As discussed in the link @fredrik provided, generally for synth circuits ferrites cannot be expected to do any significant good.
100 nF decoupling caps can, and should be used. But on the other hand the amount of difference they make can be small. I ran out of 100 nFs once and built a module without them and it seemed to work fine… but I did install them once I got more.
Seeing the first image make me wonder if its better to not short all GND pins to use different grounds for different modules.
If I’m not mistaken it wouldn’t matter because everything hooked up to the same power supply shares a ground anyway
The flat ribbon cable is a really bad design choice for a power cable to start with (these small gauge wires with just little blades cutting through the insulation to make contact were designed as a cheap way to carry many, very low power, digital signals) so you want to use every single available one of those flimsy wires to minimize the risk of a single bad connection preventing the cable from working.
Plus it would be practically impossible to coordinate the use of the different pins among different module manufacturer.
And if the idc cables are to connect 2 modules of the same manufacturer (like my DIY project) is it a good thing to use differents grounds or is better to share the same?
You really do need to use all the ground pins for all the modules.
So my question here… On all the new modules I see a lowpass filter on the power input with 10R & 10uF, which according to the calculator has a cut-off frequency around 1.5KHz, well into the audio range.
If we’re trying to filter out mains noise (50/60HZ) wouldn’t we want to revise those to be 10R & 330uF (for a cut-off around 48Hz) or 10R & 470uF (for a cut-off around 33Hz)?
The 10 µF capacitors aren’t really there as a filter component, but as a bypass, propping the voltage up when the circuit makes sudden demands.
The 10R resistors aren’t really there as a filter component. Near as I can tell they’re mostly there due to cargo cultism. People like to use them as a fuse that burns out if there’s a short, but an actual resettable fuse would be better. I always omit them in favor of Schottky diodes for power reversal protection.
I don’t think they make a very good fuse. I had a problem with a stripboard circuit lately where I did use them and they flared up 4x and they still pass current. (i pulled the power real quick on it each test… so i guess in a way they are useful as “Fault Indicators”).
agreed that a resettable poly fuse would be better, if we could find one that trips at the really low wattages we’re using in these modules…
I don’t worry about reverse voltage because I use keyed connectors.
Poly fuses trip on current, not power, right? And you can get them down to 100 mA. I have a TipTop supply with some form of resettable fuses that works well when I screw up.
As for reversed voltages, you still have a problem if (1) you install your connector backwards (it’s been known to happen that it’s silkscreened backwards on a PCB) (2) your power cable is made wrong (3) your busboard is wrong (4) your power supply is wrong (5) your connection between busboard and power supply is wrong. I don’t think I’ve ever had any of the above problems but diodes are cheap insurance for if I do.
I wasn’t (and still am not) really sure if poly fuses trip on current or power. It seems like a thermic reaction (i.e. too much power through the poly fuse generates heat which sets off a reaction). I guess really i just need to buy some and do some experimenting.
Same thing: P = I×I×R. The fuse controls R.
You probably mean here, with poly fuses?
Because in general that is not necessarily true. I can imagine that responding to heat (power) will always take some time (because of thermal resistance / capacity of materials (not sure that is the right term)) whereas responding to a high current need not depend on that.
In the fuse, yes. It only sees the current that goes through it, not what’s on the other side, so from the fuse’s perspective current (A) and power (W or J/s) is the same thing. Ohm’s law.
It depends on which power you’re talking about…
10mA * 12V = 0.12W
10mA * 1200V = 12W…
The fuse “has no knowledge” about this voltage…
Exactly. We use such little power in these modules it’s hard to find a through-hole polyfuse that’s appropriate. The lowest wattage trip on digikey that i see is 60v@100ma or 6w which is half an amp at 12v and it has a 5 second trip time. Maybe that’s fine if you’re building a module that has a 5v regulator and an arduino onboard, but for simple logic modules that’s a lot of juice.
Maybe there’s a tiny product niche for someone to have a few panels of a bunch of 0805 to 2 pin pcbs made up and pre-populated by JLCPCB with PRG21AR220MS1RK (16V, 75mA hold, 195mA trip) to sell on Tindie.
No, you’re misreading the spec. 60 V is the max voltage drop the fuse can survive, it trips at 100 mA no matter what voltage you’re running through it as long it’s within bounds. There’s no voltage in P = I×I×R.
Fredrik:
If you look at the datasheet for a polyfuse, what you’ll discover is that the resistance of the polyfuse changes as it’s temperature increases. What makes the temperature increase is wattage. 100ma at 1v is going to be 0.1W (P=IxV) (OHM’S LAW!!!) which will not have much of a heating effect and will not raise the temperature of the polyfuse to a degree that will change it’s resistance plonker-all. 100ma at 60v would be 6W which will have much more of a heating effect and change it’s resistance much more quickly.
P=V x I
P= (V x V) / R
and
P=I x I x R
are all valid equations for Wattage and they all involve Voltage, Current, and Resistance because they all stem from Ohm’s Law which describes the relationship between voltage, current and resistance even when you rearrange it eliminate one from the maths you yourself do on paper.
P= I x I x R doesn’t mention voltage, but using OHM’S LAW current is directly related to voltage and inversely related to resistance. IE: I = V / R, IE: given the same resistance with more voltage you will have… more current!!!
P= I x I x R is derived from:
P = V x I
and V = I x R
to make P = I x R x I
That does not remove voltage from the situation, wattage (and therefore heat which changes the resistance of the polyfuse) is still a function of voltage and current. Because Ohm’s Law.
[Edited to include the X-axis label to show that it is temperature dependent]
Yes, that’s what I’m saying. The wattage in the fuse itself, caused by the current going through it, not the voltage or wattage in the circuit on the other side. The fuse isn’t connected to ground, so it has no idea about the voltages involved, and no idea about the power consumption of the circuit it’s protecting – until there’s a short there, that is, at which point max voltage and max current limitations apply, to make sure it can protect the circuit from that short.
A max rating of 60 V for a 100 mA fuse means that it can protect the circuit from a shortage that will result in 60 V over the fuse (60 V to ground, +30 V to −30 V for split supplies, etc), not that it’ll only trip if the circuit it’s protecting is using more than 6 W (“half an amp at 12v”).
(This is the same as for any other fuse, or wire for that matter – which is why amperage tables list amperes, not volts or watts. They see the current = electrons going through them, not where they are in relation to ground.)
