Hi there! My first topic, Guerrilha Music, from Brazil.
Learning a lot from here, thank you very much! Here we go:
I'm builiding small projects into 5cmX5cm boards and fitting into small enclosures; now I'm looking for a tremolo. I've build EA Tremolo (improved) from General Guitar Gadgets schematic and I'm pretty satisfied, but I wonder if there is a way for eliminate the volume control and save some space on the board. I've search for something like that and it seems to be complicated, since the LFO section depends on the gain stage (is that right?)
The difference between the 'improved' and the original EA is, mainly, the input buffer, by mister Jack Orman. What's the difference if I build it without this first stage? Is there a volume drop reason for this buffer?
I use the Millenium Bypass for switching my effects. Can I apply this to the 'original' EA instead of grounding the second transistor?
(I gave up building Vicovibe because this one didn't work fine on breadboard and also had a weird interaction with Millenium switch)
Welcome.
(http://oi63.tinypic.com/13yi1lj.jpg)
Hi Paul.
Thanks for the reply!
So I avoid the input buffer, start with the .22uF cap and just replace the 25k Pot Volume with a 100k resistor to ground?
Can you tell me how did you get this value for the 100k resistor what does it stand for?
On the LFO part, should I go with the resistors indicated on the 'original' EA or can I go with the ones indicated at the 'improved'? Does this matter?
I would simply omit the level control and put the output right on the .47 cap.
I'm curious to what the resistor to ground is doing as well. I usually don't do such a thing when playing with circuits. I do as I explained and get 100% of it's level. I do this to almost all tremolos, phasers and vibes I work on. Anything with a lfo shouldn't be held back ;)
"So I avoid the input buffer, start with the .22uF cap and just replace the 25k Pot Volume with a 100k resistor to ground?"
I could be wrong but I think PPR has left out the part of the picture you dont need to see and concentrated on the bit you're talking about the output pot.
Leave off the volume pot and take your output directly from the 0.47 cap.
I'm guessing that the 100k to ground is likely to shave off some of the volume and to allow the cap to discharge to ground when the effect is in bypass mode.
The 100k to ground will also affect the output impedance of the effect, which is another reason its there IMO.
A pot with wires takes up less space than a single resistor ...
5cmx5cm is insanely luxurious for this circuit. I would not worry about saving space because volume pot is quite useful.
The volume pot sets the output cutoff frequency and is also a pull down resistor to hold the output at 0v and prevent pops in bypass.
The input buffer is in the original., not from Jack Orman. You can still find the original article online if you look for it. You need the FET buffer because the BJT's gain is very high so the input impedance is very low for that transistor. The improved version doesn't need a buffer because the MOSFET retains high impedance even at high gain. One less transistor. Hence 'improved.'
http://www.cackleberrypines.net/transmogrifox/rEAgeneratedTremolo/pages/projfiles/rEAgenerated_Tremolo_Design_Rationale_RevB.pdf (http://www.cackleberrypines.net/transmogrifox/rEAgeneratedTremolo/pages/projfiles/rEAgenerated_Tremolo_Design_Rationale_RevB.pdf)
The first image in that document (apart from some image processing/sharpening and a little markup to clarify the smudgy values) is a scanned copy from the original Electronics Australia article. As Jon said you can find the scans of the full article online as I did.
There is no input buffer.
The original has some 30 dB gain which may be a bit over-the-top without volume control unless this is the first thing in the signal chain fed by the guitar.
The depth of the effect is increased by this gain as it tends to depend more upon the FET and less on the emitter R. Probably the best pot replacement without changing the original would be a resistor divider on the output. Breadboard one with the pot. Adjust the pot to where the output volume sounds right when comparing FX on and bypass. Measure wiper to each leg and pick the nearest standard resistor value and replace.
As an aside I also tend to agree with Jon that a volume pot will fit in your enclosure and it is a nice feature...but it should be easy to do away with it just by replacing each leg of the pot with resistors.
I just showed the essential change with enough context to find it. Obviously you need the LFO, and the buffer may be worth having.
IMHO, if there isn't some bleed resistor after that output cap, your next question will be about POP. When bypassed, the right end of that cap charges-up (through leakage) to several Volts DC. When you engage, POP. The bleed resistor sucks the leakage to hold the right end of the cap near zero VDC when nothing else is hanging on. The value is semi-arbitrary. In this case, the specified 0.5uFd cap can't deliver full bass into a low resistance, but can with 100K hung on it.
