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I’m working on a wearable battery project and I think I need to use a regulator on the battery output.

Let’s say I were to connect the battery to an LM317 (detailed here) or an LM7812 (see figure 1) circuit with nothing else connected to the output. Would there be current flow through the circuit? Would the battery discharge more quickly than it would in storage?

LM7812 simple circuit

Figure 1: LM7812 Circuit

Furthermore, does the current draw (in amperes to clarify) from a voltage regulator depend on the requirements of the components connected? If I connected two LEDs in parallel to the output of a voltage regulator, would the current be greater than if only 1 LED was connected? Does the power draw (in watts to clarify) from a voltage regulator differ from the current, or does the current change in direct proportion to the power?

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  • \$\begingroup\$ Yes and yes. Furthermore, yes, yes and yes if the input voltage is unchanged. Main question : see Table 7.5 on the datasheet you linked. Adjust pin current is 50 to 100 uA. Worse, you need a minimum load, 3.5mA to allow it to regulate properly. (This is usually done by careful choice of adjustment resistors). 7812 is different, but no better. \$\endgroup\$ Dec 1 '20 at 19:43
  • \$\begingroup\$ If I connected two LEDs in parallel to the output of a voltage regulator Never directly connect LEDs to the output of a voltage regulator, always use a resistor in series with the LED to limit the current. You ask a lot of basic questions and that's OK, we've all had to begin somewhere. What I advise you to do is search the internet for similar projects and see what those people do, try to understand why they do things like that. Instructables.com is a good site to start. \$\endgroup\$ Dec 1 '20 at 19:51
  • \$\begingroup\$ ...linear regulators are wasteful, switches achieve much greater efficiency and modern ones are cheap... \$\endgroup\$ Dec 1 '20 at 21:16
  • \$\begingroup\$ Re, "...power draw..." Be sure you understand the difference between the power that the regulator delivers to the load, and the power that is dissipated (turned to heat) in the regulator. Suppose input voltage \$V_\text{in}\$, output voltage, \$V_\text{out}\$ and load current \$I\$. The Power used by the load is, of course \$IV_\text{out}\$. The power consumed by a linear regulator will be at least \$I(V_\text{in}-V_\text{out}).\$ It could be more than what the load consumes, depending on the voltages. switching regulators (see Chris Stratton's comment) don't have the same problem. \$\endgroup\$ Dec 1 '20 at 21:18
  • \$\begingroup\$ Btw I thought I knew the answers to all these questions, I’ve been working with electronics for years. I wanted to check because the project uses 4 LC18650s. Lithium battery chemistries are explosive and I want to be extremely careful doing any power electronics design with them. \$\endgroup\$ Dec 2 '20 at 4:44
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  1. Yes, regulators consume current even if nothing is connected to output. It requires current to operate and would not operate without current.

  2. Yes, due to non-zero current being consumed, it would drain batteries faster than when batteries are disconnected.

  3. Yes, current draw from regulator output is exactly the amount of current drawn by loads connected. The same current is drawn from batteries, plus the current required by the regulator itself to work.

  4. Yes, regulator output current would increase when connecting loads. Current needed by the regulator can vary slightly, but not so much it would matter much in general. In your case it might. Note that two LEDs in parallel on a 12V regulator output would get damaged unless the LEDs are connected with a resistor in series.

  5. Yes, power drawn from regulator is exactly the amount of power consumed by devices it powers. This does not include the power wasted in the regulator, as regulator input current and output current are approximately equal (if disregarding the current consumption of the regulator itself), but regulator input voltage is higher than output voltage. For instance if you have 24V battery, a 12V regulator, and 0.1A load, the load consumes 1.2W, but battery has to provide 2.4W, and the regulator converts the excess 1.2W to heat. Which is why linear regulators may not be suitable for every purpose.

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To add a bit of information to the existing answer (which directly answers all of the asked questions), it might be useful to know that more modern regulators (the LM317 is ancient) have stable outputs without almost any current consumption. For instance, the TPS7B81 (just as an example, there are many others) has a quiescent current of a few μA (typically around 3 μA). This allows powering very light loads (e.g. a microcontroller that is mostly sleeping waiting for external events) continuously from a battery, without draining it too much.

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