I've seen in several schematics the habit of having a small resistor (usually 47R or 100R) from the +9V connection to the circuit components that require power. Why is it used?
I'm no electronics expert but I know that a resistor reduces power by some amount, depending on the resistor value. A 100R resistor what difference can it make?
It does 2 things.
Primary - Forms a low pass filter with the power supply capacitors - helps reduce audio and radio frequency noise on the power supply.
Secondary - Protection - It will limit supply current in a short circuit fault. It may well go up in smoke, but it might have saved everything else. If it's only momentary fault, the protection may work long enough to stop any damage ever happening. A fuse on the other hand will blow - and that's the end of that until you change the fuse.
Should also be an electrolytic cap, 47u, 100uF.... to ground after that resistor which together will filter out noise from a power supply. Small value resistor so voltage drop across it is minimized. I power the LED from before filter as the LED has the potential to be the most power hungry part in the pedal.
Circuits dead quite powered by a battery can get noisy even from a great power supply, this little filter can kill the demons.
SLow typist here.
And a typical effects pedal circuit hardly ever consumes more than a couple of milliamperes (unless it's something fancy with clocks and LFOs like a chorus or a delay).
Let's say theres a 100ohm resistor installed in series there and the you've got a fuzz pedal that draws 3mA.
as per Ohm's law: 100ohms * 3mA = 0.3V
Which is how much supply voltage you "lose" over that resistor in exchange for better filtering of noise from the power supply.
Thanks a lot for the replies! ;)
1/2*Pi*R*C = cut off frequency. Larger cap, larger resistance - lower cut off. Larger resistance = bigger voltage drop across it, so if your effects consumes more current than smaller the resistance should be (don't forget about his power dissipation) and larger the value of the cap.
In some instances I've learnt it won't practically matter how large of a power filter cap you use until you start using the series resistor before it to complete the low-pass filter. Exactly like how it would be silly to add a smal value RF filter cap at the beginning of a circuit without a series resistor prior.
One particular circuit I was building (pnp with negative ground) would always be noiser on power supplies than an INTETNAL (shielded) battery. Even with a 470uf filter cap. Add a 47 ohm resistor and presto! Perfect.
So I have a follow-up question on this topic.
Some power supplies will use a diode is series with the supply as a reverse polarity protection. Depending on the diode, there is always a small voltage drop across the diode.
In that scenario, is the diode functioning like a resistor, and if so, will the diode followed by a large cap to ground form a low pass filter?
The diode won't really work as a resistor. There's a minimal amount of resistance (like any conductor) though. But I treat them as none for pedals.
The voltage lost across a diode is due to the forward voltage drop. That is, the voltage needs to be beyond a certain figure before it can start to pass through the diode. So any voltage lower than that, say 0.6/0.7V didn't get to pass and is subtracted from the output. It's still sitting there though and not burnt up like a resistor would.
A lot of circuits just leave it out anyway though :)
>M is the diode functioning like a resistor
In a nearby thread, someone discovered that his pedals average 32mA.
At 32mA the *dynamic* resistance of a solid-state diode is just under 1 Ohm.
If you are filtering hum/buzz you need your R-C product to be far-far below 100/120Hz (or 50/60Hz if your wart is real cheap). Say 10Hz. 10Hz at 1 Ohm is 16,000uFd. Rather large and costly.
If you can spare just 0.3 Volts you can use a 10 Ohm resistor and now you only need 1,600uFd. Maybe large for pedal but very common in simple hi-fi plans.
If your pedal sucks 32mA, then a 47 Ohm resistor drops 1.5V, which is a lot in a 9V world. However many-many pedals are much less than 32mA. At 10mA a 47 Ohm is only a half-volt, and you get great filtering with just 340uFd.
Isn't it more beneficial to reduce power drop from the resistor to instead use aa inductor in series at the power input to achieve same as the resistor only without the unwanted power drop?
Quote from: bluelagoon on May 24, 2023, 03:55:33 AM
Isn't it more beneficial to reduce power drop from the resistor to instead use aa inductor in series at the power input to achieve same as the resistor only without the unwanted power drop?
Resistor is usually the common and easy part in any value.
The voltage drop is only important in very fragile circuits.
Quote from: bluelagoon on May 24, 2023, 03:55:33 AM
Isn't it more beneficial to reduce power drop from the resistor to instead use aa inductor in series at the power input to achieve same as the resistor only without the unwanted power drop?
I did actually see an example of that just recently, on the Madbean/VFE relay bypass board:
(https://www.madbeanpedals.com/projects/_folders/VFE/schematics/VFE_SBv3.gif)
It's not that common though, for the reasons Steben gave: Everyone has resistors, but not many of us have inductors around, and resistors are cheaper anyway.
Also for 50Hz you need very high inductances to match the performance of a resistor. Low current inductors with large inductances can end up with high DC resistances. If the current is significant then such an inductor will be large and much more expensive. For high frequencies the inductors can help but they still aren't that small in value.