I do think the Volume pot is useful. The static (no wobble) gain of this thing is significant, hardly "unity gain". The exact gain is hard to predict. When it wobbles, the gain gets higher (and lower), which may drive a clean amp into sputtering overload. I think you should keep the pot. But that wasn't what you asked for.
Quote from: PRR on November 09, 2015, 11:50:28 PM
I just showed the essential change with enough context to find it. Obviously you need the LFO, and the buffer may be worth having.
That was clear, Paul. And I understand that may be worth having the buffer, but that wouldn't make no difference on saving space.
Quote from: midwayfair on November 09, 2015, 08:44:41 PM
A pot with wires takes up less space than a single resistor ...
5cmx5cm is insanely luxurious for this circuit. I would not worry about saving space because volume pot is quite useful.
That's right, Jon. But I also need to save space in the enclosure as well.
Quote from: Transmogrifox on November 09, 2015, 11:22:34 PM
Probably the best pot replacement without changing the original would be a resistor divider on the output. Breadboard one with the pot. Adjust the pot to where the output volume sounds right when comparing FX on and bypass. Measure wiper to each leg and pick the nearest standard resistor value and replace.
Thanks for the link, Transmogrifox. Got to study it a little bit. The one I've build had a insane gain. I used to leave the Vol pot at 10 o'clock, most of the time. I think I should breadbord one without the buffer and with fixed resistors instead of vol pot and see (hear) how it goes.
Quote from: Transmogrifox on November 09, 2015, 11:22:34 PM
The original has some 30 dB gain which may be a bit over-the-top without volume control unless this is the first thing in the signal chain fed by the guitar.
What do you mean, exactly? This is an impedance question, isn't it?
I'm constantly learning the backgrounds theory while I keep builiding, so sorry about silly questions. Impedance, for example, I still didn't figure out. LFO is another thing I must run after.
Omitting Volume pot isn't such a good idea in any pedal that is designed with an amount of gain..
Unless you intend to connect it at always the same next stage so you can "trim" it's output impedance in such a level that you don't realize a volume drop (or raise..)
But it definately will be a compromise between volume level and frequency loss.
Quote from: antonis on November 10, 2015, 08:29:55 AM
Unless you intend to connect it at always the same next stage so you can "trim" it's output impedance in such a level that you don't realize a volume drop (or raise..)
How calculate the impedance on this one? How much it will be if you set it for matching levels on and bypass? Let's presume that I want to connect it direct to an amp.
Quote from: antonis on November 10, 2015, 08:29:55 AM
But it definately will be a compromise between volume level and frequency loss.
It's because the output capacitor forms a RC Filter with the resistor?
Quote from: Guerrilha Music on November 10, 2015, 07:54:25 AM
Quote from: Transmogrifox on November 09, 2015, 11:22:34 PM
The original has some 30 dB gain which may be a bit over-the-top without volume control unless this is the first thing in the signal chain fed by the guitar.
What do you mean, exactly? This is an impedance question, isn't it?
I'm constantly learning the backgrounds theory while I keep builiding, so sorry about silly questions. Impedance, for example, I still didn't figure out. LFO is another thing I must run after.
Yes, I was thinking about how the guitar output impedance and impedance of the guitar volume control interact with the input.
PRR gave you the answer. Notice the GGG "improved" version is not quite as high gain as the original (but it's still pretty loud). You can set the gain more exactly to taste by taking that 100k that PRR showed and turning it into two so you have a divider, just like a pot does.
The gain, with FET out of the picture, is nearly 4. I do not understand why so high. (And you say the original was more??)
> set the gain more exactly to taste
For practical purpose, you could just reduce R4. R4=1.38K would be unity gain, neglecting interface loss. Allowing for in and out impedance, and that "unity gain" should err on the high side of 1.000.., R4=1.5K or 1.8K may be pretty near.
However this does not change the basic oddity: when FET starts wobbulating, gain goes up NOT down. You would probably expect gain to go higher AND lower around the no-trem gain. There are other trems which work more this way, so I'm not keen on re-thinking this one. Anyway Transmogrifox did a deep re-design, considering this point and others, and has posted very clear details of his re-interpretation. Compared to the buffered variant it is not much more parts.
But if you simply need space, yeah, just lose the volume pot. If it pops, add a bleeder.
Quote from: PRR on November 10, 2015, 05:55:22 PM
The gain, with FET out of the picture, is nearly 4. I do not understand why so high. (And you say the original was more??)
