Boss TR2 (only a little problem for Profi's)

[jjj]

Now, when I press the tremolo effect pedal, it goes on; than I have to press the pedal again to turn it off.
I want that it only triggers the tremolo effect for as long as I press down the pedal. How to override that flip-flop?

[Mark Hammer]

It can be done, but not in any truly easy way, simply because of the way that switching is done.

You can get a momentary DPDT switch from Small Bear ( http://www.smallbearelec.com/Detail.bok?no=25 ) that you could wire to either bypass the whole pedal or to accomplish what is normally done within the circuit via the FETs.

It might also be possible to defeat the LFO via a momentary switch, but unfortunately I don't know precisely how to do that.

[jjj]

THX for your kind reply...
From what I gather: the upper FET bridges the effect and the lower FET turns the effect on. That means the upper FET has to be off, while the lower FET is on and visa-versa. (?) Am I ...guessing right?

What I thought of doing is cutting the leads from the flip flop to the two gating FET's. The gates are then pulled up to +9 volts with 1 M resistors, and the new pedal switch would alternately pull one gate to circuit ground. I hoped there would be Profi around, who could tell me how to connect the inbuilt pushbutton, but it I think the FET's need to be switched over.
Here's my altered diagram...

[Enzo]

File not found

The pullup resistors turn on the JFETs. You need to have the control for one inverted from the other. Look at Q7, the right side of the flip flop. It is hard to read here, but it looks like R33 and C37 connect the collector of Q7 down to the base of Q6, the left side flip flop transistor. Remove those parts. Also remove C14 and C22. Put a piece of wire in place of C14. Now the switch directly controls the base of Q6.

I leave it to you to tweak the resistor values around the switch and the base of Q6. And the switch will be normally closed or normally open.

When the switch turns on Q6, which grounds the collector and removes voltage from the gate of JFET Q3 - the bypass JFET. At the same time that grounded collector turns off Q7 which allows the pullup resistor to apply voltage to the Gate of Q1 - the effect JFET. Release the switch, and the two transistors change their state to the opposite, so the bypass JFET turns on and hte effects one off.

Basically I allowed the flip flop parts to remain, but removed the cross connection, and made the switch direct acting.

[jjj]

Hi Enzo... angel of electronic salvation!
Jeez, I would enjoy that mission, too. I do my best helping others philosophically.

I renewed the Pic in the previous post, but I think your solution is much better. (No need to change the SW)
What I really want is that the tremolo effect is only on for as long I push the pushbutton switch.
Here is a clearer Pic of the section. I modified the circuit according to your description. Maybe I should leave C16 connected to the base of Q5, instead of the wire?
Besides, my electronics theory is pretty bad or unable to follow your description. I tried hard... but here's how I interpreted it; please put me right:
When the switch turns on the base of Q5 receives a negative voltage, grounds its collector
and this removes gate voltage from Q3 (bypass JFET) and turns off Q3.
At the same time the collector of Q5 grounds the base of Q7 and its collector and this removes gate voltage from Q3 (effect JFET) and turns off Q1.
That means both JFET's will be off!!?? My god! Then I must have misunderstood it or modified it wrongly?
http://img65.imageshack.us/img65/3983/originaldc2.png

[Enzo]

The left one is Q5, not Q6? OK, I couldn;t read it well.

When the switch is closed, how would a negative voltage get on the base of Q5? There is no negative voltage in the whole pedal.

First, JFETs are by their nature ON until voltage at the gate turns them OFF. You can verify this with an ohm meter. In fact that is how I identify their gates. I lay the part on the bench and measure resistance between the three legs. Two will show low resistance (100-200 ohms or something) and the third will seem isolated. That isolated one is the gate. it will act like a diode towards either of the other two legs.

Second, you need to understand how the bipolar transistor circuit works too. The transistor - Q5 for example - sits there with a positive voltage at its collector through R23. If the base is grounded, then the transistor is off - Q5 acts as if it were not there. So the positive voltage through R23 is there. That positive voltage can flow through R22 and R44 to other places. Turning off Q5 thus doesn't remove those voltages since they don't come up through the transistor, they are present because of the resistors. Now when we apply a positive current to the base of Q5, then it conducts and grounds off its collector. Now the lower end of R23 is grounded through the Q5. so there is no voltage on the lower end of R23. And there will be no voltage flowing through R22 or R44 either.

