Lovepedal - Englishman  [traced]

General documentation, gut shot, schematic links, ongoing circuit tracing, deep thoughts ... all about boutique stompboxes.
sinner
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Post by sinner »

Switch is 3pos on/off/on:
- position 1 - 2 x 1n4148
- position 2 - diode lift
- position 3 - 2 x BAT's

C5 is there to help to reduce high-frequency noise in a circuit, and yeah - C4 alone should be enough

miguelyogur
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Post by miguelyogur »

excuse my ignorance but then, how is a dpdt giving these 3 positions? to my understanding a dpdt is two spdt in a single switch so I don't understand how this switch works for that diagram

Is this the right switch for this project?
http://t0.gstatic.com/images?q=tbn:ANd9 ... kX8HaHXyDb

thanks

sinner
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Post by sinner »

miguelyogur wrote:excuse my ignorance but then, how is a dpdt giving these 3 positions? to my understanding a dpdt is two spdt in a single switch so I don't understand how this switch works for that diagram

Is this the right switch for this project?
http://t0.gstatic.com/images?q=tbn:ANd9 ... kX8HaHXyDb

thanks
You can go with spdt as long as it's 3 way on/off/on

I dont know what kind of switch you have there beside the type, gringo

Cheers

Barcode
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Post by Barcode »

miguelyogur wrote:excuse my ignorance but then, how is a dpdt giving these 3 positions? to my understanding a dpdt is two spdt in a single switch so I don't understand how this switch works for that diagram

Is this the right switch for this project?
http://t0.gstatic.com/images?q=tbn:ANd9 ... kX8HaHXyDb

thanks
dpdt refers to "double pole, double throw" and has nothing to do with the number of positions, at least not in a technical sense. In an on-off-on dpdt, you still have 2 throws and 2 sets of outer poles, but you have 3 different configurations you can have the throws in. It's still a dpdt.

miguelyogur
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Post by miguelyogur »

thanks, I just realized the existence of these switches with on-off-on positions, I was puzzled since so far I only used sdpt with on-on positions XD. I just built this, very very nice, specially the setting with the BAT diodes with minimum drive. I have used BAT42 instead of BAT46 since it what I had in my stock. Should I expect any difference? same thing for the 1n914 that I used instead of the 1n4148.

miguelyogur
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Post by miguelyogur »

Hello again, I am trying to understand this pedal better and I have a few problems. My computations with the gain totally closed give an input impedance of about 75k ohm and gain of 10. With the gain completely open I get about 112k ohm and 5.7.

All these figures seem a bit low to me, especially the input impedance, that I would expect to be higher than, say, 500k ohm, if I understand correctly. Is it possible that I am making some mistake?

thanks

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Post by cctsim »

The difference in the gain between open and closed gain knob seems a bit too low in my view.

According to my computations, assuming R11=0, Vcc=9.6V and beta(dc,ac)=500:

-For "open" (R7 in parallel with the 1k gain pot.)
Gain=12.7, Zin=95k

-For "closed" (R7 shorted)
Gain=1030, Zin=155k

Anyway, the CE amplifier is not famous for high input impedance.

miguelyogur
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Post by miguelyogur »

Thanks for the reply. I have tried a few times during the last days to check this, and although I obtain comparable results for the open R7 case, I am not even close for the R7 shorted case.

What I do is the following. In order to compute re for the T model I have to get IB. For that, VB is 1.58V because of the voltage divisor given by R2 and R6 (I am assuming DC=9V). Therefore VE=1.58-0.7=0.88V. The emitter resistance for the DC model is R4=330R so IE=IB=2.67mA. So re=25mV/2.67mA=9.36R.

Now for the small signal model I have Zin=R3||R2||R6||beta r3=4.61k assuming beta=500. For the voltage gain I have G=-(R1||R9)/re=350 approx.

I do a similar thing for the open R7 case (with appropriate emitter resistance) and I get Zin=103k and G=11.

I am self-learning transistor circuits, and it would be great if someone could tell me where I am wrong.

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Post by cctsim »

miguelyogur wrote:The emitter resistance for the DC model is R4=330R so IE=IB=2.67mA. So re=25mV/2.67mA=9.36R.
I don't think IE=IB is correct. It's more likle IE=(beta+1)*IB.

These are the equations I have used for the pi-model (written for use in Matlab):

Rb=1/(1/R2+1/R6);
Ie=(Vcc-Vbe)/(R4+R7+Rb/beta);
re_dash=25e-3/Ie;
rc=1/(1/R1+1/R9);
Av=-rc/(R7+re_dash);
zin_base=beta*(R7+re_dash);
zin=1/(1/R3+1/Rb+1/zin_base);

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Post by cctsim »

In my previous post, the equation for Ie should read

Ie=(Vbb-Vbe)/(R4+R7+Rb/beta);

where

Vbb=Vcc*R6/(R2+R6);

miguelyogur
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Post by miguelyogur »

Hello, thanks for the reply. Sure, I meant IE=\beta IB approx :D

Now, I am trying to follow your equations and I don't get this one

Ie=(Vbb-Vbe)/(R4+R7+Rb/beta)

For me, if R7 is shorted this should read Ie=(Vbb-Vbe)/(R4+Rb/beta)
and if R7 is in parallel with R10 this should read Ie=(Vbb-Vbe)/(R4+R7||R10+Rb/beta)

and if I am correct this would propagate to the expressions for Av and zin, where R7 should be substituted by either 0 or R7||R10 right? or maybe there is something I don't see...

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Post by cctsim »

Yes, you are right. R7 should be either 0 or R7||R10.

miguelyogur
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Post by miguelyogur »

Thanks! I think I understand this circuit now :applause:

However, coming back to the input impedance. Is not too low? I know that this is expected since it is one of the features of a common emitter amplifier, however my understanding is that guitar pickups should be interfaced with higher input impedances in the next stage... How is this not regarded as a design flaw for the Englishman, or, which is the reason for not putting an input buffer in this circuit? there should be a reason

regards

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