It's not clear what the inductor on the madbeans circuit is trying to achieve because it filters the incoming supply but not the local supplies. IMHO the layout of the tracks and caps around IC2 would be important and if done correctly you might not need the inductor.
Watch out for stuff like this in small inductors,
https://www.inductorchina.com/inductor-types-fixed-al0307-getwell.html
1mH 26 ohms
Whereas
https://www.farnell.com/datasheets/2156400.pdf
1mH 2.5 ohm
Quote from: loki on August 04, 2014, 10:41:20 AM
I've seen in several schematics the habit of having a small resistor (usually 47R or 100R) from the +9V connection to the circuit components that require power. Why is it used?
I'm no electronics expert but I know that a resistor reduces power by some amount, depending on the resistor value. A 100R resistor what difference can it make?
I had to add a small resistor to designs to protect against high frequency noise generated by many switch mode power supplies. Some of these cheap ones are really bad. You just have to be careful to power any components that change their current consumption due to switching (status LED) or modulation (tremolo, chorus, etc.) that use LEDs to either modulate a LDR or show status or both, as this will modulate the power supply if powered from the output side of this series resistor! Sometimes a series protection diode has enough variation in series resistance to cause the supply to modulate.
What about this one at around $0.40 AUD, its cheap enough, the inductance is 470uH, and its resistance is only 3.4 ohm, a much less power drop than any of the series resistors offer.
https://www.digikey.com/en/products/detail/abracon-llc/AIAP-01-471K-T/3060653
Also alluding to PRR's earlier post where the smaller the resistor the more huge the capacitor value needs be, but not with a series inductor, you get by with just a 100uF in the power supply with a 3.4 ohm inductor being used. isn't this a better deal?. since the voltage drop with the series resistor does add up, especially when you start using voltage doubling converters, you end up with twice the voltage drop. probably never too significant by any means in an effect pedal, but maybe if using a battery to power. but even for the practicality of a need for a much smaller Filter cap to achieve same end is an advantage.
> inductance is 470uH
What is the inductance of 470uH in the audio band? Anybody know math(s)?
I get 0.15 Ohms at 50Hz, 15 Ohms at 5kHz. Well, 3.5r and 18r counting resistance.
There are situations where a choke makes sense. Low-current low-production guitar pedals, it is hard to make a case.
Quote from: PRR on May 25, 2023, 12:46:56 AM
> inductance is 470uH
What is the inductance of 470uH in the audio band? Anybody know math(s)?
I get 0.15 Ohms at 50Hz, 15 Ohms at 5kHz. Well, 3.5r and 18r counting resistance.
There are situations where a choke makes sense. Low-current low-production guitar pedals, it is hard to make a case.
That was my earlier point you need enormous inductors and small inductors with even mild inductance will often end up with high resistance. The DC resistance will do more than the inductance!
Yes all good and well the mathematical equations being considered, but in all practical intent, wouldn't that 470uH inductor do a satisfactory job at filtering as much as a series resistor would do, and wouldn't it do the job without taking as much a voltage drop as the resistor would do, and isn't it also true that an inductor does not need the same exceedingly large sized filter capacitor to achieve the same result as the resistor and capacitor choice. Even though it may be overkill, if it works and has some advantage as not requiring as large a filter cap, and if it saves some extent of power loss over the resistor choice, then isn't it a viable alternative at maybe a few extra cents over the series resistor.?
The significance of the filtering provided by having a series protection diode should not be ignored or underestimated. Its effective forward resistance in combination with decoupling capacitors can provide a considerable amount of filtering of noise.
(https://i.postimg.cc/BXSDw18F/1-N5817-Characteristics.jpg) (https://postimg.cc/BXSDw18F)
The filtering effect of series diodes can be seen here compared to that of a 5 ohm and 50 ohm series resistor. In this case, there is a 9mA load and Nishicon 100uF capacitor and a Worth 1nF NPO ceramic capacitor for decoupling.
(https://i.postimg.cc/NLtKxgxS/Effect-of-Diode-on-Filtering.png) (https://postimg.cc/NLtKxgxS)
Ah, so i am guessing the 50 ohm resistor is the winner there in that last graph response.
All said and done, I have a particular power supply, if some of you more knowledgeable folk wouldn't mind please commenting on its
strengths and weaknesses, and suitability. I were mostly wondering if C1 Filter cap ought to perhaps be higher at 220uF , and if a second smaller 100nF filter cap would help in paralell with C1? Thanks
(https://i.postimg.cc/YvkHx7kT/power-supply.png) (https://postimg.cc/YvkHx7kT)
I'd make C1 470μF/16V and place 100nF ceramic cap in parallel (as close to R1 right leg as physically possible..)
D1 is for reverse polarity protection I guess? you lose about 0.3 volts here, something you don't seem to like. Maybe opt for no protection and just not reverse polarity, or choose a method without voltage loss?
