I've made a circuit that needs a power supply adjustment. It works fine: 15v and 200-300ma max 5W but its potentially unsafe. I've reduced the power supply requirement to 5V but I don't know what max current to give it. The circuit is just 2 parallel circuits each with a ballast resistor. So what power supply characteristics should I keep in mind about the replacement ps? The schematic below has 5V for the voltage source now, it was 15.

 

You have to remember that most wallwart supplies give a "rated voltage" at a certain "rated load". Your load is very light. Thus, a "5V at 500ma" wallwart (for example) may be as high as 8 to 10V at your 34ma load.
You can load down the wallwart with an external resistor that draws 100ma or so, like a 50 ohm 2W or 3W resistor (it will get warm/hot).
Wallwart supplies are not well regulated, the voltage changes a lot with load.
Nothing wrong with a 15V supply either, provided you change the ballast resistors to make the same load current.
Always measure the voltage of any wallwart supply under the expected load....
You could also use any USB charger as a power supply, those are regulated to 5V (approx) and generally can supply at least 500ma.

 

I have found the opposite that modern/current Wall Warts are now SMPS and quite accurate at the output and under load, The older version that were simple linear supplies and were quite high at open circuit/no load..

 

The power supply only supplies the current drawn by the circuit. If the circuit needs 10mA then the PSU supplies 10mA. If it needs 400mA and the PSU is rated for 300mA you will typically see the PSU voltage drop as it tries to supply more current than it is rated for, or the current limiting fuse/breaker pops saving the PSU from overheating and emitting its magic smoke! From your circuit, you have 2 - ~16mA ckts so 40 -50mA PSU capacity is needed and allows a bit of overhead.

 

Yep. Max is right. The vast majority of switching power supplies are well regulated. If the supply claims to be 5 volts, it will be close to that across its rated range.

Switchers are also pretty good at taking care of themselves. If you exceed the rates current draw, the voltage falls off quickly.

One thing to watch out for: check the polarity! Don't assume the red wire is positive. If you cut the connector off a center-negative supply, chances are the red wire is negative. Why? The color code of the pre-built cables matches the typical cenrer-positive configuration. When building a center-negative supply, the connections at the power supply are reversed.

 

15VDC @ 250mA (average) is 15V x .25A = 3.75W
Your load is (15 ÷ .25) 60Ω
Drop the supplied voltage down to 5V assuming 60Ω gives you a current of 83mA. 83mA x 5V = 450mW. That's milli-wats, not watts.

If the resistance (load) doesn't change but you change the voltage you will change the amount of current your circuit draws. That in turn changes the wattage. Further; you report 15V at 5W, that comes to 333mA. Not 200 to 300 mA. To accurately predict your circuit we need very specific information. Voltage supply, Forward Voltage (Vf) of each LED and the resistance in each parallel circuit. Your diagram shows 150Ω and a Vf of about 1.24V.

(5V - (1.24Vf + 1.24Vf)) = 2.52 supplied volts.
2.52V ÷ 150Ω = 16.8mA. (PER PARALLEL CIRCUIT).
Since there are two parallel circuits exactly the same the total current drawn at 5V would be 33.6mA and a total wattage being supplied from the WW (wall wart) would be (5V x 0.0336A) 168mW. That's a far cry from your 5 watts you mention.

You need to start with the desired current for the LED's then calculate the correct resistance for the voltage supplied. And don't forget to calculate for the wattage the resistor has to dissipate. Or you can burn that out too.

 

You could also use any USB charger as a power supply, those are regulated to 5V (approx) and generally can supply at least 500ma.

To use the USB charger as a power supply could this be done using a breadboard and by stripping back one end of the dongle to expose the USB ground and positive? Want to make a PCB eventually. Attach these leads to the + and - rails of the BB.

 

Before you strip anything - do the math. That way you're not stripping something useful down in to useless stuff. Re-read my post, as I've edited it since posted.