In my suggested circuit, if you close the switch, the base of Q5 becomes grounded.

By the way, look on the circuit board. Are diodes D1, D2 drawn backwards? All the circuits work to positive, and how would a positive current flow through the diodes as drawn. But lets ignore the diodes for now.

When base of Q5 is grounded, the transistor turns off. That allows the voltage through R23, R22 to get to the gate of Q3. That positive voltage at the collector of Q5 now also runs throughu R44 to the base of Q7, which turns it on. A turned on Q7 thus grounds its own collector. That action grounds the lower end of R24, thus killing any positive voltage there, so no such voltage gets to the gate of Q1. Remember the voltage at the gate turns a JFET OFF.

SO with the switch closed, Q5 is off so Q3 stays on. (There is no turn off voltage at its gate) The bypass route is open. Also since Q5 off makes Q7 turn on, Q7 on turns Q1 off. And the effect route is now blocked. The switch open reverses the two conditions. So switch open turns on the effect.

If you want to step on the switch for effect to happen, then either wire up the switch for normally closed, step to open. Or reverse which side of the flip flop to control. In other words instead of removing R33, we'd remove R44, and we'd conect to switch to the base of Q7 instead of Q5.

The two caps C16, C22 were there to provide a trigger pulse to the flip flop. But we disabled the flip flop to make it a stable inverter. The switch needs to be wired directly to provide continuous control. In my original idea, I replaced ONLY C16 with a wire, C22 is removed and nothing in its place. (In the previous paragraph, if we changed sides, then C22 only wou0ld get the wire and C16 wouold be vacant.)

[jjj]

Thanking you for the lesson!!
Jeez, I spend hours to search for transistorized flip-flop circuits to learn more about it. I'll copy your good advice and peruse it thoughtfully. You are a great electronics teacher... I'm quite interested in these things. I called the negative polarity or ground "negative voltage". Also, I totally confused the operation of transistors and FET's. Apart from the 4066 accordion bass, I didn't do much electronic tinkering the last years. Computer music robbed me of my hobby, but I have no problems to solder and wire things together according to a schematic. There I gained quite a bit of practice.

So, the circuit alterations I displayed in the last circuit are OK?
How about the C16 (220p)? Should I just bridge it or leave it in the circuit?

I now have to check the puschbutton how it switches...

[Enzo]

I don't know if the resistor values are the best around the switch and the base of the Q5, I would have to have the thing in front of me. I offer my suggested mod as a concept only, and not a finished process. You may have to make some alterations.

The switch circuit was isiolated from the flip flop by the two caps. We remove the caps and replace the one with a wire. In my example it was the cap on the left C16. Your last drawing already shows C16 as bypassed with the red wire.

[jjj]

I don't know if the resistor values are the best around the switch and the base of the Q5, I would have to have the thing in front of me. I offer my suggested mod as a concept only, and not a finished process. You may have to make some alterations.

It might even work as is, because that's how it worked as flip-flop, too.
The trouble is, a year ago I packed all my electronics parts away, since we thought we had a buyer for our house... but then the sale failed and now I'm stuck and don't know what is where; a case of Murphy's law!
I only roughly scribbled the box number and contents onto paper. Yet, the solder, iron, tools and multimeter are accessible.

The switch circuit was isolated from the flip flop by the two caps. We remove the caps and replace the one with a wire. In my example it was the cap on the left C16. Your last drawing already shows C16 as bypassed with the red wire.

OK, I'll try to do that today and let you know it worked out.
First of all I have to check the pushbutton if it's N/O or N/off type.

I gratefully appreciate your advice. "Dios te paga", (God will repay you) they say here and I do believe in good Karma, don't you?

[jjj]

The PB-Sw is NO-off !! So I reverse the story.


