Measuring current consumption on a split rail. - Page 1
Forum section: electronics / beginners

author: paulca
author_rank: Super Contributor
timestamp: February 26, 2018, 11:36:53 am
content: Are there any tricks to measuring current on a +-15V split rail? If I measure current no the common will that give me total current?Or do I just measure both + - and add them?

author: StillTrying
author_rank: Super Contributor
timestamp: February 26, 2018, 12:07:59 pm
content: "If I measure current on the common will that give me total current?"That would give youthe differencebetween the +/- rail currents."Or do I just measure both + - and add them?"That's the only way.

author: newbrain
author_rank: Super Contributor
timestamp: February 26, 2018, 08:20:25 pm
content: Quote from: paulca on February 26, 2018, 11:36:53 amMeasure both.But adding them make very little sense.Simple proof: consider your + 15V and -15V with a perfectly symmetrical load with 1A current draw.As StillTrying says, the current in the common rail will be the difference 1A-1A = 0A.If no current is flowing in the common rail, I can cut it without altering the circuit behaviour.So, finally, my circuit is drawing 1A (across 30V, of course).What if I had summed the current in the two branches?2A, nowhere to be seen.

author: BrianHG
author_rank: Super Contributor
timestamp: February 26, 2018, 10:57:23 pm
content: Quote from: StillTrying on February 26, 2018, 12:07:59 pmOnly if you measure each current and convert it to Watts, then you can add the 2 wattages together. Once you work your way backwards from wattage to current over the total supply voltage, you will see that the current will be the average of the 2 currents measured, not the sum.Example #1:measure 1.0 amp on +15v = 15 wattsmeasure 1.0 amp on -15v = 15 wattsTotal power = 30 watts.-15v to +15v = 30v30 watts / 30v = 1 amp total.Example #2:measure 1.5 amp on +15v = 22.5 wattsmeasure 0.5 amp on -15v = 7.5 wattsTotal power = 30 watts.-15v to +15v = 30v30 watts / 30v = 1 amp total.or, 1.5 amp + 0.5 amp = 2 amp2 amp /2 = 1 amp average... (This average only works since your +&- supply are equal in voltage, otherwise you need to calculate everything in wattage, then work your way back to current as in the above example)See, all 3 calculations give you the proper 1 amp.Now, if what would happen if we summed the current, using example 1:1 amp @ 15v + 1 amp @ 15v = (ignoring the voltage and just adding the 2 currents together) = 2 amp.2 amps across 30v = 60 watts???This is wrong, this circuit isn't consuming 60 watts, so, just adding the 2 currents together is wrong.I hope I got this right...

author: Brumby
author_rank: Unknown
timestamp: February 26, 2018, 11:40:00 pm
content: For measuring power supplied, it is perfect, but for measuring current, I don't see how it is helpful.What value is there in saying a current (however qualified) of 1A when one rail is 1.5A and the other 0.5A?Quote from: BrianHG on February 26, 2018, 10:57:23 pmThis part you got wrong. You added the currents AND the voltages. It would have been less incorrect to say 2A across 15V.

author: Nusa
author_rank: Super Contributor
timestamp: February 27, 2018, 12:54:24 am
content: If we'd been given the circuit attached to the power supply, the answer might have been trivial. But since we haven't the best answer is to look to Kirchoffs laws to analyze the situation. Which, if properly understood, probably would have prevented the question from being asked.Here's a 2-video sequence that uses a similar problem (+24 and -12 instead of +/- 15):If you watch that and need help understanding how to reduce a matrix by hand, this video will do. It's another analysis, but skip to 4 minutes if you just want the math part.

author: Brumby
author_rank: Unknown
timestamp: February 27, 2018, 02:51:00 am
content: Quote from: Nusa on February 27, 2018, 12:54:24 amI don't see how having the circuit would make any difference.The question was about measuring - not analysing:Quote from: paulca on February 26, 2018, 11:36:53 amThe fundamental question is:For whatpurposedo you want to know this "total current"?Only after this is clearly stated can we evenbeginto consider if a single number answereven exists.

author: not1xor1
author_rank: Frequent Contributor
timestamp: February 27, 2018, 06:02:32 am
content: Quote from: paulca on February 26, 2018, 11:36:53 amIMHO, while you might measure and calculate total power usage, "total current" has no real meaning.You might calculate total power / total voltage to get an equivalent total current, but I think that would have no real purpose.You should just rather check how much current flows through the positive and negative rails just to see if it is within safe limits.From the sum of the two currents you can calculate the current flowing through the central rail.Of course one of the two values is negative so if there is no current through the central rail, the sum is zero.

author: Nusa
author_rank: Super Contributor
timestamp: February 27, 2018, 07:41:34 am
content: Quote from: Brumby on February 27, 2018, 02:51:00 amIt certainly could. For instance, if the circuit isn't even connected to the common rail, then you simply measure the current on one side and you're done. Don't have to measure the other side, as it's the same, nor do they get added up.Quote from: Brumby on February 27, 2018, 02:51:00 amThere were THREE questions asked. Those answers are not really, probably not, and no, but i think my general-purpose answer was better.Your question is an improvement, but asking for the purpose is nearly the same thing as asking for the circuit so you can determine whether the answer is trivial or not.

author: paulca
author_rank: Super Contributor
timestamp: February 27, 2018, 07:42:48 am
content: The circuit has 2 dual op amps and associated passives. It is powered by a switch mode isolated spilt rail PSU which is being fed from my bench supply. It is showing to be consuming too much current when I compare the consumption with and without my circuit.However I can't be sure exactly how much the split rail supply is wasting without measuring what is actually being consumed by the circuit.The bench supply shows the quiescent current of the split rail PSU as 30mA. Which is close to the datasheet figure of 25mA. When the circuit is connected it shows as consuming 76mA. A simple subtraction and a adjustment for 84% efficiency stated in the datasheet for the PSU would be 46ma * 0.84 = 38.64mA ... which is far too high for 2 dual op amps.However the datasheet for the PSU only gives one efficiency figure for full load and it itself is rather confusing. It claims at +-200mA it uses 597mA supply current.I figured it was a better idea to actually measure the current in my circuit to see if I am indeed pulling too much current or if the PSU has poorer efficiency that stated at a low load.The op amps have a rated maximum of +-10mA.

author: Zero999
author_rank: Super Contributor
timestamp: February 28, 2018, 04:55:12 pm
content: To get the total power dissipation, any two of the three power supply currents need to be measured, whether it be the +V & 0V, +V & -V or -V and 0V. As long as the current in two of the power supply conductors is known, the current through the other conductor can be calculated, using Kirchhoff's current law.