> set the gain more exactly to taste
That is true with the FET out of the picture and it looks like the question was answered later in your post as you mentioned gain goes UP as modulation depth increases...but another way to look at is is that the gain is high, and the high gain goes DOWN as modulation increases.
In the original the min depth setting was a preamp with no modulation. This equates to 0 Vgs, which causes the FET to go to a low RdsON state. The original circuit uses a 12k in the collector, so the gain with a FET with RdsON of 100 (for example) could be as high as 12k/(180+100) = 43 = 33 dB. By the same reasoning the GGG version with 4.7k comes out at 20-25 dB depending on the FET.
The GGG version is approximately the same situation. Ultimately it is a high gain preamp with modulated attenuation when you consider all depth settings max out at max gain. LFO being AC coupled to the FET gate, it's like a 1/2 wave rectified LFO shape where only the excursions below 0V result in a substantial gain change.
Anyway I hope this dialogue between myself and PRR is working out to be educational to somebody. I think PRR answered the OP question in the first post while the rest of the responses have helped to explain somewhat more how this little gem works :)
Either one is pretty loud without reducing the collector resistor or application of a resistor divider or volume pot.
> equates to 0 Vgs, which causes the FET to go to
Duh. Overlooked that tidbit, but of course ON is the natural rest-state of a JFET without trickery.
Although I believe your queries have already answered..
Quote from: Guerrilha Music on November 10, 2015, 09:31:44 AM
Quote from: antonis on November 10, 2015, 08:29:55 AM
Unless you intend to connect it at always the same next stage so you can "trim" it's output impedance in such a level that you don't realize a volume drop (or raise..)
How calculate the impedance on this one? How much it will be if you set it for matching levels on and bypass? Let's presume that I want to connect it direct to an amp.
You just have to calculate it's impedance as an "intermediate" circuit which adds or subtracts nothing..
It's output impedance should be exactly the same as the impedance of previous connected one..
(actually you have to replace PRR's suggested R12 with a trimpot - forming a voltage divider, as well said by Transmogrifox - and adjust it at the point that your tremolo doesn't alter the volume when ON..)
Quote from: Guerrilha Music on November 10, 2015, 09:31:44 AM
Quote from: antonis on November 10, 2015, 08:29:55 AM
But it definately will be a compromise between volume level and frequency loss.
It's because the output capacitor forms a RC Filter with the resistor?
Yeapp..
Same thing also happens with the original Volume pot..
(actually, there are formed 2 RC filters: 1 "fixed" with RC value = Cout X total Pot value & 1 "variable" with RC value = Cout X Pot value between upper lug & wiper..)
Quote from: Transmogrifox on November 09, 2015, 11:22:34 PM
There is no input buffer.
Me: Open mouth, insert foot!
I used a trim in a mini build I made:
(https://jonpattonmusic.files.wordpress.com/2012/05/mini-tremolo-guts.jpg)
I don't think the layout's in my perf library, though -- I basically squished the Runoff Groove layout down a little. A 3362 trim pot is quite small, and only takes up 3x3 holes on the perfboard (think three small box capacitors right next to each other).
The volume is slightly different at different depth settings, but not nearly as bad as some other tremolos.
I like your yellowish tantalum caps, John.. :icon_wink:
Do they behave good in LFO ..??
(I allways had an innate dislike for their kind..) :icon_redface:
It's certainly not a case of saving space only, I'm really learning a lot from the discussion about how this thing works!! Thanks PRR, Transmogrifox, etc for all efforts.
I would like to put another queries, Antonis, if you all dont't mind.
This schematic
http://www.home-wrecker.com/eatremolo.html
discussed here
http://www.diystompboxes.com/smfforum/index.php?topic=80479.0
seems like a little different input stage. The rest of the circuit looks like the same.
Is this a MOSFET transistor, right? Does it work similar to JFET ones, in this case?
What is that Trimpot, 47uF and 1M resistor (along the Gate?) standing for?
Quote from: Transmogrifox on November 10, 2015, 11:03:46 PM
Quote from: PRR on November 10, 2015, 05:55:22 PM
The original circuit uses a 12k in the collector, so the gain with a FET with RdsON of 100 (for example) could be as high as 12k/(180+100) = 43 = 33 dB. By the same reasoning the GGG version with 4.7k comes out at 20-25 dB depending on the FET.
Could you please set the names on this formula?
Is it similiar for a BJ calculation?
Quote from: Guerrilha Music on November 11, 2015, 02:23:35 PM
I would like to put another queries, Antonis, if you all dont't mind.
