Would like to build a CS-3 in a 1590a

[dean owens]

I have a 1590a enclosure and if I could stuff a CS-3 (I'd even be fine with a CS-2) into it then I could fit my TU-2 on my pedalboard.

Here's my background.  Years ago I took electronics in high school and have tinkered with/fixed various electronics things since then.  About 10 years ago that turned to building my own tube amps and that's the world I've been in.  I can read schematics and know how to solder.

Here's where I'm lost.  I'm just not used to solid state stuff.  I was going to wet my whistle by doing the Tap Tempo Tremolo from here but I'm waiting on Taylor to get some more boards in.  So I've never build a pedal before.  For me, the schematics are drawn a little different and the signal doesn't always seem to flow the way I'd think.

So, a one or two of you kind people take someone under your wing?  I'm not looking for someone else to do my job.  I'm looking for someone who will answer some questions as I have them... recognizing that some of them could be dumb    So will one or two of you help me out?  Here's where I'm starting.

I'm looking over a schematic I found here and I'm having a hard time following it.


First, I'm guessing everything between D7 & D8 is the buffer.  I'm not against buffers but if I can cut it out that's less I have to put in the chassis.
- Assuming it's the buffer, what can I leave out?
- What would the centers of Q4 and Q5 attach to?
- If I put a 3PDT switch in it's place I know I'll basically bypass everything from R1 to R31.  I just need to know what to leave out.

Second, I'm used to seeing triangles as grounds.  I'm thinking in this schem that the triangles are the positive connections and the single horizontal line is ground.
- That being the case, do you run a single ground buss and positive buss that every connection runs to?
- Is the ground grounded to the chassis at all?  Obviously it is if I use switchcraft type jacks but what if I use isolation jacks?

I'm curious when it comes to my layout.  With tube amps there is a logical way you should go that will keep the signal flowing one way and keep too many wires from crossing to keep noise down.
- How important is layout in solid state stuff?
- Even if I get everything hooked up where it should be, do I risk having a noisy pedal if I don't lay it out right?

For anyone who has made it this far and is willing to help me out, thank you.  I appreciate it.  I hope there's one or two of you that'll help me out.

[ashcat_lt]

IC1a is the input buffer.  Q1 is the output buffer.  All that stuff above Q5 is the flip-flop switching circuitry.  If you're going with latching switching I think you can just cut from right before C14 to right before C15.  Connect the wiper of the Level pot to C15 without the transistor in the way.

You're correct about the horizontal line being ground.  Somebody else will probably have more info re: routing of power and signals, but I don't think it's usually quite so critical in low voltage pedal design as in amplifiers.  You will want the metal chassis to connect to ground somehow for shielding purposes.

[R O Tiree]

The bit between D7 and D8 is a flip-flop which not only lights the status LED (from D9 to GND), but also switches the FETs (Q4 and Q5) to route the "dry" and "effect" signals around.  The diodes at their gates are kind of counter-intuitive, because they look reversed... that's intentional.  The FETs would switch in nano-seconds, if they were allowed to, and the reverse leakage current through the diode and the 1M resistor is enough to soften the switching time so there are no unpleasant pops.  At least, that's my understanding of it.  You'll see this arrangement in almost every Boss pedal (I've never seen one that doesn't have it, but that doesn't mean there isn't one) and also Ibanez.

IC1a is the input buffer.  Unity-gain, non-inverting.  The back to back diodes are there to limit large input signals from overloading the rest of the circuit.

IC1b sets a pretty much rigid "virtual ground" of 4.5V.  R2 and R28 divide the 9V-GND in half, but it wouldn't be very rigid.  Feeding that voltage into the +ve input of the opamp, and then connecting output to the -ve input means that the voltage at the output is absolutely stable.  You'll see little black up-arrows with "4.5V" printed next to them all over this diagram and, as you correctly assumed, they are all connected to that source at the output of IC1b.  You're also correct in your assumption that the GND connections are those little horizontal bars.  Non-standard for the 21st century, but still quite common when this circuit was designed.

