How to correctly measure L7805ACT regulator output?

:smiley: :smiley: Hello, everyone !
I am trying to measure the output of a L7805ACT voltage regulator (datasheet) with a multimeter, without any additional parts (no caps, etc, just the regulator). I’m using a regulated wall adapter power supply with settable output (5-12VDC). The adapter’s connector can be inversed so + becomes -, but I paid attention to connect it the right way.

The multimeter I’m using is an EXTECH EX330.

Here’s the regulator’s pin layout:
attachment 1

set multimeter to VDC setting
attach COM probe to regulator’s GND (middle pin)
attach red probe to output pin (3rd)
connect wall adapter to input pins (1 and 2)
switch the wall adapter on

attachment 2
With the first regulator I tried, the multimeter displays the input voltage and the regulator quickly heats. No voltage drop whatsoever.

At first I thought 12V input is too much, 7V have to be dissipated as heat and that the overheating is normal, so I set the input to 7.5V, but the results were the same.

Changed the wall adapter, same thing. Measured the adapter’s output power with the same multimeter, 7.5V exactly as set.

Tried with another regulator, 7.5V input, and now the measured output voltage starts from about 7V and rapidly drops to 0 (didn’t really waited to get to 0, don’t want to see the regulator burning).

Although the multimeter shouldn’t be the problem, as even without connecting it the regulator still heats, I thought to change the multimeter batteries. Low battery indicator was showing on the multimeter, although the batteries were only half discharged. I changed them anyway. Same result.

So what am I doing wrong? Does the regulator need a load on the output, like a resistor? This might sound stupid, but does the multimeter draw too much current from the regulator?

I know this is a basic task, but it got me a little frustrated after possibly destroying 2 brand new regulators bought from [url][/url].
Any suggestions will be very helpful ! !

well my first thought would be the pinout on your particular 7805 variant is “not standard” and you have the input current attached to the output. (having said that I have not ever seen one that isn’t input-gnd-output when viewed from left to right with the thing lying flay on the heatsink and legs to you, like Intelligent Power and Sensing Technologies | onsemi ). Then I’d check that you really really have the + and - outputs of the power supply the right way around.

Let me ask you this - does your multimeter show you the correct voltage when you measure the output of the wall adapter? So when you set 7.5v does your multimeter confirm that? And if you reverse the + and - it shows -7.5?

When a multimeter measures voltage, it has a large input impedance and doesn’t short circuit the circuit under test, so it’s not going to cause a high drain situation. What you have described doesn’t sound like a high drain, unless there’s something else shorting in your circuit. How are you holding all this together? Are you using a breadboard?

You need the capacitors. These devices tend to be unstable without them and this is probably what you are seeing.

I found this handy little explanation on electronics stackexchange.

[b]First of all, these capacitors aren’t there for smoothing the ripple, but to maintain stability of the regulator.

The 78xx regulator works roughly this way. There is a bipolar transistor placed between IN and OUT pins in the regulator, you can imagine that as a variable resistor. You could just place a fixed resistor there instead (leaving GND pin open) and calculate its resistance as R = (VIn-VOut)/IOut. The pity is that you generally know neither IOut nor VIn, as both may vary as the circuits works. So you need a mechanism that would set the resistance according to changes of these variables. This mechanism is called negative voltage feedback. There is a complex circuitry in the regulator IC that measures output voltage (voltage between OUT and GND pins) and compares it to an internal stable voltage source (again, for now, don’t care where this voltage comes from). If the regulator detects a voltage drop on the output (i. e. you connect another LED on the output), it opens the transistor more, lowering its resistance and delivers more current to the load. When you put the additional load away, the voltage would rise and the regulator closes the transistor, cutting the overvoltage away.

An ideal regulator wouldn’t require any of the capacitors, but there are some properties of the real circuit design that make it unstable (oscillations of voltage would appear on the output). That’s why you need to place a correct cap on both input and output; just follow the datasheet and (important!) place the capacitors as close to the IC as you can.[/b]

Dave is correct! you MUST have input and output caps on regulators!!!
Typically a 0.1uf on both the input and output sides. Then a larger one (47uf-470uf) to handle the larger load responses, because these linear type regulators are not that fast at handling quick load responses. Thats where the stored energy of the larger cap comes into play.

I’m also interested to understand why as a newbie you decided to come to the NETMF forums for a purely electronics question, rather than an electronics one? I’m not suggesting you’re not welcome or anything, but the audience here is diverse enough to answer this question, but it’s still not what I see as a core question that you’d guarantee could be answered.

My best guess is he trying to promote the website, otherwise why would he provide the link where he got the regulator?