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A regulated power supply RAC02-055C/277 provides 5V to an Arduino driving 2 relays. The power supply is rated to provide 2W of power.

Connecting a multimeter between one of the leads of the power supply to the +5V bus (which connects to Arduino 5V pin and the +5V pins of the relays), and another multimeter across the +5V and GND rails, the current reads 98.6 mA and the voltage is 4.969 V.

When one of the relay is turned on, the readings are 209.4 mA and 4.911 V.

When both relays are turned on, the readings are 301.9 mA and 4.869 V.

How can we prevent the voltage output from the supply from dropping/changing whenever the relays are turned on and off? The +5V rail is used to bias a input signal, so a constantly changing +Vcc will give inaccurate readings.

Nyxynyx
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  • How are you measuring the current? – Ignacio Vazquez-Abrams Mar 28 '14 at 03:48
  • @IgnacioVazquez-Abrams Using the multimeter, with the red leads connected to the `400 mA` jack of the multimeter, using the `mA` measurement mode in `DC` setting. The multimeter is connected between the +5V output lead of the power supply and the +5V rail on the breadboard (which is connected to Arduino's 5V pin) – Nyxynyx Mar 28 '14 at 03:58
  • Also, what is your voltage input? What is the exact part (regulator) you are using? There are 2 available with 5V output, the first can't handle either load and neither can handle your total current load. When a load nears its max, the voltage level will drop. – Kurt E. Clothier Mar 28 '14 at 03:58
  • It's not a good idea to directly use a voltage rail as a reference voltage for exactly this reason. – tcrosley Mar 28 '14 at 03:59
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    My last comment was a bit off about handling the total load, I was adding your loads together... but, the data sheet does give a +-5% output voltage tolerance. 5% of 5V is .25V, meaning it could go down to 4.75V and still be within spec of the datasheet for normal operation. – Kurt E. Clothier Mar 28 '14 at 04:04
  • @KurtE.Clothier The input voltage is 110VAC. I'm using the `+Vo` and `-Vo` outputs of the power supply. I'm using Fluke 179, so the burden voltage for the 400 mA input is 2 mV/mA. Does it mean if I were to use the 3W version (RAC03-05SC/277), the voltage drop will be lesser? – Nyxynyx Mar 28 '14 at 04:11
  • It's possible, considering an output is usually more stable under less load. But like I said before, the output you describe is within the given tolerance in the datasheet, so it is within spec. – Kurt E. Clothier Mar 28 '14 at 04:41

3 Answers3

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According to the datasheet, the voltage drop is within spec:

"Load Voltage Regulation 10% to 100% full load ±6% max." If "full" load is 400mA for a 2W, 5V output version, then you are getting up to 75% load by then. (4.869V / 5.0V ) * 100 gives us 97.38% of load voltage regulation, which is still within reason.

I suggest you use a more constant reference.. Some people use a reference (zener) diode or maybe another small linear regulator that goes from 5V to 3.3V but its more of a reference than a 'power supply' so to speak. Basically you need something to act as a stable reference, and not ever-changing due to such large changes in load as one or both relays turn on.

Edit: To use a resistor and zener diode as a voltage reference, see this question and it's answer by Anindo for how to do it, and calculate the required resistance to make it work. Using a Zener diode as voltage reference

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KyranF
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  • you can put a resistor in series with a zener diode to ground from your regulator output, and use the voltage over the zener as the reference to your biasing circuit. A really simple example is shown here: https://www.circuitlab.com/circuit/7f3ndq/zener-diode-voltage-reference/ – KyranF Mar 28 '14 at 04:30
  • I'd imagine you are meaning to use a Zener shunt regulator to drop the 5V output down to 3.3V for the reference rather than to replace the power supply regulator... – Kurt E. Clothier Mar 28 '14 at 04:39
  • @KurtE.Clothier I tried a LM317T regulator to output a 2.0V, connecting its input pin to the *changing* +5V rail from the power supply. I do not have any zener diode with me now. However I notice it drops from 1.90V with no relays turned on, 1.63 V with one relay on, down to 1.496 V with both relays on. Does it mean this regulator is not suitable? – Nyxynyx Mar 28 '14 at 04:53
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    Sorry, the drop in LM317T's output is because the power supply was accidentally disconnected and Arduino was powered by USB instead. LM317T gives a steady 1.999V to 2.0V when the relays are turned on and off. Will be using this 2.0V as the reference voltage. – Nyxynyx Mar 28 '14 at 05:05
  • LM317 can be tricky to get right...I'd suggest something with a constant output. – Kurt E. Clothier Mar 28 '14 at 05:05
  • @KurtE.Clothier Will a 2V zener diode be simplier? How do you decide when to use a zener diode and when to use a regulator like LM317. – Nyxynyx Mar 28 '14 at 05:06
  • well the regulator is more expensive, and larger. A 2V zener is probably $0.20 and only needs a simple resistor as it's current source to maintain the voltage over the zener. Look around for examples. The Art of Electronics by Horowitz and Hill (1980) on page 331+ (chapter 6.14) discusses zener diodes as the most basic voltage references, and their temperature considerations. The book suggests that below the 6V mark, they are very stable with regard to current changes. A simple resistor and zener reference is definitely a good option for your case @Nyxynyx – KyranF Mar 28 '14 at 05:31
  • @Nyxynyx also the LM317 will suffer the same load and line issues! Where the Zener will be stable under changing load/line for most cases – KyranF Mar 28 '14 at 05:33
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    @Nyxynyx a simple reference voltage generator only needs a few hundred nA or ones of mA. A Zener is great for that. A full regulator like the lm317 is more for a regulator power source. It can do both, but a lm317 is overkill for what you need. – Passerby Mar 28 '14 at 05:49
  • See discussion here about the same topic, and the use of them as voltage references. The answer for the question by Anindo shows how to calculate the required resistance to give appropriate results. This is based on the datasheet though, so make sure you look at that. http://electronics.stackexchange.com/questions/57977/using-a-zener-diode-as-voltage-reference I have edited my original answer with this suggestion – KyranF Mar 28 '14 at 06:07
  • You could also check out this guide I made... its for a Zener Shunt Regulator, but the principle is pretty much the same. http://www.instructables.com/id/Zener-Diode-Shunt-Regulator/ – Kurt E. Clothier Mar 28 '14 at 15:47
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The biggest things for load regulation are distribution loss (conductive or copper loss), and specified load regulation. In this case the output is dropping about 2% over your test load range. This is within spec for the part (+/-6% from 10% to 100% load). For conductive loss supplies frequently will include remote sense lines, but this model doesn't have any remote sense. It looks like the only thing you can do for a DC drop if you plan to continue to use this supply would be enhanced copper thickness or shorter distribution (and even then it sound like it could still fall short just by performing to spec).

gsills
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Your power supply has an output voltage accuracy of 5%, or +- 0.25V for a 5V supply. It is behaving within its specified limits!

You should use a separate power supply for the bias voltage. If the relais stay on for only short periods of time, it might be possible to decouple the bias voltage using capacitors & a resistor in the line charging it.

EvertW
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