CMOS Power Supply (Split/Bipolar Supply)

[karczilla]

Hi all, I have some questions for you lovely mad scientists.

I plan to begin a large CMOS project I planned a year or so ago but life took over and never started it.

From what I've seen online and read in CMOS Cookbook, 4000 series operates at 3-15V (given a reference GND of 0V).

I'd like my lunetta modules to run on a split/bipolar supply -5/5+ where -5 would be the CMOS circuit's GND (and logic 0/low) and 5+ would be VDD (and logic 1/high). So -5 and 5+ would be 10V peak to peak, under the constraints. (powering CMOS at VSS=-5, VDD=5+)

I'd just like to confirm that split supply is okay for the 4000 series chips.



This would allow me to interface with bipolar signals (from other synths) and also allow any output to be used as an input (audio, logic, CV, etc). Also opamps that require bipolar supply would work out of the box (do I need headroom for the op amps, should I also create a -9/9+V bus for op amps?).

I can use this circuit (sorry at work and can't upload things)
https://www.amazon.com/reader/0750699434?_encoding=UTF8&query=split%20supply (page 44)
to protect the inputs to only -5/5+.

It would clamp the signal, but I can run it through a simple attenuater (pot voltage divider) before input to CMOS module to prevent clipping the signal, if necessary. AC coupling the input would be smart, would the voltage drop remove too much of the signal to trigger a high logic though? (especially if i put limit/mix resisters on the outputs and/or inputs?).

Plan is to use a circuit based on the MFOS Wall Wart PSU with a 5-15V AC 1A power adapter (LM7805/LM7905 max voltage is 18, max input current is 1.2A).

Any issues or suggestions regarding my thinking and planning?

Thanks much for any help provided!

[PHOBoS]

I'd say yes, a bipolar supply should work fine as long as you stay below the maximum supply voltage. If you are going to use a supply based
on the MFOS wallwart one I wouldn't go any higher than 9V AC in (6V seems to be ideal) because of heat dissipation.

Headroom for opamps depends on the the signals you want to process. If you want to process the +/-5V signals then yes, you'd be much
better of by powering the opamps with a higher voltage.

I don't have the CMOS cookbook (although I might actually have a PDF somewhere) and the preview doesn't show me page 44 so
I can't see what circuit it is. But a voltage drop caused by a diode shouldn't be a problem and because CMOS chips are high impedance
putting resistors in series with in-/outputs usually doesn't cause any problems either.

As for AC coupling, well,.. CMOS is pretty much all digital so you are working with squarewaves and in that case I would recommend
using comparators for digital inputs. If you make the treshold level variable you can use all kinds of voltage levels and signals. If you do want
to use an analog signal (e.g. for CV in of a CD4046) then AC coupling can be used.

[karczilla]

This is what I meant by protection by diodes clamping the signal if more than -5/5+ signal (found you posted this after i posted this question).

This is what the CMOS Cookbook basically illustrated on how to interface CMOS with signals higher/lower than the CMOS circuit VDD/VSS voltages.


For the op amps- I think i'll have to create another bus that runs higher than the CMOS chips.

For AC coupling. I mean removing the DC bias so all signals are normalized around the reference ground (0V). I can assume that most of my signals in the lunetta will be like this, but basically I want to be able to patch into MFOS bipolar gear I'm finishing/will start (which will all be +/-12V).

[blue hell]

[karczilla]

Thanks for the helpful information guys.

I guess +/-5V for CMOS and +/-9V for OpAmps will be my target. Patching out to other gear, the signals may need to be amplified, that's fine.

Now that being decided, what would be the best way to create this quad power circuit? (+/-9V, +/-5V, GND)

The MFOS WallWart circuit calls for the regulator to match the AC adapter (12V AC for
+/-12V output), but the data sheets specify
"The input voltage must remain typically 2.0V above the output voltage even during the low point on the input ripple voltage."
https://www.fairchildsemi.com/datasheets/LM/LM7805.pdf
What do you recommend?

BUT-
PhoBos commented that the regulator could heat up a lot given a higher voltage (i would attach heat sinks anyway..but idea below to lower heat disipated from the 5V regulators...):

If i use 12V AC and regulate down to +/-9V and +/-5V...

