New booster circuit

Started by Godslildrummer, May 27, 2019, 12:21:02 AM

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Godslildrummer

So I'm pretty good with math and really good with signal flow (that's my job, I do AV design and install) so I thought I was pretty good on designing my own pedal as I have repaired and modded several. Only what I made what not exactly what I expected to make and I don't understand why.

So let's start from the beginning. I wanted a boot pedal. So I started with a lpb1 schematic. I studied how the first resistor couples the power and blocks DC current, and the second the same. How the listed 0.01uf Cap along with the first R2 created a high pass filter. I learned about how the R2 and R1 affect forward bias and what common emitter topology was and the difference in npn and pnp transistors. I learned about how R3 and R4 affect gain and how allowed to much voltage lessens available power to the base pin causing clipping.

I decided to make my own circuit. I changed all the transistors, the transistor and the caps too. But not at random. I found the formulas and did math, LOTS of math. Then I built it. But just virtually using every circuit. And I saw some issues so I changed it again and again until it was perfect. It had 24db of gain taking a 250mv input signal to 3.74V output. I then bread boarded it. And it worked. But not like I had hoped.

The volume pot worked, the sound was nice and clean, but instead of making a boost, I apparently made an attenuation pedal. Here's the thing, I don't know why! I did the math, it looked great on paper but didn't work as planned.

I checked the battery (it's brand new) 8.7 volts. I tried a 9v ps. It worked fine (but I did blow my led) it measures 9.2v.

The circuit clearly works. No problems with connections on my breadboard. But apparently my math is off and I must have also missed something in Everycircuit.

I am attaching a screenshot of it. The only substitute I made from the screenshot was r1 called for a 300k resistor. I didn't have one so I used a 270k and a 43k resistor in series. I checked emitter and collector pins and swapped them just in case (sounded awful wasn't it,) swapped back and signal was fine still just low.

Please help. Since I changed everything I'm not sure how to begin diagnosis. I am using all +/-1% resistors, 0.015uf metal film caps, a 2n5089 transistor and a 100k pot.

Original schematic:
https://www.coda-effects.com/p/lpb1-circuit-analysis.html?m=1

My circuit on Everycircuit:


Expected signal gain:




Please help!

amptramp

What is your signal source?  If it is a signal generator then you have to add its output impedance in series with the generator.  If it is a guitar, you have a complex impedance dominated by the inductance of the pickup and whatever position the guitar volume control is at.

For a midband signal where the input capacitor doesn't matter, the input impedance consists of the parallel combination of 1 megohm, 300 K, 100 K and hfe * 150 ohms.  This is going to be rather low for a pickup that may have 3 to 5 Henries of inductance.

Also, your simulation shows the output volume pot at the midway point in resistance.  If the pot was all the way up, the gain would be higher.

Godslildrummer

#2
So when I breadboarded it I was using my telecaster as a signal and a little amp as a test. When it was in my circuit it was quieter. Straight tele to amp about twice to three times as loud. Volume and tone all the way up on the tele for all tests. Pot on breadboard was checked at full sweep. Turned down beautifully but max volume was less than half of regular guitar output.

Quote from: amptramp on May 27, 2019, 01:17:09 AM
Also, your simulation shows the output volume pot at the midway point in resistance.  If the pot was all the way up, the gain would be higher.

If you look, the output signal on the Sim was taken after the cap but before the pot resulting in full signal strength. I even took the pot out of my breadboard and went straight out to see if it affected it but it did not.

antonis

Quote from: Godslildrummer on May 27, 2019, 12:21:02 AM
the sound was nice and clean,
I'd take your word for that only in case of not permanently "bottomed" transistor..  :icon_wink:
(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..)
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Gus

Did you read this?https://www.diystompboxes.com/smfforum/index.php?topic=29816.0

Check you resistor values with a meter.
Measure the transistor voltages.

The LPB textbook circuit has a number of threads about it over the years at this forum.

Godslildrummer

 :icon_confused: I did read that. I wasn't looking to necessarily debug it as it did work as much as understand why my theory didn't match my application. I did this at my in law's house and don't have a meter with me. I will get values later this evening when I get home. I have searched but all I found were classic circuits and tweaks. Nothing really explains why it works the way it does and especially to the extent of the changes I made since it's not anywhere near textbook anymore. I have a tendency to jump in and when it doesn't work I can usually fix it. It's when it does work, just not as expected that I have problems.

