I'm trying to create a circuit with two loads in parallel ( essentially piggy backing one circuit on the power supply of the other ). Circuit 1, the main circuit requires ~900ma at 12v. Circuit 2, requires ~70ma at 5v. To step down I'm using a buck regulator that provides 300mA at 5v. The main supply is 12v DC rated for 900mA, but in reality is providing more like 1.3A. If I connect the circuits in parallel without balancing with resistors, all the current flows through circuit 1 ( its load will consume all available current ).

Now, correct me of I'm wrong, but I think I should add a resistor to each of the two circuits to manage current. So circuit 1 should get 12v / 0.9a = ~13 ohm. For circuit 2 though, I'm umsure how a buck regulator functions in this context. Should I add a 70 ohm ( 5v / 0.07a ) resistor after the buck regulator, or a 170 ohm ( 12v / 0.07a ) before the regulator. Or would both have the same effect?

Or perhaps I have it all wrong, so any advice appreciated. Pseudo circuit attached:

enter image description here

======= UPDATE 1 ======= Circuit 1 is a terminal for a scale unit. So logic board, power to a high capacity battery and a display, and excitation for the load cell itself. All encapsulated.

Circuit 2 is a rs232 converter and bluetooth chip. I took the 1.3A reading across Vin and Gnd with just the terminal and buck converter in parallel. I'm not sure how to resolve this difference, essentially the DC transformer supplied with the terminal is rated at 12V 900mA so that was my assumption on the load of the terminal. I thought 70mA was probably not going to go missing.

======= UPDATE 2 ======= Based on Peter and Photons advice, and the discovery that I had misused my multimeter and blown the circuit by connecting the two probes in parallel (across the circuit) and not in series, I have now managed to connect everything up correctly and it works fine (as per the diagram with no additional resistors).

I'm not sure what the actual load from Circuit 1 is until I replace the multimeter's fuse, but it is less than the 900mA the supply is rated for, so the addition of ~70mA on Circuit 2 is not a problem and works fine.

  • \$\begingroup\$ It's very unlikely what you're suggesting will work. Unless circuits 1 and 2 are purely resistive, you aren't telling us enough to know how the current draw will be reduced by adding resistors in series with their power connections. But the most likely scenario is that before their current draw drops enough, you will reduce their available power supply voltage below the minimum spec. \$\endgroup\$ – The Photon Jan 15 '14 at 21:53
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    \$\begingroup\$ If you can say what are circuits 1 and 2 (digital logic? op-amps? rf amplifiers? motors? solenoids? ???) we can maybe give you a more complete explanation of whether this will work and why not. \$\endgroup\$ – The Photon Jan 15 '14 at 21:55
  • \$\begingroup\$ Also, your numbers don't add up. 900 mA + 70 mA * (5 / 12) * (efficiency factor) is nowhere near 1.3 A. Where are the other 350ish mA going? \$\endgroup\$ – The Photon Jan 15 '14 at 21:58
  • \$\begingroup\$ thanks photon, and sorry for lack of details, thought simpler might be better. \$\endgroup\$ – user35616 Jan 15 '14 at 22:37
  • \$\begingroup\$ Have edited and added further details \$\endgroup\$ – user35616 Jan 15 '14 at 23:03

You said you measured the 1.3 Amps across Vin and GND. This is NOT the way to measure the current capability of a power supply - it does give you the short circuit current, but that is usually not a useful value. If the existing power supply claims to be 12 volt and 900 mA, you should believe that current rating, and not attempt to draw more currrent.

To measure current, you must connect your meter in series with the circuit - you break the circuit to do this.

You should measure the actual current drawn by the 12 volt load. If it is less than 900 mA, then the difference is the current you have available to power your step-down converter.

As others have said, the 300 mA rating of the stepdown converter is the maximum it can supply, not what it will actually draw. Since the stepdown ratio of the converter is 5/12, we can expect the current drawn from the 12 volt supply to be a bit more than 5/12 of the load current - perhaps 35 mA. (As a rough approximation, we can assume that the power into the converter equals the power out, plus some losses in the converter.)

The total current drawn from the 12 volt supply will then be whatever you measure for the scale, plus the 35 mA or so for the step-down converter and 5 volt load.

  • \$\begingroup\$ Thanks Peter. After retrying to measure and getting nothing, I spotted that was my mistake. Of course I shorted the circuit through the multimeter, and read the spiked reading just before the fuse blew... I have since rewired the circuits on the breadboard from scratch, without resistors and all works perfectly. \$\endgroup\$ – user35616 Jan 17 '14 at 17:13

This is not a complete answer but meant to clear up a couple of misconceptions that appear in your question.

  1. If the load on your 5 V converter is 70 mA, the converter will only produce 70 mA. The 300 mA is a maximum rating that doesn't affect how much current is drawn from the 12 V supply.

  2. If you reduce the input voltage to the buck converter, it will actually increase the current it draws. This is because in order to produce Vout from Vin with conservation of energy you have

$$I_{out} \cdot V_{out} = E \cdot I_{in} \cdot V_{in} $$

(\$E\$ is an efficiency factor)

Since the regulator is trying to keep Vout the same, and the load will require the same current as long as Vout is the same, Iin will have to go up when Vin goes down.

  • \$\begingroup\$ So, if the load is 70mA at 5v after the buck, then the buck will draw 29mA ( X E which is ~90% ) at 12V from the main supply? \$\endgroup\$ – user35616 Jan 17 '14 at 17:38
  • \$\begingroup\$ You'll have to divide by E, not multiply (because power in > power out); but otherwise, yes. \$\endgroup\$ – The Photon Jan 17 '14 at 18:05

A buck regulator (normal ones, anyway), maintain a fixed voltage at the output by providing whatever current is required to maintain that voltage. It is, as are most power supplies, designed to approximate a voltage source.

If your main supply is rated for 900mA but you are drawing 1300mA from it, you are overloading it. This is bad and nothing is guaranteed to work under these conditions.

You need a bigger main supply. Your 12V circuit already requires the full rated current of your power supply. How can you expect it to also supply the 5V circuit if it's already loaded to the maximum?

  • \$\begingroup\$ Thanks phil. see above for details regarding current discrepancy and what's on the ends of each circuit. I'll remeasure when I get a chance. I had assumed that the terminal wouldnt require the full 900mA from the default transformer supplied with it. \$\endgroup\$ – user35616 Jan 15 '14 at 22:49
  • \$\begingroup\$ So does that mean the regulator will draw 300mA ( what it is rated for ) regardless of what is attached to Vout? That, and poor readings, might explain my ~300mA overloading. \$\endgroup\$ – user35616 Jan 15 '14 at 22:55
  • \$\begingroup\$ @user35616 no, a buck regulator draws only enough current from the input such that input power equals the output power, plus losses. See electronics.stackexchange.com/q/34745/17608 and electronics.stackexchange.com/q/85450/17608 . If you measure current higher than that, you probably have a fault in your circuit or have misread the specifications. \$\endgroup\$ – Phil Frost Jan 15 '14 at 23:10
  • \$\begingroup\$ Thanks. That and Photons explanation make sense of what the current draw should be. I'm going back and checking both circuits separately now. \$\endgroup\$ – user35616 Jan 15 '14 at 23:31

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