OK, as I said, that circuitry around Q2 and Q3 is a flip-flop.  When Q3's collector (top pin in this diagram, denoted as Q3C hereafter) is "high", the Gate of Q5 is also high.  This means that Q5 is "turned on" really hard, so the resistance of the Drain-Source channel is very small indeed, so the buffered input signal travels straight through to the emitter-follower output buffer (Q1).  In addition, Q3C is almost at 9V potential, so no current can flow through the LED>Zener (D9)>Q3collector>GND, so the LED is off.  Lastly, Q2C is in the opposite state to Q3C, so it's low - almost GND potential.  That means that Q4's Gate (Q4G) is also low compared with its Source (Q4S) so the D-S resistance is insanely high... therefore, no signal can appear at the output from the "compression" portion of the circuit.  The signal from pin1 of IC1a is stripped of its DC bias by C14 and then those two 1M resistors (R25 and R26) apply an identical 4.5V bias to both the Drain and Source of Q5 - the only variable is now the AC signal, so that's DC "pops" when Q5 is switched taken care of.

Neat but not gaudy, so far, then.

Moving onwards, let's flip-flop it...  Q3C goes low, so the LED turns on.  Q5G also goes low, so it cuts off the direct path from the input buffer to the output buffer.  Q4G goes high, so your compressed signal gets through to the output buffer.  Time to look at the rest of the circuit, then...

First, a few words about the THAT2159... aaaaaaannnd the phone just went and I have been called in to work, so I'll pick this up later, unless someone else explains the rest of this in the meantime.

Cheers.

[R O Tiree]

OK, where was I?  The THAT2159.  I can't find a datasheet for it, but I'm assuming that it's part of the THAT2150 family of Voltage Controlled Amplifiers (the pinout seems to confirm this assumption).  It takes a current in at pin 1, spits a current out at pin 8 and the gain is controlled by the voltage differential between pins 2 and 3.  If pin 3 is higher than pin 2, the gain cuts back and, if it's lower, it boosts.  It was designed to operate around GND, with equal and opposite polarity supply rails at up to +18V at pin 7 and -18V at pin 5.  Can't do that in a 9V pedal (without using a charge-pump, for example, or 2 batteries...) so that's why they've put that little circuit fragment around IC1b to supply a really stable 4.5V reference voltage (mid-way between +9V and GND).  Because of the circuitry around it, the 2159 thinks that 4.5V is actually GND, the +9V supply is really +4.5V and true GND looks like -4.5V.   Even though the input and output are currents, this family of VCAs cannot handle much of it without the distortion products becoming unacceptable (?) hence the input buffer with its back-to-back diodes to limit the max signal going through the 33k into pin 1 of the THAT 2159.  Add to that the odd behaviour of this chip... Look at IC2a.  There's no input resistor!  The neat trick here is that it's actually the 33k R9 that is acting as IC2a's input resistor - bizarre.  Effectively, the 2159 "secretly" alters the current out and is transparent to the input of IC2a.  It's all done with (current) mirrors!  So, I think you're stuck with the requirement for an input buffer of a similar design, for all those reasons - input current limiting, guaranteed input impedance and input signal level limiting.

As to the output buffer... You might be able to lose it, I guess.  It was almost certainly included to tidy up the signals from Q4 and Q5.