Quote from: Clint Eastwood on May 25, 2023, 07:20:11 AM
D1 is for reverse polarity protection I guess? you lose about 0.3 volts here, something you don't seem to like. Maybe opt for no protection and just not reverse polarity, or choose a method without voltage loss?
D1 is not needed for reverse polarity as long as the Zener diode and resistor R1 are high enough wattage. The Zener will provide both overvoltage and polarity protection with R1 providing the current limiting and noise filtering in conjunction with the decoupling capacitors.
Quote from: antonis on May 25, 2023, 07:01:18 AM
I'd make C1 470μF/16V and place 100nF ceramic cap in parallel (as close to R1 right leg as physically possible..)
Actually, for better noise suppression, the 100nF would be better placed as close as possible to pin 1 and 8 of the charge pump IC, not R1.
2W for R1 should be the minimum power rating..
As for the Zener, any DO-41 package (like BZV85) should easily handle more than 212mA (the current through Zener and R1 in case of power reverse polarity conection..)
Thanks all for the advice, and Clint, I'm not totally against voltage drops, just would prefer not if there were ways to avoid. I know if you get that desperate you can put two 5817 diodes in parallel, then that effectively halves the voltage loss of just a single diode on its own, Sometimes if you are making the pedal as optionally battery powered you hope to save on voltage drops, but like RG and others keep commenting, the battery powered effect pedals have pretty much gone the way of the dinosaur!
The idea to drop the 1N5817 sounds good. Just wondering what size voltage rating would be needed for the zener and the 39 ohm resistor? Thanks
Quote from: FSFX on May 25, 2023, 07:33:52 AM
Actually, for better noise suppression, the 100nF would be better placed as close as possible to pin 1 and 8 of the charge pump IC, not R1.
In such a case, Vref shouldn't be RF (or any HF) filtered..
(of course, they all depend on layout configuration..)
Hi Antonis, its like you read my mind on that last question I had, posted before I asked. Cheers.
Quote from: bluelagoon on May 25, 2023, 07:48:44 AM
I know if you get that desperate you can put two 5817 diodes in parallel, then that effectively halves the voltage loss of just a single diode on its own,
:o :o :o
They should share CURRENT, not voltage.. :icon_wink:
P.S.
Actually, diode forward voltage drop DOES lower with smaller currents but in a, IMHO, consideration unworthy manner..
(https://i.imgur.com/N0dcN8V.png)
You can presicely estimate the forward voltage drop difference by solving Shokley diode equation for different given currents.. :icon_wink:
(which, in your case of 2 parallel diodes is just 17.3mV..)
Quote from: antonis on May 25, 2023, 07:49:41 AM
Quote from: FSFX on May 25, 2023, 07:33:52 AM
Actually, for better noise suppression, the 100nF would be better placed as close as possible to pin 1 and 8 of the charge pump IC, not R1.
In such a case, Vref shouldn't be RF (or any HF) filtered..
(of course, they all depend on layout configuration..)
I really don't think you understand this at all. That is a charge pump and so needs the 100nF as close to it as possible. Please read all of the stuff published by Maxim and others to understand the issues with charge pump input noise..
Quote from: FSFX on May 25, 2023, 08:06:27 AM
I really don't think you understand this at all.
I think we're dealing with entirely different purposes cap implementation.. :icon_wink:
Quote from: antonis on May 25, 2023, 08:17:01 AM
Quote from: FSFX on May 25, 2023, 08:06:27 AM
I really don't think you understand this at all.
I think we're dealing with entirely different purposes cap implementation.. :icon_wink:
Not really, it is all about the correct way and best practice for reducing the noise on the power to the circuit.
Quote from: antonis on May 25, 2023, 07:01:18 AM
I'd make C1 470μF/16V and place 100nF ceramic cap in parallel (as close to R1 right leg as physically possible..)
+1 agree. 39R/100u is a cutoff at 40Hz. Since this is only a -6dB/oct filter, that's barely any cut for 50/60Hz mains hum.
Changing to 470u would push that cutoff down another couple of octaves, and that improves the filtering quite a bit.
100n ceramic cap in parallel is always a good idea for the amount that it costs.
Quote from: ElectricDruid on May 25, 2023, 01:20:42 PM
Quote from: antonis on May 25, 2023, 07:01:18 AM
I'd make C1 470μF/16V and place 100nF ceramic cap in parallel (as close to R1 right leg as physically possible..)
+1 agree. 39R/100u is a cutoff at 40Hz. Since this is only a -6dB/oct filter, that's barely any cut for 50/60Hz mains hum.
Changing to 470u would push that cutoff down another couple of octaves, and that improves the filtering quite a bit.
100n ceramic cap in parallel is always a good idea for the amount that it costs.
In practice it will be better that that because the hum from a rectified output is actually 100Hz/120Hz - I used loose 50Hz/60Hz numbers earlier on as well.
A big cap will help stop noise from the circuit itself getting in the audio. That's more to do with current pulses on the circuit side and the size of the cap. Layout can make a big difference as well, especially with DC/DC converters.