sums with the bias voltage, so when the input signal is below 0V, the voltage at "B" drops.  Less current flows through the Base-Emitter junction of Q1.  This causes less current to flow from through "C" to "E".  The voltage at "C" is higher.  When the input voltage goes above 0V, it sums with the 4.5V bias voltage.  This causes more current to flow from "B" to "E", which in turn allows more current to flow from "C" to "E", and the voltage at "C" goes down.

Those changes at "C" are an AC signal.  It is larger than the input signal, because the amount of current from "C" to "E" is larger than the current from "B" to "E" by whatever the h_fe of Q1 happens to be.  It is also proportional to the input signal.  And because a rise in V_B causes a fall in V_C, it also has its polarity inverted.*  The DC portion of the signal gets blocked by C2, and you have an AC signal that varies from (let's pretend) +4.5V to -4.5V.

But you are correct, the input signal does get sent to ground - we just make sure it does some useful work (turning Q1 on and off) before it gets there!


Gabriel






*(Some call this "out of phase", but this is untrue.  Phase is a measure of time, not polarity, and while that doesn't make any difference with the sine waves we all learn this stuff with, it DOES make a difference with real musical signals, which are almost always a long way from being actual sine waves!  Still, about half the time if you hear someone say something is "out of phase," what they actually mean is that its polarity is reversed.  Don't worry about it right now, but try to learn the difference at some point, because you will run into situations where it actually matters.)

[Venusblue]

Alright, Now that makes a lot more sense. I guess i'm having a hard time understanding where the flow goes. I'm assuming it's like water or electrons in all the chemistry classes I never paid attention in, and that it goes to the path of least resistance.

I think that it goes in all directions and filters through the resistors, if that's right. I could be wrong again, but I still feel like i've learned a great deal and i feel like i know the LPB just about inside and out.

Yeah, I always hear Brian May talk about switching his pickups on together and out of phase, but i figured he meant "In parallel" kinda.

Thank you very, very much for explaining all of this to me.

[G. Hoffman]

Alright, Now that makes a lot more sense. I guess i'm having a hard time understanding where the flow goes. I'm assuming it's like water or electrons in all the chemistry classes I never paid attention in, and that it goes to the path of least resistance.

Yup, the electrons follow the path of least resistance.  But the input signal causes the resistance of Q1 to go change, so the electrons encounter more or less resistance.  At some point, they all go down Q1, and other times none will go down Q1.  But there is a range where some go to ground, and the rest goes to the output.  Keeping it in this range is why we need bias voltage on the Base of Q1, and that is where it is useful.


Gabriel

[GibsonGM]

It can be very confusing to understand "transistor action", for sure.  I mean, the word transistor came from 'transferred resistance', or transferred varistor.    As a signal goes positive and the transistor amplifies it, causing a larger current to flow C>E, there is a voltage DROP across the collector resistor (in this configuration).   That drop allows an increase in current to the load. 

Just start reading up on how transistors work, you will eventually sort of 'get it', and don't need a ton of math - unless you want it - to work into practical understanding.

Small signal goes into the base and thru the base-emitter junction to ground.   Same time, the very small current always flowing from collector to emitter is 'modified', or modulated, by that input signal and develops a voltage across the collector resistor, equalling a current flow from C to E which is larger than the input.  The output is taken from the collector, and the signal which is now amplified is inverted.  That is amplification, very simplified. Like a valve on a water pipe.  There are more intricate terms and math to explain it, but for the beginner that should suffice.... 

[PRR]

> So now i read that the DC never actually goes through Q1. Um i dont understand.

Neither do I. DC does flow through the transistor. From Collector to Emitter. Another smaller DC current flows Base to Emitter. These two currents are "coupled" in the "transfer resistance" (obsolete terminology) which is the Emitter.

Say 1mA of steady DC. The base-signal modifies this more and less, say 0.9mA to 1.1mA. The 0.1mA change is the output. We like voltage-output, so a load resistor turns current to voltage. Our next box don't like DC, so a cap blocks it.

[PRR]

http://www.falstad.com/circuit/e-ceamp.html

Yes, not all PCs will run these Java applets. It is worth finding one that will.

This is an animated electronic amplifier stage very similar to the one you cite.

Yes, you have to really watch the Base flow to see anything move.... as you say, most of the input power is lost along the way. But that tiny Base flow, times the large DC flow through Collector and Emitter, means larger output power.

[G. Hoffman]

http://www.falstad.com/circuit/e-ceamp.html

Yes, not all PCs will run these Java applets. It is worth finding one that will.

This is an animated electronic amplifier stage very similar to the one you cite.

Yes, you have to really watch the Base flow to see anything move.... as you say, most of the input power is lost along the way. But that tiny Base flow, times the large DC flow through Collector and Emitter, means larger output power.

A picture is worth a thousand words, or something like that.  I totally forgot about that site - good call.


Gabriel

[Venusblue]

Thank you everyone, very much. The drawing did help me a lot, too.

I once read a thing about transistors saying "They are very, very complex to understand... but easy to use, so it doesnt really matter"

Everything makes more sense now though, Thank you.