 

15VDC @ 250mA (average) is 15V x .25A = 3.75W
Your load is (15 ÷ .25) 60Ω
Drop the supplied voltage down to 5V assuming 60Ω gives you a current of 83mA. 83mA x 5V = 450mW. That's milli-wats, not watts.

If the resistance (load) doesn't change but you change the voltage you will change the amount of current your circuit draws. That in turn changes the wattage. Further; you report 15V at 5W, that comes to 333mA. Not 200 to 300 mA. To accurately predict your circuit we need very specific information. Voltage supply, Forward Voltage (Vf) of each LED and the resistance in each parallel circuit. Your diagram shows 150Ω and a Vf of about 1.24V.

(5V - (1.24Vf + 1.24Vf)) = 2.52 supplied volts.
2.52V ÷ 150Ω = 16.8mA. (PER PARALLEL CIRCUIT).
Since there are two parallel circuits exactly the same the total current drawn at 5V would be 33.6mA and a total wattage being supplied from the WW (wall wart) would be (5V x 0.0336A) 168mW. That's a far cry from your 5 watts you mention.

You need to start with the desired current for the LED's then calculate the correct resistance for the voltage supplied. And don't forget to calculate for the wattage the resistor has to dissipate. Or you can burn that out too.

 

It works fine: 15v and 200-300ma max 5W but its potentially unsafe.

What's potentially unsafe about 15V at 300mA? 15V isn't normally considered a lethal voltage.

For 5V @ 34mA, I'd use a surplus USB charger.

Unless you're using old wall warts with power transformers that could be unregulated, most are going to be regulated switching supplies.

 

So do your recommendations allow for the 5V circuit you are referring to also include a "load resistor"? This was mentioned above. Is the load resistor distinct from the ballast one?

 

What's potentially unsafe about 15V at 300mA? 15V isn't normally considered a lethal voltage.

Well, that's what I said when it was pointed out to me. I got the WW because it was <500ma. It prompted me to have a second look. Truth be told I have learned a lot from that point moving forward.

 

For 5V @ 34mA, I'd use a surplus USB charger.

What's potentially unsafe about 15V at 300mA? 15V isn't normally considered a lethal voltage.

For 5V @ 34mA, I'd use a surplus USB charger.

Unless you're using old wall warts with power transformers that could be unregulated, most are going to be regulated switching supplies.

What's potentially unsafe about 15V at 300mA? 15V isn't normally considered a lethal voltage.

For 5V @ 34mA, I'd use a surplus USB charger.

Unless you're using old wall warts with power transformers that could be unregulated, most are going to be regulated switching supplies.

surplus USB charger example like this?

 

Well, that's what I said when it was pointed out to me.

Why did they say it was potentially unsafe? Were you going to touch the circuit with your tongue or insert probes into body tissue?

I got the WW because it was <500mw.

Your circuit will only draw the current it was designed to. Think about a typical electrical circuit in your house. If you're in the US, that would be 120VAC at 15A. Do you give that a second, or any, thought when you plug in your wall wart? The wall wart is only going to take what it needs.

It prompted me to have a second look. Truth be told I have learned a lot from that point moving forward.

Unless you're getting that information from a reputable source or something that's peer reviewed, you can't believe everything you hear or read.

 

surplus USB charger example like this?

That's a cable. A charger has a voltage regulator in it:

 

You have to remember that most wallwart supplies give a "rated voltage" at a certain "rated load". Your load is very light. Thus, a "5V at 500ma" wallwart (for example) may be as high as 8 to 10V at your 34ma load.
You can load down the wallwart with an external resistor that draws 100ma or so, like a 50 ohm 2W or 3W resistor (it will get warm/hot).
Wallwart supplies are not well regulated, the voltage changes a lot with load.
Nothing wrong with a 15V supply either, provided you change the ballast resistors to make the same load current.
Always measure the voltage of any wallwart supply under the expected load....
You could also use any USB charger as a power supply, those are regulated to 5V (approx) and generally can supply at least 500ma.

