Modifying my cheap Chinese power supplies...

vigilante398 said:
Here's an example:

Let's say you have a 300mA load on your 9V regulator. The datasheet doesn't have the efficiency vs output load graph like I wanted to see, but let's assume 85% efficiency, which is a reasonable approximation for this thing.

With the stock power supply:

Vin = 15V
Vout = 9V
Iout = 0.3A

15V * I_in = 9 * 0.3 + loss
15V * I_in = 2.7 + loss

Loss with 85% efficiency of a 2.7W load is going to be 47.6mW, which gives us

15V * I_in = 3.176
I_in = 212mA

So let's look at the same thing with an 18V supply:

18 * I_in = 3.176
I_in = 176mA

Thanks for coming to my TED talk :p
Click to expand...

So for us Dummies out here, even on the 15 vDC supply, a 300 mA tap won’t yield 300 mA and when we step up the voltage of the supply, it yields even less.
 
Big Monk said:
So for us Dummies out here, even on the 15 vDC supply, a 300 mA tap won’t yield 300 mA and when we step up the voltage of the supply, it yields even less.
Click to expand...
Incorrect, you're looking at it backwards. I'm saying that when you're getting 300mA @ 9V to your pedals, you're only sucking 176mA @ 18V from your 18V wall wart.
 
I keep trying to find better ways to explain it but they feel convoluted, I really just need a big whiteboard and for you to come over so I can explain it in person :P
 
I think it helps to look at it in terms of power. Power = voltage * current. I'm going to pretend it's perfectly efficient for the sake of explanation.

input power (from the wall wart to the regulator) = output power (from the regulator to the pedals)

I'm going to stick with a 300mA @9V, so we know that our output power is 2.7W (9 * 0.3). Since they are equal, we know that input power is also 2.7W, regardless of what the input voltage is. The input voltage does not change the output power. So if input voltage is 15V, we know that input current is 2.7/15 = 180mA. We "gain energy" (for lack of a better expression) when we step down voltages so we only use 180mA @ 15V to generate 300mA @ 9V. If we increase the input voltage, the output power stays the same, which means the input power is still 2.7W, so the input current becomes 2.7/18 = 150mA.

That's all I've got, hopefully my ramblings make sense, but also let me know if they don't.
 
vigilante398 said:
I think it helps to look at it in terms of power. Power = voltage * current. I'm going to pretend it's perfectly efficient for the sake of explanation.

input power (from the wall wart to the regulator) = output power (from the regulator to the pedals)

I'm going to stick with a 300mA @9V, so we know that our output power is 2.7W (9 * 0.3). Since they are equal, we know that input power is also 2.7W, regardless of what the input voltage is. The input voltage does not change the output power. So if input voltage is 15V, we know that input current is 2.7/15 = 180mA. We "gain energy" (for lack of a better expression) when we step down voltages so we only use 180mA @ 15V to generate 300mA @ 9V. If we increase the input voltage, the output power stays the same, which means the input power is still 2.7W, so the input current becomes 2.7/18 = 150mA.

That's all I've got, hopefully my ramblings make sense, but also let me know if they don't.
Click to expand...

This am the previous posts made major sense.

Let me see if I can summarize in a sentence: switching from 15 vDC to 18 vDC supplies means the same current delivered to the taps but with less current drawn from the input to the regulator.

Did I get it?
 
Big Monk said:
This am the previous posts made major sense.

Let me see if I can summarize in a sentence: switching from 15 vDC to 18 vDC supplies means the same current delivered to the taps but with less current drawn from the input to the regulator.

Did I get it?
Click to expand...
Yup, nailed it!
 
@vigilante398

So jumpering where the 100 mA ferrite beads are gets me essentially unlimited current draw (depending on the pedal load) and so long as combined I don’t exceed the supply and regulator rating.

What purpose do they have? It’s easy enough to get some 300 mA ones to match the other 9 vDC taps and just run an 18 vDC supply.
 
fig said:
noise filtering?
Click to expand...

I think that’s what we discussed earlier.

If that’s the case, I think replacing the 100 mA units with 300 mA is a cheap solution when co pared to just jumpering them.

So then all I’m going to do is swap 3 ferrite beads, grab a 12 v zener with a higher power rating then currently installed and grab an 18 vDC/2 A Wall wart from Amazon.

This way I can have an 18 vDC tap, firm up the 12 vDC tap, and have 5 9 vDC/300 mA taps instead of 2 in case I run more than 2 high load digital pedals.

Turns my $25 power supply into a $40 power supply but it will be an improved version.
 
thewintersoldier said:
I can not believe we are on page 4 already...
help-us-somebody-help.jpg
Click to expand...

C’mon!

Where is your curious spirit?! We just figure out how to make something cheap and available a little better! Well @vigilante398 did at least!

You elitist power supply snobs must be stopped…😂 (as I look at premium power supplies on Reverb…)
 
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