This question arose from a conversation over the weekend and we tied ourselves in knots trying to get to the bottom of it. I didn’t have time to put a simple circuit together so it would be great to get some expert thoughts on this.

Given a circuit as follows: circuit with a voltage convertor in the middle, a load on the right and supply on the left, both with voltmeter and ammeters, an open switch connects the two grounds

With a 12v supply attached we get the following readings: V1 - 12v A1 - 50mA V2 - 5v

(1) With SW1 open, assuming the buck convertor is 80% efficient, what is the expected value of A2?

(2) With SW1 closed, same question/assumptions.

(3) Why?

(4) What is the correct way to talk about the difference between measurements at A1 and A2. Is it correct to say that the excess current (46mA) flows into the negative terminals of the buck converter?

Thanks in advance!


This is a simplified version of a real circuit that promoted some discussion. The original components were as follows:

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    \$\begingroup\$ Closing the switch makes no difference. Why? KCL. Also, closing the switch does not put A1 and A1 in series. \$\endgroup\$ – Elliot Alderson Aug 13 '18 at 11:43
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    \$\begingroup\$ and also: almost all buck converters have input - and output - shorted. This means SW1 is always shorted. If this is a buck converter where the + sides are shorted then also short them on this schematic to make things more clear. \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 11:48
  • \$\begingroup\$ Thanks. What is KCL? Also why are they not in series? Because of the buck convertor ground connections? \$\endgroup\$ – chrismilleruk Aug 13 '18 at 11:58
  • \$\begingroup\$ @Bimpelrekkie not sure about the ones I have. I’m guessing whatever is standard. I could take a closer look later today if that helps. \$\endgroup\$ – chrismilleruk Aug 13 '18 at 12:01
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    \$\begingroup\$ What is KCL It is Kirchhoff's Current Law. Very basic stuff anyone dealing with electronics needs to understand. Go read: en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 12:08

Forget about the SW1 switch open or closed it doesn't matters. Because the input and output are not completely isolated(means -ve will be common for input and output).They are already interconnected inside the buck converter itself. The answer to your first question is that in theory ,you can take 96mA output from A2 but it still purely depends on the load.

Buck converter efficiency will vary depending on the load. If you have lighter load on the output side,the efficiency will be much good. If you have more heavy load,then it's efficiency will start to degrade.The degraded performance is due to dissipation of heat due to heavy loads. So practically it not the case that you'll always have constant efficiency irrespective of the load at the output of the buck converter.

  • \$\begingroup\$ Ok, that’s what I was expecting. Interesting point about the efficiency, I didn’t know that. \$\endgroup\$ – chrismilleruk Aug 13 '18 at 12:04
  • \$\begingroup\$ Yup, Kindly go through the datasheet of LM2596,which is a general buck converter. I'll also recommend you to go through 34063 dc-dc regulators,so that overall you'll get more information. \$\endgroup\$ – Aadarsh Aug 13 '18 at 12:16

Allow me to redraw that schematic into a more realistic/practical situation:


simulate this circuit – Schematic created using CircuitLab

Here the buck converter is a "positive" regulator meaning that in- and out- are shorted in the converter as well.

(1) The input power is 12 V * 50 mA = 0.6 W, at 80% efficiency that would mean

80% * 0.6 W = 0.48 W at the output.

At 5 V that would be 0.48 W / 5 V = 96 mA => Iload = 96 mA

This means the load is then equivalent to 5 V / 96 mA = 52.1 ohm

(2,3) When SW1 is closed a typical Buck converter will switch off. That is because when SW1 closes 12 V is connected directly to the output making the output voltage larger than 5 V. A buck converter designed for a 5 V output voltage will then decide that it does not need to do anything (the output voltage is high enough) so it will switch off (or at least, not provide more power to its output).

Assuming the Rload is still 52.1 ohm and that this buck converter also switches off and consumes no current: Iswitch = Iload = 12 V / 52.1 ohm = 0.23 A. Do note that the load now gets 12 V, not 5V.

  • \$\begingroup\$ Thanks. This makes sense but I’m not sure why you moved the current sensors and switch to the top rail. Doesn’t this change the constraints of the question? \$\endgroup\$ – chrismilleruk Aug 13 '18 at 12:39
  • \$\begingroup\$ Let me reverse the question, why do you have the current meters in the negative line? I place them in the positive rails because that allows me to have one common ground everywhere which is 1) easier to understand 2) common convention 3) prevents grounding issues. \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 12:53
  • \$\begingroup\$ No this does not change anything. You were using a buck converter with a common positive rail, which is extremely uncommon. It will be a challenge to implement the system like that. I just changed the ground reference, this does not influence the calculations I did. \$\endgroup\$ – Bimpelrekkie Aug 13 '18 at 13:00
  • \$\begingroup\$ The question is an (over-)simplification of a real circuit in a breadboard. The components in use are cheap little pcbs. For reasons unknown, the ammeter (A1) only functions on the return side / -ve supply terminal. A2 doesn’t exist since it is the focus of the question and SW1 is analogous to “should I short the -ve supply and -ve load terminals on the buck convertor”. \$\endgroup\$ – chrismilleruk Aug 13 '18 at 13:18
  • \$\begingroup\$ You mean that "in- and out- are shorted", don't you? \$\endgroup\$ – Elliot Alderson Aug 13 '18 at 14:31

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