This schematic
http://www.home-wrecker.com/eatremolo.html
discussed here
http://www.diystompboxes.com/smfforum/index.php?topic=80479.0
seems like a little different input stage. The rest of the circuit looks like the same.
Is this a MOSFET transistor, right? Does it work similar to JFET ones, in this case?
What is that Trimpot, 47uF and 1M resistor (along the Gate?) standing for?
MosFet input stage is a replacement for the original combination of input buffer and main gain stages (FET Buffer & BJT Amp)..
It sets input impedance a little higher than original (500k instead of 333k) and keeps the same gain (about 3.4 when Q2 is Off)
(actually, it sets total stage's impedance more higher because it "saves" the extra loss between the 2 stages due to BJT's voltage divider bias..)
100k biasing trimpot serves for adjusting MosFet Drain voltage at 4.5 - 5 Volts (it forms a voltage divider from Vcc/GND to Gate)
1M resistor simply connects trimpot to Gate keeping input impedance high...
(input impedance is the parallel combination of 1M anti-pop resistor and Gate to Power Supply/GND resistor)
47μF stands for bias stability purposes..
An approximation that can help is all of these 3-terminal devices (MOSFETs, JFETs, BJT, IGBT, Triode) work the same way in a high-level behavioral sense: You have one terminal connected to a higher potential and another connected to a lower potential. A third terminal changes the amount of current that flows through the other 2 terminals.
A conversion of control voltage to current is termed "transconductance". The details are more complicated, but as long as the device is relatively high gain they all behave nearly the same in a gain circuit such as found in the EA trem.
The thing that simplifies this and makes these devices interchangeable in a functional sense is the use of feedback. When transconductance (device gain) is high compared to the resistor in the emitter/source/cathode to ground, then you get a voltage gain very near to unity (gain of 1) at the emitter/source/cathode. And for clarity, this is a feedback configuration. By increasing voltage on the gate, the current wants to increase...but..because current increases it makes the voltage on the emitter increase which wants to shut itself off and restrict current. It balances itself at a tightly controlled constant Vbe (or Vgs) relationship that stays about the same for small fluctuations so that the emitter (source) has to follow the base (gate) voltage fluctuation at a nearly 1:1 ratio.
By this principle if you apply 10mVpp at the FET gate (for example), you get something very near to 10mVpp at the source. The resistor in the source will conduct current in proportion to the voltage applied from the gate. Since in a FET current can't come from the gate, there is only one place for the current to come from: the drain. In a BJT this relationship still holds approximately true because it will draw a much higher ratio of current from the collector than from the base (by a factor of hFE, or "beta").
So if you have, say, 1.2k in the source, and 10 mVpp at the source, then the AC portion of the current is 10mV/1.2k = 8.3 uApp through the source resistor.
This 8.3 uApp has to go somewhere so it is supplied through the drain. If you attach a 1.2k resistor in the drain, you end up with a unity gain inverting amplifier (pushing up at the source causes the current to pull DOWN at the drain).
If you add a 12k resistor in the drain, you get 12k*8.3uA = 100 mV, which is a gain of 10. As you can see, this is the same as 12k/1.2k.
As a general rule you can approximate gain as Rdrain/Rsource, or Rcollector/Remitter, or Rplate/Rcathode as long as that resistor is decently large (>1k for most transistors). I'll add in the case of tubes, though, there aren't very many gain circuits in which this assumption is useful -- for example Rcathode would have to be above 50k before this is a reasonable approximation for a higher gain triode like a 12AX7.
As the emitter/source/cathode resistor gets relatively small, then the transconductance (gain of the device) starts to have much more effect and this approximation becomes less true.
In the case of the EA tremolo, the control FET acts as a variable resistor, so the change in a device will have a stronger effect on the maximum gain (when looking into the 180 ohm series low control JFET RdsON).
That said, as long as you get the circuit biased properly for the different devices you could use a MOSFET, a Triode (tube), a JFET or a BJT and get approximately the same result.
The BJT is the most different because it is slightly different in how it works. It takes some current on the input, so it reflects back a smaller equivalent resistance from the emitter if the emitter resistor is small. In that sense, an EA trem with a BJT and no buffer reflects a variable resistance back to the base of the transistor, and that resistance is changing with the LFO.
:icon_biggrin: Well said...!!! :icon_biggrin:
Just a note to:
<As a general rule you can approximate gain as Rdrain/Rsource, or Rcollector/Remitter, or Rplate/Rcathode as long as that resistor is decently large (>1k for most transistors)>
If there is a negative feedback loop (which is true in most cases..) the Gain is proportional to Rfeedback/Rbase, as long as the ratio of Rf/Rb is much lower than Rcollector/Rbase and Rf is much higher than Rc..