Quote from: antonis on May 27, 2019, 05:13:13 AM
Quote from: Godslildrummer on May 27, 2019, 12:21:02 AM
the sound was nice and clean,
I'd take your word for that only in case of not permanently "bottomed" transistor..  :icon_wink:
(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..)

I don't understand what you mean by "bottomed" and could you expound on the later part? The way I understood it was I only needed a .6v signal on my base pin and everything else I should send to my collector to get the highest gain possible.

merlinb

Quote from: Godslildrummer on May 27, 2019, 04:30:41 PM
:icon_confused: I did read that. I wasn't looking to necessarily debug it as it did work
'work' is not the same as 'makes some noise'. Dude, you've wired it up wrong. Go check. You probably got the emitter and collector the wrong way around or used a 270R resistor instead of 2.7k or something.

"No I triple checked and it's definitely not wrong"

I do this for a living. I triple check stuff all the the time. And it's STILL turns out to be wrong on the fourth check.

Danich_ivanov

#7
Collector voltage is lower than the base, which is why it doesn't work. To fix this, that 100k at the base should be somewhere around 30k to get about 1/2 vcc at the collector. 1/2 vcc will give more or less the same waveform at the output - clean boost.

PRR

My idiot says it should work. 250mV peak input gives an asymmetric clipped 3V peak output, which is surely "louder".

As others are saying: it isn't connected according to plan. DC voltage checks are an excellent first step.

However in "pedal world" it is good to be sextuple-sure your IN and OUT are the right way around. Going into the output, and out the input, sometimes "passes signal" (especially in simple amps) but doesn't give Gain.

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PRR

Quote from: antonis on May 27, 2019, 05:13:13 AM....(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..

Ah, I see what you did there. Yes, 1/4 of 9V on Base and 150r emitter would be 11 mA. But base current is very significant for these values. How can you guess at a glance? Divide the resistance at the base by hFE, and compare to Re. Quickly: 100k/100 is 1,000r, which is much higher than 150r, so you know the base resistors are not only "significant", they dominate the DC bias. Even if you run the math for 300K||100K and run hFE from 100 to 300, the base resistors dominate. (A super-jellybean with hFE >500, both resistances have equal effect.)
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Godslildrummer

#10
Quote from: Danich_ivanov on May 27, 2019, 06:49:16 PM
Collector voltage is lower than the base, which is why it doesn't work. To fix this, that 100k at the base should be somewhere around 30k to get about 1/2 vcc at the collector. 1/2 vcc will give more or less the same waveform at the output - clean boost.

So my measurements for the transistor are Base: 0.62v Collector: 7.34v and Emitter was 0.05v.

IDK enough to know if a 0.05v on the emitter is ok?? But this should be in forward bias right?

Also I went through and double/triple checked resistors and their placement. They are as follows:

R1-300k ohm from 9v to P2 of C1/base of 2N5089/P2 of R2. Actual measurement is 297K ohm from 2-150K ohm in series.

R2 - 100K ohm from P2 of R1/base of 2N5089/P2 of C1 to ground. Actual measurement is 100.4K ohm.

R3 - 2.7K ohm from 9v to collector of 2N5089/P1 of C1. Actual measurement is 2.64K ohm from 2.2K ohm and 500 ohm in series.

R4 - 150 ohm from emitter of 2N5089 to ground. Actual measurement is 144.8 ohm.

R5 - (for bleed) 1M ohm from input/P1 of C1 to ground. Actual measurement is 997K.

Next are my voltage readings:
R1 -  P1 5.11v and P2 (pwr) 7.90v

R2 - P1 0.59v and P2 (gnd) 0v

R3 - P1 (pwr) 7.90v and P2 6.97v

R4 - P1 0.04v  and P2 (gnd) 0v

R5 - P1 0v and P2 0v

Transistor B - 0.62v C - 7.34v E - 0.04v

C1 - P1 (input) 0v and P2 0.62v

C2 - P1 0.04 and P2 0v

My guitar has a resistance of:
Neck 5.91K ohm
Both 3.16K ohm
Bridge 6.54K ohm
(All readings taken at end of cable that goes into pedal input)

Battery is 9.37 no load, 8.93 under load.

