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MFC4040

Started by digi2t, March 30, 2022, 11:55:44 AM

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digi2t

Phil and I are futzing with a USS1 clone, and I was wondering what can replace the the MFC4040 flip flop?

I was thinking that perhaps the 7474 could be up to the task, as it is in the HiFli. I found this design on the net...



but I'm not sure if the reset would have to be pulled high or low in this case? Looking at the MFC4040 datasheet, it seems to me that the trigger is active high, so I'm assuming that the set reset inputs on the 7474 would have to go to the positive rail, instead of the negative as in the image above?

Or, would using a 4013 be a better choice here?

I did find a Canadian supplier with MFC4040's in stock, so building one isn't the issue. We'd just like to be able to build in an alternative scheme, like R.G. did for the Ludwig LFO.
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ElectricDruid

If you've got 5V power, 7474 would probably do. If not, 4013 is probably better since it can run at 9V.

Kevin Mitchell

#2
Somebody say flip-flop? While it's fresh on my mind;
Tom's got the right idea.
Besides the obvious supply voltage advantage, the 4013 is already commonly used for sub-octave pedals like the EH Micro Synth & EQ Bit Commander to name a couple. If it's used simply as a frequency divider I believe it wouldn't matter what polarity the trigger looks for - assuming the trigger signal has been appropriately shaped.

Also, GO MAN GO! After looking at the schematic I'm already giddy about seeing PCBs for something like this.
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Govmnt_Lacky

How many of the flip flops need to be accounted for in the design?

(My question is based on the Google search for the MFC4040 which appears to be a tiny little single flip flop device)
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digi2t

Quote from: Kevin Mitchell on March 30, 2022, 02:00:22 PM
Somebody say flip-flop? While it's fresh on my mind;
Tom's got the right idea.
Besides the obvious supply voltage advantage, the 4013 is already commonly used for sub-octave pedals like the EH Micro Synth & EQ Bit Commander to name a couple. If it's used simply as a frequency divider I believe it wouldn't matter what polarity the trigger looks for - assuming the trigger signal has been appropriately shaped.

Also, GO MAN GO! After looking at the schematic I'm already giddy about seeing PCBs for something like this.

Phil's got the schemo done, and I need to comb through it next. The 4040 was a bit of an "if factor", but we'll figure something out. Hopefully, if it comes together, we'll have a board set in a couple of months.

Quote from: Govmnt_Lacky on March 30, 2022, 03:00:21 PM
How many of the flip flops need to be accounted for in the design?

(My question is based on the Google search for the MFC4040 which appears to be a tiny little single flip flop device)

Just the uno for the sub section.

I think I read that the 4013 is good to 20v supply, so it's a prime candidate here. The 7474 needs 5v, which means adding a zener, as done in the Hi Fli. Hey... one less part is one less part.
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Rob Strand

It's worth checking if the logic on the pins is positive or negative (especially clock, set and reset).   One of the commerical octave units sub'd a 74xx74 for a 4013 and the circuit was completely wrong afterwards.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

anotherjim

74 series TTL controls like Set & Reset and Clock are nearly always active when Low. 4000 series are active when High.
As you wouldn't want a divider flip-flop permanently disabled, the Set & Reset will be tied to the required power rail which for a 4013 will be 0v.
The active clock edge for the flip-flop may not matter purely for audio. You cannot hear the difference in phase.
There are single D-type parts out there in 8pin SOP -  but 74 series only. There's even one I saw once in a 5pin SOP internally configured as a clock divider - you only have the Q output.




ElectricDruid

Quote from: anotherjim on March 31, 2022, 05:05:04 AM
The active clock edge for the flip-flop may not matter purely for audio. You cannot hear the difference in phase.

4013 also has both Q and ~Q outputs, so you get to choose later if necessary.

anotherjim

True you have 2 output phases but some things might be fussy about when the states change compared to some other event. An example is a data latch sampling a data bus, the clock edge is chosen to give the best chance of stable data levels e.g. data changes on +edge then latches on -edge. This automatically gives time for the data bus levels to settle and cover propagation delays.


