How to detect 5V?

Question

Hi everyone,

I'm rather new to electronics but would love to be able to work on projects at home! (now I only do electronics at school)

As a first project, I would love to convert an old Power Supply Unit into a power bench!

I read that PSU have a line called Power Good, which outputs 5V if the PSU is working correctly. I thought it would be cool to connect an LED to this line, so I could see if everything is OK!

For this I first thought of a simple LED+resistor, but then I figured this would make the LED emit light (less bright) even when there is eg. 3V on the line.

So I was thinking of a simple circuit with a zener diode with a breakdown voltage of 5V, but I'm not sure whether this would work as expected...

Since I pay for the components myself, I wanted to be sure this would work, before buying everything!

Can anyone tell me if this idea would work, and if not how to do it otherwise? I've looked on the internet and the main solution I find is with a comparator. I cannot use this because if the PG Line is wrong, I cannot know if the 5V line of the PSU is actually 5V...

What I want is an easy little circuit that has a LED emit light when there is 5V and not emit light when there is no 5V. (doesn't have to be completely exact) As far as I have found, this power good line varies between 0-5V (does not go higher).

Thanksfor your help!

Schematic of my idea:

shabaz · Accepted Answer

Hi,

Are you sure the power good line isn't a logic level signal (i.e. two states, high and low)? Usually, they output something undefined when the PSU output voltage is very low (e.g. just powering up), then they indicate power bad for the milliseconds that the supply is still coming up, then they indicate power good, then they remain there unless the output voltage is not within some threshold in which case the signal indicates power bad by changing logic level again.

So, if you have enough power to drive your circuit from the main supply, then you can trust the logic level of the power good signal.

You could just drive an LED from the main PSU using a resistor and a transistor. Connect the Power Good output via a resistor to the transistor base. Or (even more straightforward, it should only have an extremely dim state during the undefined stage when the voltage is very low), just connect the LED directly to the Power Good line through a resistor.

Jan Cumps · Answer

If the measure needs to be precise, there are dedicated ICs: 'Supply Voltage Supervisors'. They are 3-pin devices that can directly drive an LED if voltage is above a certain threshold. For 5 V monitoring, there's the TPS3809I50D (trips at 4.55 V). The EEVBlog µCurrent uses a 2.64V one of that family as battery monitor:

jc2048 · Answer

"It seems I was over-complicating things in my head!" But that's good. It lead to you asking questions, forming a specification for what you wanted to do, and then looking for a means to do it. You are learning to be an engineer. Your second circuit with the zener wouldn't work as you've drawn it. The zener doesn't allow current to flow until the voltage reaches 5V, so with only 5V as the source it never starts to conduct. If you make the zener about 2V, then you've got 2V for the zener, 2V for the LED and about 1V for the resistor. Current would start to flow when the input got up to 4V, and the brightness would then steadily increase as the voltage went to 5V [the brightness of an LED is proportional to the current]. Here's what it looks like in a simulation with a 2V zener. I've got the input (VG1 - a voltage generator) ramping up from 2.5V to 5V and then back down again. You can see the current (measured by AM1 - an ammeter) which flows through the LED start to increase just below 4V. The peak current you can adjust with the resistor (a lot of LEDS have reasonable brightness at just a few milliamps, so don't feel you have to drive it at 20mA) If you think about it for a moment, you'll see that instead of the zener you could use a second LED. (One of the great things about designing is that there are always several ways to do things.) Wouldn't make sense for a commercial design, but a quick fix if the only components you have are LEDs and resistors. When I said "you've got 2V for the zener, 2V for the LED and about 1V for the resistor," I was informally giving you one of Kirchhoffs laws. If you don't mind a bit of theory, have a look at Kirchhoff's circuit laws https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws specifically the voltage law. It looks a bit scary, but really it's just a bit of applied common sense. In English, he's just saying we can't get something for nothing. [Technically, Kirchhoff's voltage law doesn't apply to this circuit because the zener and the LED both have some capacitance that can store energy, but once everything settles, after a fraction of a second, the capacitances are no longer relevant and we can treat it as though Kirchhoff did apply.] Could I suggest that you get one of the free simulators that are on offer? That way you can learn to design for yourself without having to buy lots of components. But don't just rely on simulation; when you get a circuit that works well in simulation, build it and measure it and you'll learn what the weaknesses of simulation are and how the real world doesn't always conform to the models you are given (Mother Nature is a somewhat contrary old lady at times). TINA-TI SPICE-Based Analog Simulation Program http://www.ti.com/tool/tina-ti LTspice IV http://www.linear.com/designtools/software/#LTspice

How to detect 5V?

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