Is 1/2 watt essential here or is this just per original spec?
Since I'm not sure what that BS250 is doing, (I'm not familiar with the device) My gut response is you need the 1/2w.
Because math, because Madbean.
Brian wouldn't have spec'd a 1/2 unless it was needed. Let's see if we can get him involved?
Madbean
Madbean
Madbean
In case Brian's busy: It's a 1/2W so that you don't burn up the resistor when a lot of current is pulled through it (the Aquaboy Deluxe had this issue). You can use a larger value, but then you're losing more voltage, particularly with the charge pump.
The BS270 is for polarity protection. I think it's from one of RG Keen's designs. Circuit Salad uses it sometimes, too.
Quote from: midwayfair on May 11, 2014, 08:47:07 PM
In case Brian's busy: It's a 1/2W so that you don't burn up the resistor when a lot of current is pulled through it (the Aquaboy Deluxe had this issue). You can use a larger value, but then you're losing more voltage, particularly with the charge pump.
The BS270 is for polarity protection. I think it's from one of RG Keen's designs. Circuit Salad uses it sometimes, too.
Jon, Maybe you can fill in my seemingly failing math?
R=V
2/P
9
2/.25 = 324 or ~330Ω
9
2/.5 = 162 or ~180Ω
It can't really be because of the normal current draw. The charge pump can only source about 20mA so give 40mA to say it's only 50% efficient. Pretend like the VB and LED need another 60mA and you still don't get enough current through the 10Ω to need that much power (right?).
But, look at in the situation where the input voltage is over the zener voltage on the input zener. The zener can pull a ton of current in that case and the 10Ω CLR would need to be pretty beefy.
Quote from: Clayford on May 11, 2014, 09:21:51 PM
Quote from: midwayfair on May 11, 2014, 08:47:07 PM
In case Brian's busy: It's a 1/2W so that you don't burn up the resistor when a lot of current is pulled through it (the Aquaboy Deluxe had this issue). You can use a larger value, but then you're losing more voltage, particularly with the charge pump.
The BS270 is for polarity protection. I think it's from one of RG Keen's designs. Circuit Salad uses it sometimes, too.
Jon, Maybe you can fill in my seemingly failing math?
R=V2/P
92/.25 = 324 or ~330Ω
92/.5 = 162 or ~180Ω
that's correct but that's a single resistor from V+ to GDN, it's only valid when there is 9volts across it which is rarely true when used as a power supply filter. In actuality, the voltage across it changes as the current draw changes.
luckily, there are several formulas for power: P = IV = I^2 / R. The easiest one to use is the last one since it relates it only to current.
Thank you.
ggarms - I noticed you posted in the parts jar section. My local electronics store surplus (supposedly) has 1/2W 9.1Ω or 12Ω in stock. Would you like me to grab them for you?
I don't see how the maximum current draw with a 9.6V or so power input can be up to the point where you need anything over a 1/4W resistor though.
P = I V, V = I R ==> P = I ( I R) = I^2 R. So, if you assume that the pedal draws 100mA, which has to be more than it can actually be using, then P = 0.1A^2 * 10Ω = 0.1W is an upper limit to the power that can be dropped on that resistor. So, 1/4W resistor should be fine. But, if you did put more than 12V, then you could end up in a situation where you need a 1/2W resistor. If you compute the value of CLR you need for that zener voltage regulator section assuming a 15V input and a 12V zener with a 40mA load, you get a 60Ω resistor @ 150mW. So, the smaller 10Ω would be over or way too close to the limit for a 1/4W.
I think that if you don't exceed the limit of that zener protecting the charge pump, then you could use a 250mW resistor in that spot and be safe.
I added the 10R resistor based on a suggestion made a while ago by a forum member concerning Zener protection diodes. If an over-voltage diode should fail, it will short the power to ground, so the suggestion was made to put a current limiting resistor there. I decided to use the lowest value I had, which is 10R, and since I also had some 1/2W I went with that thinking it would be better for noise and/or heat dispersion.
Honestly, the only math I ever invoke on pedal design is for filters so if the math shows a better solution than what I put there, by all means. You could use a 47R or 100R 1/4 resistor in place of the 10R without any negative effects, I think (a slightly higher voltage drop). I think the I recorded the voltage drop over the BS250 and 10R combo at around 100mV.
Quote from: madbean on May 12, 2014, 01:58:23 PM
I added the 10R resistor based on a suggestion made a while ago by a forum member concerning Zener protection diodes. If an over-voltage diode should fail, it will short the power to ground, so the suggestion was made to put a current limiting resistor there. I decided to use the lowest value I had, which is 10R, and since I also had some 1/2W I went with that thinking it would be better for noise and/or heat dispersion.
Honestly, the only math I ever invoke on pedal design is for filters so if the math shows a better solution than what I put there, by all means. You could use a 47R or 100R 1/4 resistor in place of the 10R without any negative effects, I think (a slightly higher voltage drop). I think the I recorded the voltage drop over the BS250 and 10R combo at around 100mV.
Since the Zener and CLR are there for protection and you aren't really supposed to run it at over 12V, the design as you have it makes sense to me. If I actually planned on running it with an input above 12V and I were going to use the Zener as a regulator, I'd move the CLR to about 60Ω, because the added voltage drop wouldn't make any difference. But, if I were going to use it as it's designed to work at 9V, then moving to a 60Ω resistor would increase the voltage drop up to at least 600mV and I'd consider that to be too much. Either way, I'd stick with the 500mW resistor, because 150mW doubled is over 250mW and if you are using it in the protection scheme, it's going to need to be at 500mW too.