0
\$\begingroup\$

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.

\$\endgroup\$
7
  • \$\begingroup\$ How are you measuring the current? \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Mar 28, 2014 at 3:48
  • \$\begingroup\$ @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) \$\endgroup\$
    – Nyxynyx
    Commented Mar 28, 2014 at 3:58
  • \$\begingroup\$ 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. \$\endgroup\$
    – Kurt E. Clothier
    Commented Mar 28, 2014 at 3:58
  • \$\begingroup\$ It's not a good idea to directly use a voltage rail as a reference voltage for exactly this reason. \$\endgroup\$
    – tcrosley
    Commented Mar 28, 2014 at 3:59
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Kurt E. Clothier
    Commented Mar 28, 2014 at 4:04

3 Answers 3

Reset to default
1
\$\begingroup\$

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

\$\endgroup\$
11
  • \$\begingroup\$ 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: circuitlab.com/circuit/7f3ndq/zener-diode-voltage-reference \$\endgroup\$
    – KyranF
    Commented Mar 28, 2014 at 4:30
  • \$\begingroup\$ 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... \$\endgroup\$
    – Kurt E. Clothier
    Commented Mar 28, 2014 at 4:39
  • \$\begingroup\$ @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? \$\endgroup\$
    – Nyxynyx
    Commented Mar 28, 2014 at 4:53
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Nyxynyx
    Commented Mar 28, 2014 at 5:05
  • 1
    \$\begingroup\$ @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. \$\endgroup\$
    – Passerby
    Commented Mar 28, 2014 at 5:49
0
\$\begingroup\$

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).

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.