(same for Rdrain & Rgate..)
Thanks again!
I'm trying to follow everything been discussed by now!
I think there's a lot of information that's gone be very useful, for me, at least in the future..
Seems like I'm not on that level yet, so I think it's better understanding the basics a bit more about a single BJT and how it works, in boosters and simple drives, like Electra Distortion, for example.
I need to understand how to bias a transistor properly to get a 1:1 gain first, and then go into clipping territory. Voltage gain and current gain also are situations not clear for me.
Altought I might be going away of the initial subject of the post I would like to ask two or more things about how the audio signal (AC signal) relates with the DC signal while entering a simple transistor circuit.
Thinking about a circuit like LPB Boost (no distortion for now):
After the input cap (that blocks DC current), the signal coming from the input relates (mixes, add, what happens?) with the DC coming from 9V power supply to ground. Then the magic happens around the transistor and resistors along it. Finally, the signal goes through the output cap (that also blocks DC current).
I've read that the small flutuactions (voltage? current?) in the base causes big flutuations (voltage? current?) on the collector, but still don't understand what happens around and how the audio signal goes from in to out after that interaction with DC.
It's hard to formulate my doubts! Looks like I got dozens of pieces of a puzzle, I can glimpse the final result, but still can't arrange everything together.
Quote from: Guerrilha Music on November 13, 2015, 09:49:15 AM
the signal coming from the input relates (mixes, add, what happens?) with the DC coming from 9V power supply to ground.
Better to read few things about BJT as a current source..
(before some guys come and slap your tonge with a rule (or similar..) :icon_biggrin:
quickie: You may substitute "mixes, add, what happens" with "disturbance" :icon_wink:
Signal goes to transistor base and is further "lost" via emmiter to ground - no original signal anymore..!!
It have annoyed enough Collector's serenity so Collector reacts in a rude manner, resulting in an exravagant punishment..
Emitter is busy to maintain a voltage difference accepted by Base so it can pass the original signal but Emitter's resistor doesn't like the whole situation so it raises Emitter's voltage to anticipate Collector's rudeness..
There is a no man's land between Collector and Emitter which plays the role of boxing bug..
Sorry guys but it's quite late here.. :icon_redface:
In some post recently PRR gave a good analogy about flooding your basement with a 4" pipe and draining it with a 4" pipe, controlling the drain with a flap and watching the water level in your basement go up and down. I thought is was a good analogy -- worth a read if you find it. Don't search more than a month back.
It's not too far removed from the idea of sound in air. Atmospheric pressure is nearly constant at a certain altitude...changes very slowly compared to sound waves. This could be called the "DC" air pressure.
When sound waves travel through the air, they make small fluctuations in the air pressure which makes your ear drums move back and forth. An audible sound wave would be the "AC" part.
Your speakers rest in a center position. When electric energy is applied to the coil, it makes them move forward and backward. If you had your speaker forced as far forward as it can go, then an electric signal pushing it forward would have no effect (except it might blow up your amplifier). The reverse moving electric signal would pull the speaker backward, then it would return to full forward. This would sound horrible.
The speaker being pushed full forward can be compared to applying an AC signal to a BJT with its base grounded (not biased). In that state the collector voltage is stuck high and can't go any higher. When you give some voltage at the base, it begins to pull down and causes the voltage at the collector to move down.
Bias is to make the collector voltage stay somewhere near 1/2 the supply voltage so there's room for it to move up and down.
The wave doesn't move through the transistor. Its movement is only copied (mirrored) at the collector...and that's what these circuits do. It's like, if the guy across the room moves his hand up 1 foot, I move my hand down 3 feet. The guy watching me through the window in the next room moves his hand up 2 feet when I move mine down 1 foot, etc.
The electrons in your guitar pickup don't touch the speaker coil any more than the guy in the other room ever touches my hand. It's just a big game of copy-cat from one transistor circuit to the next and we get distortion when the first guy moves his hand 1 foot and the 3rd guy can't move his hand up 6 feet -- it stops as high as he can reach at about 2 feet above his shoulder.
So, to bias his hand at 1/2 the supply would be to hold it straight out parallel to the ground so he has room to move it up 2 feet and down 2 feet.
Maybe that helps. Something maybe easier to follow than the more technical stuff.