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So I've got a circuit that requires both 5V and 3.3V so I bought a 5V 2A AC/DC converter from a store, which I will add resistors to later for the 3.3V. Before I connect it to my circuit, I wanted to verify that the DC output was correct, so I plugged it in, dialed my multimeter like so:

enter image description here

For 20V max DC voltage. Then I attach the common to one of the contacts on the power supply (shouldn't matter which because I'll either get a positive or negative voltage) and the 10A "hot" end to the other contact. When I make contact, the power supply's LED, which is on when plugged in and not connected, goes out and I get a 0 reading on the multimeter.

What is happening? Is my setup correct? Should I add be adding resistance anywhere? Is the multimeter pulling too many amps? Do I need a different multimeter? What is the best way for me to test my 5V 2A DC power supply?

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    \$\begingroup\$ @Tuskiomi - Don't be a dingus. Your sentence sounded like sage advice until you added the unnecessary insult. The :p neither ameliorates the behaviour nor improves your image. Overall a useful comment ruined. \$\endgroup\$ – Russell McMahon Jun 29 '16 at 14:32
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    \$\begingroup\$ It's worth pointing out that the 10A socket bypasses the switch, so the 10A-COM connection is always there, regrdless of the switch position. \$\endgroup\$ – Neil_UK Jun 29 '16 at 14:39
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    \$\begingroup\$ @tuskiomi It's not clear that you understood my point. The usual response to "you are behaving unacceptably obnoxiously " is not "don't concern yourself about it", as you effectively responded. Why should your response reduce anyone's concern? Of the various meanings available for dingus the only one that fits your response is "an unintelligent person; "idiot"; "moron"." [Online slang dictionary]. It's a little thing in internet behaviour terms. but goes against the ethos which this site aims at. I suggest you take the point and don't instead try to test my assertion :-). \$\endgroup\$ – Russell McMahon Jun 29 '16 at 15:20
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    \$\begingroup\$ @DmitryGrigoryev Once one knows a little on a subject it can be hard to appreciate how inobvious and difficult even apparently simple things can be to newcomers. The fact that the question IS being asked when the picture provided "clearly" carries its own answer indicates what difficulty the OP is facing. Note Neil_UK's comment which adds information that many newcomers would be unaware of. Q&A like this can form part of a valuable learning resource. \$\endgroup\$ – Russell McMahon Jun 29 '16 at 15:24
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    \$\begingroup\$ "which I will add resistors to later for the 3.3V" - this is not how you make a power supply, you need a regulator. \$\endgroup\$ – pjc50 Jun 29 '16 at 16:09
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You have your multimeter leads plugged into the amps socket. It will act like a short circuit.

If you are trying to measure voltage then plug the red lead into the red socket marked "VmA" (in the middle).

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  • \$\begingroup\$ But that other "FUSED" socket reads "200mA 250V Max" for DC, does that mean I cannot test it since the power supply outputs 2A? Should I use a different multimeter? \$\endgroup\$ – David Rhoderick Jun 29 '16 at 16:19
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    \$\begingroup\$ If you are measuring VOLTS, the meter will have a high input impedance (probably 10Mohms) and thus take virtually no current, whatever the supply can output. So use the VOLTS (etc) socket. Only use the 10 Amp range for measuring Amps ( > 0.2A) - it has a separate socket because it's unswitched - such a high current would burn the switch. \$\endgroup\$ – Brian Drummond Jun 29 '16 at 16:32
  • \$\begingroup\$ @DavidRhoderick - In case Andy isn't around, I'll respond by saying that you're confusing what you are trying to measure. You are trying to measure the output voltage, meaning you put the multimeter in parallel with the output. For that voltage test, it doesn't matter whether the 5V PSU is capable of supplying 2A or 200A. As Andy said, put the red lead into the "VmA" socket, leave the switch at position in your photo, and you can measure the output voltage of the PSU. Measuring current becomes a different (and longer) reply and a different test configuration. Don't confuse the two. \$\endgroup\$ – SamGibson Jun 29 '16 at 16:33
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    \$\begingroup\$ For the mA range the max current is that of the fuse but when you have the dial selecting voltage the inputs are high impedance such as 10 Mohm and only a micro amp (or thereabouts) will flow into the meter. \$\endgroup\$ – Andy aka Jun 29 '16 at 16:35
  • \$\begingroup\$ @DavidRhoderick I don't mean this to be rude, honestly, but you lack sufficient basic electricity knowledge to safely use that multimeter. Fixing the few defects in your knowledge we might happen to identify is not going to be sufficient to produce a safe and sane result. \$\endgroup\$ – David Schwartz Jun 29 '16 at 18:24
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To help with the second part of your question:

I will add resistors to later for the 3.3V

Don't do that, if you want to convert the 5 V input from your PSU into 3.3 V. Attempting to do so causes all sorts of problems, as the 3.3 V "output" will then depend on the current being drawn, which will vary depending on what the 3.3 V circuitry is doing.

Instead you should use either a 3.3 V linear (e.g. LDO - low dropout) or buck (switch mode) regulator, connected to the 5 V supply, to produce a regulated 3.3 V supply for those components which require that lower voltage.

Linear regulators require simpler associated components (typically 2 capacitors for fixed-voltage types), tend to be cheaper, have relatively little ripple on the output, but produce more heat (especially as output current increases). Buck regulators are generally the opposite of those characteristics. Both types are available on pre-made PCB modules, if you don't want to build them yourself. There are several similar questions about choosing and using such modules on EE.SE.

Bottom line: Don't attempt to use resistors to "convert" power supply voltages.

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  • \$\begingroup\$ I'm only trying to use 5V for a relay and 3.3V for a 250mA peak WiFi module, should I still avoid using a 5V linear regulator and a voltage divider for 3.3V output? \$\endgroup\$ – David Rhoderick Jun 29 '16 at 19:04
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    \$\begingroup\$ @DavidRhoderick You can't use a voltage divider for that. Impossible, unless it draws a very constant 250 mA in which case it would be possible to calculate a divider. For a WiFi module, forget it. You need a 3.3 volt regulator. \$\endgroup\$ – pipe Jun 29 '16 at 19:19
  • \$\begingroup\$ @DavidRhoderick - I completely agree with pipe (+1). Since a WiFi module does not take a constant current (in fact it varies significantly during use) you can't use a voltage divider as you suggested, to provide the 3.3 V supply. That is what I was saying about varying current causing varying voltage drop (i.e. varying 3.3 V "output" from a voltage divider). You can research the topic of "Ohms's law" for more background about the relationship of voltage, resistance and current. HTH. \$\endgroup\$ – SamGibson Jun 29 '16 at 19:52
  • \$\begingroup\$ Thanks @SamGibson, I was just looking at LDO implementation and it doesn't seem to difficult to accomplish. I will also look into buck regulation. Thanks for the answer! \$\endgroup\$ – David Rhoderick Jun 29 '16 at 20:51

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