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?

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    \$\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 Jan 28 '18 at 10:35
  • \$\begingroup\$ @Transistor you are also correct! Goes down to 16V \$\endgroup\$ – Arturs Vancans Jan 28 '18 at 10:50

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.

| improve this answer | |
  • \$\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 Jan 28 '18 at 15:34

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