3
\$\begingroup\$

I have a circuit of 6 solenoids connected to Darlington Array (http://www.ti.com/lit/ds/symlink/uln2803a.pdf) and driven by 18V (2x9V batteries) as can be seen below:

enter image description here

From what I understand, each load should draw 200mA (18V/88Ohm) and when all of them are switched on at the same time, the total draw should be 1.2A. They are all in the active state for a brief moment (20ms).

I wanted to confirm my calculations by measuring the total current draw with a multimeter connecting it in series with the current source.

The measurements showed a maximum of 47mA draw instead of 1.2A.

What could be the reason why the actual measurement does not match the theoretical?

\$\endgroup\$
2
  • 3
    \$\begingroup\$ Is the voltage still 18 V with all six solenoids turned on? You will find that your batteries have internal resistance. The terminal voltage will drop as current increases. \$\endgroup\$
    – Transistor
    Commented Jan 28, 2018 at 10:35
  • \$\begingroup\$ @Transistor you are also correct! Goes down to 16V \$\endgroup\$ Commented Jan 28, 2018 at 10:50

1 Answer 1

7
\$\begingroup\$

Your meter is not fast enough to capture the 20ms current pulse it is averaging it over a period of time to give you the reading you see.

If you have a scope monitor the voltage across a small resistor connected in series with the voltage source to see the real current.

\$\endgroup\$
1
  • \$\begingroup\$ And do not forget that, depending on solenoid inductance, time constant might as well be so long w.r.t 20ms Ton not to let build significant current. \$\endgroup\$
    – carloc
    Commented Jan 28, 2018 at 15:34

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.