No matter what I did I couldn't get a voltage reading from playing my guitar.

As far as placement, everything in in the right place. As I said before it works just not like it should. I don't know enough about this 2N5089 and information is limited where I can find it. I know the pin out is correct as I've both looked it up and tried it. When I reverse the C and E it cuts in and out horribly. All I can assume is it is not being pwered correctly due to my changes. But I don't know why. It seems to work in theory.

Quote from: PRR on May 27, 2019, 09:27:22 PM
Quote from: antonis on May 27, 2019, 05:13:13 AM....(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..

Ah, I see what you did there. Yes, 1/4 of 9V on Base and 150r emitter would be 11 mA. But base current is very significant for these values. How can you guess at a glance? Divide the resistance at the base by hFE, and compare to Re. Quickly: 100k/100 is 1,000r, which is much higher than 150r, so you know the base resistors are not only "significant", they dominate the DC bias. Even if you run the math for 300K||100K and run hFE from 100 to 300, the base resistors dominate. (A super-jellybean with hFE >500, both resistances have equal effect.)

I don't understand how amp draw affects this at all. All I found information for was for voltages. I do understand the relationship with each other (P=IxE, V=IxR etc) but not how it applies here. What am I missing? Every time I changed the resistor one place it would change my voltage at another. I don't want to go lower with R2 value as this with the .015uf caps makes a high pass filter as low as ~10Hz. The one in the lpb1 was as high as ~40Hz.

I don't want to change parts at random because I want to understand why It does what it does.

willienillie

QuoteR3 - 2.7K ohm from 9v to collector of 2N5089/P1 of C1. Actual measurement is 2.64K ohm from 2.2K ohm and 500 ohm in series.

I think you mean C2, not C1.

QuoteR1 -  P1 5.11v and P2 (pwr) 7.90v

R2 - P1 0.59v and P2 (gnd) 0v

Don't these two resistors meet at Q1b?  They should share a voltage there.

QuoteBattery is...8.93 under load

Where's 7.90V coming from?

Something is hooked up wrong.  Can you can post a couple pics of your breadboard?

bluebunny

#12
There's definitely something wrong.




  • The top end of R1 and R3 connect to the battery.  They should be 9.37V.
  • The bottom of R1, top of R2, one end of C1 and the base of the transistor are all connected.  They should have the same voltage.
  • The bottom of R3, the collector and one side of C2 are connected.  They should have the same voltage.
  • The output end of C2 is at 0V, despite there being a 250K pot in the way.

Please show us pictures of what you've built.

Edit: ooh! synchro-post!  ;o)
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Ohm's Law - much like Coles Law, but with less cabbage...

willienillie

Great minds simul-post alike!

antonis

Quote from: Godslildrummer on May 28, 2019, 02:19:14 AM
Quote from: PRR on May 27, 2019, 09:27:22 PM
Quote from: antonis on May 27, 2019, 05:13:13 AM....(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..

Ah, I see what you did there. Yes, 1/4 of 9V on Base and 150r emitter would be 11 mA. But base current is very significant for these values. How can you guess at a glance? Divide the resistance at the base by hFE, and compare to Re. Quickly: 100k/100 is 1,000r, which is much higher than 150r, so you know the base resistors are not only "significant", they dominate the DC bias. Even if you run the math for 300K||100K and run hFE from 100 to 300, the base resistors dominate. (A super-jellybean with hFE >500, both resistances have equal effect.)
I don't understand how amp draw affects this at all.

You should read about Thevenin's equivalent resistance (which is the parallel combination of voltage divider resistors IN SERIES with transistor Base) and Emitter reflected resistance [which is hFE X Emitter total resistance (Emitter resistor + Emitter intrinsic resistor).
Actual Base voltage is the voltage on divider node minus voltage drop on Thevenin's resistance, which voltage drop is proportional to current flowing through it.
That current is proportional to DC current gain (hFE) of the specific transistor for a given Collector working current - the higher the hFE the less the flowing current hence the less the voltage drop on Thevenin's resistance resulting in a more stable Base bias..
(the famus 1/10 rule of thumb..) :icon_wink:

So, as Paul already told you above, you have to know your specific transistor hFE to calculate the exact Emitter voltage and find the respective Collector current..