You need to start with the desired current for the LED's then calculate the correct resistance for the voltage supplied. And don't forget to calculate for the wattage the resistor has to dissipate. Or you can burn that out too.

Following up with the recommendations, I changed the diagram for the inclusion of a load resistor ( is it in series?) and updated the parallel resistors and voltage source = 5v. I also calculated the power for the resistors: the load was P = .084W and the ballasts were both .029W. I also calculated the total circuit Power P = .145W

 

No. You don't need a "load resistor" with a regulated power supply.

If you have an old cell phone charger, check the rating printed on it but it will almost certainly be 5 volts. If it has a cable, you can cut off the plug. USUALLY red is positive and black is negative, but this us not a certain thing. Even the cheapest voltmeter will let you check polarity and voltage.

If you have an extra USB wall wart (thrift stores are a good source if you don't), it will be 5 volts.* Get a USB cable, cut the "wrong" end off and you're set....but again, checking the polarity is highly recommended.

 

No. You don't need a "load resistor" with a regulated power supply.

If you have an old cell phone charger, check the rating printed on it but it will almost certainly be 5 volts. If it has a cable, you can cut off the plug. USUALLY red is positive and black is negative, but this us not a certain thing. Even the cheapest voltmeter will let you check polarity and voltage.

If you have an extra USB wall wart (thrift stores are a good source if you don't), it will be 5 volts.* Get a USB cable, cut the "wrong" end off and you're set....but again, checking the polarity is highly recommended.

Jon I don't completely understand. What part of the USB 5V charger do I connect to the breadboard?

 

I'm not exactly understanding what you're trying to do. It appears you have an LED and resistors in series to power.

A DC power supply has positive and negative (or common) connections. If you have a power supply like this picture (available cheaply at thrift stores), you can cut off the barrel plug, strip back the insulation and you'll find either 2 wires with separate insulation or a center wire with a layer of wire wrapped around it. One of these wires is positive, the other negative but the color of the wires may be meaningless.




A USB power supply usually has a USB A connector.



You'll need a USB cable to use on of these. Presuming your breadboard doesn't have provision to plug in a USB cable, you'll have to cut off the 'wrong' end of the cable. You'll find 4 or 5 conductors in the USB cable. With a little luck, there will be red and black conductors, often heavier than the other conductors, along with green and wires. The red and black are probably power, with green and white being data. But again, there is no guarantee and colors may be different or even switched.

 

No. You don't need a "load resistor" with a regulated power supply.

I respectfully disagree. What you said would be true IF your PS was a constant current device. With a CC device you get varying voltages designed to maintain the current. Cell phone chargers are not CC, they're CV (Constant Voltage). If the rated output current is exceeded the voltage WILL drop. But since the TS wants to drive some LED's at less than 30mA, a cell phone charger should be well equipped to handle that load.

As mentioned before, the math is critical. In your latest illustration you show a 50Ω resistor in series with two parallel circuits containing 75Ω resistors. The math gets a little tricky with a series parallel circuit but I'll try to manage the whole thing as a single series setup.

The two parallel 75Ω circuits combine to create a single series resistance of 37.5Ω. That in series with the 50Ω resistor creates a total resistance of 87.5Ω. The math works out this way:

(5V - 2.46Vf) ÷ 87.5Ω = 29mA. Your diagram confirms it, saying the current is 29.1mA whereas I get 29.0285mA. I wouldn't round that up to 29.1mA though. IF you didn't have any resistance in your circuit you'd burn the LED's out in short order because, again, the PS is not CC, it's CV.

I don't completely understand. What part of the USB 5V charger do I connect to the breadboard?

The green and white wires are used for data, not power. The red and black wires give you the 5V you're looking for. This case the red wire is 5 volts positive and the black wire is the negative. Between red and black you'll find a potential of 5 volts. I hesitate to call the black wire -5V because some people get confused and think that there's a potential of 10 volts. No. Just 5 volts. So use the red and black wires and trim and isolate the green and white wires so they don't short to anything.