Now the rest of the circuit, because you mentioned that solid-state is not your bag - IC2b is a neat little trick that leaves bass freqs unchanged, but varies the gain of treble freqs from 0.3 to 3 .  Swamping of treble freqs in these simple, all-pass compressors is always a problem, and this is how Boss tried to sort it out.  The next bit is elegant and simple, once you get your head around it... Look at Q8/R14/D5 - it's a half-wave rectifier, essentially.  Every time the signal at the base of Q8 gets above about 0.6V, it turns the transistor on and it will pass current to ground.  IC4a is an inverting 1:1 buffer and the same trick is played with Q6/R18/D4 on the "other half" of the waveform.  OK, 9V feeds, via R2 and the 250k "Attack" VR to fill up C8 (4µ7F).  So, with very small signals, Q6 and Q8 don't get turned on and C8 sits at a high voltage.  This keeps the Drain-Source resistance of Q7 very low in comparison to R13 (10k), so the junction of R11 and R13 is sitting at very nearly 9V.  The other end of R11 is connected to pin 2 of the 2159 and the 1k R10 biased at its far end at 4.5V.  It can be seen that this is a 1:15 voltage divider between 4.5V and 9V, so pin 2 sits very slightly higher than pin 3.  Remember earlier that I said that if pin 3 is lower than pin 2 the gain goes up?  It turns out that 122mV difference gives about 20dB gain, so 300mV is going to give even more.  It's logarithmic, and I can't be @rsed to work it out... sorry.  But it's a lot!  What about if we go the other way?  Suddenly hit this thing with a large input signal, round the circuit we go and Q6 and Q8 begin to conduct on every half-cycle each, so the charge in C8 dribbles away to GND.  The voltage at Q7G will drop and the Drain-Source resistance goes up to a very high value.  The bottom end of R13 is now at very nearly 0V.  R11 and R10 now act as a 1:15 voltage divider between 4.5V and GND, so pin 2 of the 2159 is sitting very slightly lower than 4.5V... so pin 3 is about 300mV higher than pin 2 and the gain drops.  Equilibrium will be found very quickly and, as the input signal dies away, the gain will gradually ramp up again (C8 filling up under the control of R2 and the "Attack" pot, which is not really "Attack" at all, but "Recovery" - lots of compressors mis-label this control).  Finally, IC4a and the "Sustain" control.  You can see that this works like a variable version of the stable voltage reference I mentioned in the previous post.  Wind the Sustain pot up towards the 9V end and things work as I've just described.  Wind it down towards the bottom end and you introduce a limit to how far up C8 can charge.  The diode at D6 ensures that the voltage at the top of C8 can never go below 4.5V, so you'll always get a bit of compression.

To finish answering your second question, the following explanation is to be found in probably 99% of stompbox pedals.  What you want to avoid is draining the battery when you're not using the pedal.  Somehow, you need to break the circuit.  The output jack socket is a standard mono one, usually.  The input socket is usually a stereo one.  Funny, that, because guitar jack plugs only have a tip and a long sleeve - no ring terminal.  OK, here's the trick - If you look at the bottom right corner of the schematic, you'll see that battery -ve is connected to "A".  Look up in the top right of the diagram and you'll see that the "ring" terminal of the socket is also connected to "A" and the "sleeve" terminal is connected to all the other ground points.  So, with no jack plug connected to the input, the battery "A" terminal on that diagram has no physical connection to the other ground points.  Plug your guitar in, and suddenly the long sleeve on the plug connects "A" to "GND", the circuit is complete and the pedal starts working.  Neat, hey?  So, isolation jacks will work, but you'll need to star-ground everything from the sleeve of the input socket and connect batt -ve to the ring terminal.

As to layout, it's nowhere near as critical as it is with tube amps, but it makes sense to have nice, fat traces for GND and +9V, and try to make the circuit flow linearly.  A neat feature of the 215X family is that it automatically cuts the output when the input signal is 0, so any mush is a result of the rest of the circuit, not the chip.

[dean owens]

wow.  Thanks.  that's a lot to read.  thank you for teaching me.  i'll read over it a few times.  if i have any questions i will be back.  before i do anything i'll draw up my own schematic without the buffer and flip flop switch and my layout.  i'll post them here to have you guys make sure i haven't missed anything.

again, thank you.  i greatly appreciate you two taking the time to teach me.

[R O Tiree]

No problem, Dean.  Post back if you have more questions and I'm sure there'll be a very fast answer.

[Heliarc]

I just wanted to register to say that explanation was fn awesome R O Tiree ! I know nothing about circuits but I do make music, and find this stuff fascinating. Maybe I'll get around to piecing a box together one day. I'd like to have a few diy modular units for FX to run my synth through, Lord know I have enough junk electronics boxed up. I just got a boss ds-1 40th anniversary in the mail yesterday and it's great. Guitar sounds better than the synth thru it though 😄