P.S.
It's not unusual for many people to use the well established, by many textbooks, 4 resistor bias CE amp without taking in mind resistors values ratio - in the mean of they tend to consider that bias method "independed" from hFE and not "immune" to hFE variations only.. :icon_wink:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Godslildrummer

Quote from: willienillie on May 28, 2019, 02:49:18 AM
QuoteR3 - 2.7K ohm from 9v to collector of 2N5089/P1 of C1. Actual measurement is 2.64K ohm from 2.2K ohm and 500 ohm in series.

I think you mean C2, not C1.

QuoteR1 -  P1 5.11v and P2 (pwr) 7.90v

R2 - P1 0.59v and P2 (gnd) 0v

Don't these two resistors meet at Q1b?  They should share a voltage there.

QuoteBattery is...8.93 under load

Where's 7.90V coming from?

Something is hooked up wrong.  Can you can post a couple pics of your breadboard?

I did mean C2 not C1, thanks!

And R2 is measuring the same, I must have bumped another pin while measuring. I checked again and it is the same.

And as far as the 7.90v apparently my battery is draining as it took me over two how's to take these measurements as one of my kids was trying to eat dog food and the other was screaming because I made her wear pants (Mr Mom) :icon_rolleyes: I checked again and battery is at 8.4v no load so that's probably where it came from as it was one of the last ones I took.

I will post pictures later after my phone charges as it is so low I can't take pictures.

Quote from: antonis on May 28, 2019, 04:29:01 AM
Quote from: Godslildrummer on May 28, 2019, 02:19:14 AM
Quote from: PRR on May 27, 2019, 09:27:22 PM
Quote from: antonis on May 27, 2019, 05:13:13 AM....(your bias scheme calls for 10mA Collector current resulting in 27V voltage drop on Collector resistor..

Ah, I see what you did there. Yes, 1/4 of 9V on Base and 150r emitter would be 11 mA. But base current is very significant for these values. How can you guess at a glance? Divide the resistance at the base by hFE, and compare to Re. Quickly: 100k/100 is 1,000r, which is much higher than 150r, so you know the base resistors are not only "significant", they dominate the DC bias. Even if you run the math for 300K||100K and run hFE from 100 to 300, the base resistors dominate. (A super-jellybean with hFE >500, both resistances have equal effect.)
I don't understand how amp draw affects this at all.

You should read about Thevenin's equivalent resistance (which is the parallel combination of voltage divider resistors IN SERIES with transistor Base) and Emitter reflected resistance [which is hFE X Emitter total resistance (Emitter resistor + Emitter intrinsic resistor).
Actual Base voltage is the voltage on divider node minus voltage drop on Thevenin's resistance, which voltage drop is proportional to current flowing through it.
That current is proportional to DC current gain (hFE) of the specific transistor for a given Collector working current - the higher the hFE the less the flowing current hence the less the voltage drop on Thevenin's resistance resulting in a more stable Base bias..
(the famus 1/10 rule of thumb..) :icon_wink:

So, as Paul already told you above, you have to know your specific transistor hFE to calculate the exact Emitter voltage and find the respective Collector current..

P.S.
It's not unusual for many people to use the well established, by many textbooks, 4 resistor bias CE amp without taking in mind resistors values ratio - in the mean of they tend to consider that bias method "independed" from hFE and not "immune" to hFE variations only.. :icon_wink:

Thank you! I will definitely read up on this!

As far as hfe, I'm not sure how you find that. Here is the data sheet if someone can decide it.

https://www-digchip-com.cdn.ampproject.org/v/s/www.digchip.com/datasheets/parts/datasheet/343/2N5089_amp.php?amp_js_v=a2&amp_gsa=1&usqp=mq331AQA#referrer=https%3A%2F%2Fwww.google.com&amp_tf=From%20%251%24s&ampshare=https%3A%2F%2Fwww.digchip.com%2Fdatasheets%2Fparts%2Fdatasheet%2F343%2F2N5089.php

Apparently there is a difference in hFE and hfe? I feel like I won't know enough to continue this conversation until I read up on Thevenin.