Processaurus

It is good practice to tie every input on the unused flip flop to a logic level- as I understand, if they float it can change states randomly and cause noise, or be damaged by rising above the power supply voltage w static, etc. Ground or V+, often doesn't matter, ground is fine.

I'd say use it as a 2nd octave down, but it always sounds pretty bad.

digi2t

Yeah, I already knew about tying the unused pins to ground deal, so that will be accounted for when I test. I'll be making daughter boards on vero for the tests so I can just swap them in and see what sticks.
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digi2t

OK... the USS-1 clone is not going down without a fight. Got the waveshaper, phaser, filter, and sample/hold working. But, the envelope is still MIA. As for the subharmonic, I had a bum transistor that flummoxed me for a while, and the MFC4040's that I ordered I don't believe are 4040's, so I resorted to jumpering to a 4013 on the breadboard.

Initially, I wired it up like this...



but I wasn't getting any divided frequency action. Then I tried the set and reset pins high (to +12v), but no dice either. For shits and grins, I connected them to the input, like this;



Lo and behold, divided frequency output appeared. WTF???

On the downside, there is a considerable volume drop at the output of the 4013. I managed to snag a USS-1, and I probed it at the output of the 4040, and there is no such volume drop. Not sure yet what the dillio there is yet. Any ideas?
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Slowpoke101

#12
Have a look at the following PDF. It contains some info on the MCF4040 which may help explain why it is does not behave exactly as a digital divider. It is more of an RTL device than a TTL or CMOS. It is behaves more closely to the divider in the Shin Ei OB-28 Octave Box - which can be a lot of fun when you get it to work.

https://worldradiohistory.com/hd2/IDX-Consumer/Archive-Radio-Electronics-IDX/IDX/70s/1973/Radio-Electronics-1973-04-OCR-Page-0055.pdf
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Rob Strand

#13
QuoteInitially, I wired it up like this...
That first circuit should work.  It has been used in many circuits.

My suspicions would be:
- the input clock is glitching
- there is ground bounce on the 4013
- the heavy load on the output of the 4013 is upsetting the way the 4013 works.
   (perhaps a quirk of the brand of 4013 you have).

I'd be thinking of perhaps putting a resistor (say 1k to 100k) between the transistor flip-flip on the USS-1 and the 4013 clock.

QuoteLo and behold, divided frequency output appeared. WTF???
WTF indeed.

QuoteI probed it at the output of the 4040, and there is no such volume drop. Not sure yet what the dillio there is yet. Any ideas?
Well, I can elaborate on how the 4040 works:
- the clock is on negative edge
- the Q output changes state when the clock falls to about 0.7V
- Slowpoke101's link shows the internal circuit.  You can see
   it presents a resistive load.   I have a suspicion the resistor
   connections around the first transistor provide some hysteresis
   to prevent the clock from glitching.  (not confirmed)
- The output is a 2k pullup resistor and hard transistor pull-down.

Compare this to the 4013:
- the clock is positive edge.  I don't think that's an issue in this circuit.
- the threshold will be a higher voltage.   That doesn't seem to be an issue either.
- At 9V, the CMOS output impedance pulling up is usually 1k ohm or less.
  So it's weird the CMOS gate output is pulling down.

Check the power and ground pin connections to your 4013.
See if you can probe the 4013 power pin with an oscilloscope.

Try another 4013, perhaps a different brand or batch.

If you are desperate you can put a pull-up resistor on the output of the 4013 to provide more
current drive.  Don't over do it as the output low voltage will creep up and at some point
that will cause problems.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

anotherjim

A slow switching edge to the clock pin can indeed screw up the 4013 divider. As it's buffered logic, slow input can cause jittering multiple clock edges when there should be only one.  Ideally, a Schmitt trigger should be used to tidy up the clock. If you only need one of the flip-flops, I have a hack to make a Schmitt trigger with the unused one.