All done, but, now the effect runs continuously; even without pressing the PB and no change when I press the PB.
The LED doesn't light up.
Voltages= Diodes (cathode to ground) D3= 0V1 // D2= 2V4 ///// Base of Q7 has 0V35 // Base of Q5 has 0V6
I want the effect to be only on when I press the (N-off) PB; if PB not pressed bypass should be on and effect off.
I found a 500k and a 100K Pot... and some old PCB, where I can harvest resistors etc.
Besides, I found a BOSS error: R25 appear twice (one (1M) near P1 bottom right and (56K) near collector Q7).

[Enzo]

Why did you add the extra blue wire to the LED? The circuit should be OK without it.

[jjj]

Because I reversed the lot. Before the LED was on the other side, but this side is now dead. You think it might somehow be the cause of the problem? Is my brief description of PB on/off correct?
The problem now is that I cannot turn it off... Maybe the voltage has to be increased or R25 (now 56K) reduced?

[Enzo]

Take that off. All I have been doing here is making your flip flop into a basic switch. The LED ALREADY was following the JFET action. Leave the LED alone and it should be what you want it to be. The only thing we are reversing now is wwhich way the circuit operates when the switch is idle and when it is pushed. That doesn;t change the relationship between the JFETs and the LED.

The LED should stay connected to the collector of Q5.

Before we reversed it, the only problem we should have had was the thing was on until we pushed the switch, and then it went off. That is reverse of what you want, but it would prove the circuit worked, right? Then we move the control to the other side to reverse the switch action.

We haven't changed the circuit or voltages from before modification, when it worked. WHy should we think those voltages are wrong now? Only thing we changed is the connection of the switch to the base of Q5.


Trooubleshoot it, don't start guessing. COnnect the switch to the base of Q7 like the drawing. Doesn;t matter at this point if the switch works backwards. Monitor the voltage at the base of Q7, and click the switch off and on. Does the voltage at that base move up and down with the switch action? This is what I mean about tweaking the resistors, maybe R25 would be better as 100k. But we need to see what is working. If that base voltage doesn;t budge, wwell then nothing else will work either.

What voltage is at the collector of Q7? WHen the base of Q7 is grounded, the collector should rise to some positive vooltage. When the base is pulled up above half a volt or so, the collector should drop to about nothing. Until we get that much, nothing further will work.

Oh you can verify things. If the collector of Q7 is +V, then the collector of Q5 should be about zero. If you ground the collector of Q7, then the collector of Q5 should rise to +V. That verifies the inverting action.

[jjj]

The LED should stay connected to the collector of Q5.

Done that! Now the LED burns again OK. See circuit:

Monitor the voltage at the base of Q7, and click the switch off and on. Does the voltage at that base move up and down with the switch action?

Yes! SW not activated: Base of Q7= 0V35 // SW activated= 0V

What voltage is at the collector of Q7? When the base of Q7 is grounded, the collector should rise to some positive voltage. When the base is pulled up above half a volt or so, the collector should drop to about nothing.

SW activated and deactivated: Collector of Q7= both states invariably show 4V38
When the PB is not activated ground should kill the +V, but it does not!!

Verify: If the collector of Q7 is +V, then the collector of Q5 should be about zero. If you ground the collector of Q7, then the collector of Q5 should rise to +V. That verifies the inverting action.

Collector of Q5= both states invariably show 0V4

My conclusion: it must be that the +V at Q7 is too low to (when PB is not activated). The changes we made now require more +V at the base of Q7, when the PB is not activated.
So, I suppose lowering R25 (1M) to 100K or so might make it work? I have got a 300K Pot. I can solder it parallel with R25 (1M) and then turn it slowly 2/3rd down or have I not learned a thing from you great electronics teacher?

[jjj]

Yep, it was that! Now it works fine. The moment I swapped R26 (1M) for a 500K Pot the tremolo the bypass turned on and the effect off and when I pressed the pedal the tremolo came in. Great... but as usual one never gets enough of good things:
Now I wished the tremolo would not come in that sudden, but gently or with a gentle volume rise/ delay. Like that: <
Maybe that can be done by just adding a little Cap or Electro and an empty up resistor on the right spot? Do you think that is possible? Where would be the best spot? (In the effect path, between R24 and Q3 ??)