...to do this, can I take the input directly from voltages present on rectified AC signal after the large buffer caps to create all 4 regulated voltages?

Or should I regulate to +/-9V and THEN regulate down off those two +/-9V outputs to +/-5V?

[PHOBoS]

For total heat dissipation I don't think it would matter a whole lot how you do it. When regulating all voltages from the rectified voltage
the 5V regulators get warmer, when regulating 5V from 9V the 9V regulators will heat up more. For stability and to seperate the digital
noise from the opamps I would say use them all seperate, so from the rectified voltage. This might also be somewhat more efficient.

As for the voltage, 12V AC would give you around 16.3V rectified which is a bit high. It wouldn't damage anything but as you know a lower
voltage means less waste in heat. 9V AC would give you around 12.0V rectified which is pretty much perfect.

[karczilla]

Thanks for the help guys.

PHobos, I was going to ask where you got 16.3V from but decided to go back and reread Ray's project documentation and of course he explained the whole reasoning

[PHOBoS]

yep, Vpp = Vrms * √2 and because it is single wave rectified you need to subtract the voltage drop caused by 1 diode, which is about 0.7V.
You can find more info on RMS voltage here.

note: If it was double wave rectified you would have to subtract the voltage drop caused by 2 diodes. Another difference is that single
wave rectification leaves a larger ripple voltage so you have to use larger capacitors than with double wave rectification.

[karczilla]

Hi, I ask someone to please take a look at what I am thinking for my dual bipolar power supply. It is attached below.

Any comments/suggestion? Power supply is critical and I don't want it failing/shorting other modules I spend hours on. That would suck


Are the two sets of 3300uF buffer caps too much? Should I only have one set and regulate the two bipolar supplies off that one set?

Are resistors needed for my power bus? I guess the diodes will protect the regulator, but want to be sure.

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[karczilla]

v2 with one set of large buffer caps

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[PHOBoS]

One set of capacitors should be more than enough. Especially since you are mostly working with digital chips which hardly use any current.
Use some smaller caps (say 10uF) for seperate circuits and a 100nF per chip and you should be good. I wouldn't use any resistors for the power
rail but if you want to use those to force inputs low it should work. However, you might want to use those GND connections for something else too
and then it would be more useful without resistors. As for the regulators, as long as you connect them right (keep in mind that the negative ones
have a different pinout and don't connect them to 1 heatsink or at least not without insulation) it would be very hard to destroy them.

[karczilla]

Thanks PhoBos!

[karczilla]

PhoBos, how can I protect my modules if the AC adapter is opposite of what it should be? If the center is negative (rather than positive as it should be, and normally is)..the GND bus would become an unregulated AC signal and if some other gear was plugged into my common ground bus in order to patch in/out of the lunetta, would it blow?

[PHOBoS]

With a single AC signal there is no positive or negative (unless you are looking at something like mains voltage where there is a neutral
that is actually connected to earth at the powerplant or at least at some point). All that happens is that the polarity alternates between them
but it doesn't matter which one you use as GND. And you do need an actual AC adaptor/wallwart to work with this kind of bipolar PSU, not one
that gives out a DC voltage. So it is just a transformer without rectification.

What you do have to watch out for is that if you have a metal case and the AC input is not isolated from it that you use the connection that makes
contact with the case as GND.

[PHOBoS]

Here's a photo that might help you out a bit. As you can see they are both labeled as AC adaptor. But the left one actually has an AC output
(which is what you need) while the right one is DC. You will also notice that the left one doesn't have an image of the connector on.

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[karczilla]

I understand that I need to use an AC-AC adapter, I was just worried about the plug being connected backwards



But now thinking about it and your explanation, i guess it wouldn't matter with AC.

Thanks for the help [/img]

[karczilla]

here is the final working design and build (minus more power bus connectors to add)

all files https://github.com/ckarcz/Modular-Lunetta/tree/master/L000-Power

more to come this winter

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[Grumble]

I would place a resistor between cathode D2 and the + of the three large caps C1, C2 and C3 and a resistor between anode D1 and the - of C4, C5 and C6.
These resistors should have an impedance of just a few ohms and their purpose is to lessen the current from the power adaptors when power is switched on.