Rob Strand

#15
This is my what I'm taking as the schematic,



I haven't simulated the rise and fall times but the clock is at least coming from a "digital' circuit.


A quirk of the transistor circuit compared to say the boss flip-flip is the trigger is only going to one
of the bases.   Maybe someone can check it.  I'm at the end of a day with 3 or 4 hrs sleep.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

digi2t

Quote from: Rob Strand on August 17, 2022, 03:26:04 AM

I'd be thinking of perhaps putting a resistor (say 1k to 100k) between the transistor flip-flip on the USS-1 and the 4013 clock.

I'll throw a trimmer in there tonight and give it a go.

Quote
If you are desperate you can put a pull-up resistor on the output of the 4013 to provide more
current drive.  Don't over do it as the output low voltage will creep up and at some point
that will cause problems.

Any recommendations on the resistor value? Or, do I start with a 1M trimmer and slowly wind downwards?
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Digital Larry

Quote from: ElectricDruid on March 31, 2022, 07:28:45 PM
Quote from: anotherjim on March 31, 2022, 05:05:04 AM
The active clock edge for the flip-flop may not matter purely for audio. You cannot hear the difference in phase.

4013 also has both Q and ~Q outputs, so you get to choose later if necessary.
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Rob Strand

#18
QuoteAny recommendations on the resistor value? Or, do I start with a 1M trimmer and slowly wind downwards?
After looking at the circuit I don't think it's going to help.  I think there's something wrong in the circuit.

Some basic ideas:
- If R145 isn't pulling up then perhaps the simple diode "VCA" D14/D15 isn't letting enough signal through.
- If there are incorrect parts in the filter R174 and beyond it could be filtering so much there is not signal left.

Possible areas for VCA Issues
- Circuit around Q36: Q36 not turning on, or not turning on Q37
- Circuit around Q37: wrong part values, C36 upside down, general connection issues.

If you have a real unit then try to find where the clone and the original deviate.
- Compare the voltages around Q36 and Q37.
- Compare the levels at each point through the filter R147

If you are still stuck try lifting the C36 side of R145 and connect it to V2.
That should let full signal through D14, D15.

That D14/D15 VCA is very rough and ready as the input signal from the MC4040 is very large for such a circuit. (Probably works OK because we are using square waves.)


Back to the 4013:  One thing that stood out to me this morning is R167.   That's going to reduce the output swing at the collectors Q44.   At any one time either Q43 or Q44 is on that means either R168 or R171 are pulled to ground and that means that R167 (1K) is always dividing the supply voltage at the top of R168/R171.  The output will only be 0.69 times the supply voltage and maybe that's upsetting the 4013 when noise is present, making it glitch easier.   Normally you might expect the 4013 to work under these condition but maybe not in practice.   So what you could do is drop the value of R167 to say 220 ohm.   Make sure you are still getting signal on the collector of Q44.  The MCF4040 doesn't have this issue since the switching threshold is about 0.7V.

FYI for others:

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

digi2t

Well...

I was doing more research on 4013 dividers, and lo and behold, I came across Fredrik's Parasit Studio site. He has a page covering just this stuff. Funny enough, in his experiment project, he states that there should be a 10k resistor between the 2 and 5 pins. I found this curious, since every other schemo I've looked at to date shows a direct connection. Anywho, I connected pins 4 and 6 to ground, as they should be, and replaced the 2 and 5 jumper with a 10k resistor.

CAP'TIN... THERE BE SUB-HARMONIC HERE!!

It's not only working, but it's also copping the same performance of the original. So, consider this issue resolved. But, I would like to know what exactly that 10k resistor is doing. Knocking back the second output signal enough to keep it from glitching?

So now, the only outstanding thing is the envelope. Right now, when I turn it on, audio goes silent. I'm checking voltages now, and I see that the pins that should be flipping between 0v and 12v with transients, are not. This in turn, from the way I see the schemo, isn't sending any current to the 3080, so no attack and decay action. Already found one bum transistor in the sub-harmonic section, combing through this section now.

I'm sooooo close...
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