Thanks for the help guys!

antonis

You don't need to be familiar with Thevenin theory 'cause you only need to know the equivalent resistance of voltage divider 2 resistors..
(and there is the risk of messing around with Norton theory, too...)  :icon_redface:

ALL hFE's (DC curent gain, beta(β), hfe, hFE, hFE etc) should be considered equal for small signal model..

You can estimate it from the table below or more precisely from the below graph..

"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Danich_ivanov

Quote from: Godslildrummer on May 28, 2019, 02:19:14 AM
Quote from: Danich_ivanov on May 27, 2019, 06:49:16 PM
Collector voltage is lower than the base, which is why it doesn't work. To fix this, that 100k at the base should be somewhere around 30k to get about 1/2 vcc at the collector. 1/2 vcc will give more or less the same waveform at the output - clean boost.

So my measurements for the transistor are Base: 0.62v Collector: 7.34v and Emitter was 0.05v.

IDK enough to know if a 0.05v on the emitter is ok?? But this should be in forward bias right?

Also I went through and double/triple checked resistors and their placement. They are as follows:

R1-300k ohm from 9v to P2 of C1/base of 2N5089/P2 of R2. Actual measurement is 297K ohm from 2-150K ohm in series.

R2 - 100K ohm from P2 of R1/base of 2N5089/P2 of C1 to ground. Actual measurement is 100.4K ohm.

R3 - 2.7K ohm from 9v to collector of 2N5089/P1 of C1. Actual measurement is 2.64K ohm from 2.2K ohm and 500 ohm in series.

R4 - 150 ohm from emitter of 2N5089 to ground. Actual measurement is 144.8 ohm.

R5 - (for bleed) 1M ohm from input/P1 of C1 to ground. Actual measurement is 997K.

Next are my voltage readings:
R1 -  P1 5.11v and P2 (pwr) 7.90v

R2 - P1 0.59v and P2 (gnd) 0v

R3 - P1 (pwr) 7.90v and P2 6.97v

R4 - P1 0.04v  and P2 (gnd) 0v

R5 - P1 0v and P2 0v

Transistor B - 0.62v C - 7.34v E - 0.04v

C1 - P1 (input) 0v and P2 0.62v

C2 - P1 0.04 and P2 0v

My guitar has a resistance of:
Neck 5.91K ohm
Both 3.16K ohm
Bridge 6.54K ohm
(All readings taken at end of cable that goes into pedal input)

Battery is 9.37 no load, 8.93 under load.

No matter what I did I couldn't get a voltage reading from playing my guitar.

As far as placement, everything in in the right place. As I said before it works just not like it should. I don't know enough about this 2N5089 and information is limited where I can find it. I know the pin out is correct as I've both looked it up and tried it. When I reverse the C and E it cuts in and out horribly. All I can assume is it is not being pwered correctly due to my changes. But I don't know why. It seems to work in theory.

Oh, apologies then, i thought you have 27k at the collector, with 2.7k it should be about right. Definitely something somewhere is incorrect, i would simply try another transistor in there.

bluebunny

I'm not sure that replacing a transistor will account for three different voltages being measured at its base, and three different voltages at its collector!  :o
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Ohm's Law - much like Coles Law, but with less cabbage...

Godslildrummer

Ok! So I went to take pictures of the breadboard and I took like 20 of them checking each pin at every angle and there it was. P1 of C2 was connected to the emitter instead of the collector! It works! I had checked it like 10 times but I guess I just needed to one more! Thank you for yalls help!

Now this leads to my next question, is there somewhere I can learn all of this and not piece bits and pieces together from different websites and stuff? Some free (or cheap) course or a good beginner website for pedal theory or a book you recommend. I was so sure I didn't have something right I doubted my theory instead of my application. Like I still don't fully understand hfe and how to read the transistor data sheets. And if I move into a tremelo or delay I'm sure I would be lost.

Any help is much appreciated! Thanks again!