basics of adding parallel loads to a circuit - Page 1
Forum section: electronics / beginners

author: carpetpaul2129
author_rank: Contributor
timestamp: October 23, 2022, 10:55:01 am
content: As ever, I'm a slow learner, forgive me, but perhaps you can help my understandingTheres probably quite a lot of questions below, if you're interested, please just pick out aspect that you feel you can comment on, many thanks as alwaysI'm slowly working on building some 18650 battery packs, and my side project is pondering the basic electrical conceptsI have a bench power supply set at the 12v and 2amp that my battery charger requires. I noted that the cells seem to get quite warm during discharge/charge (does seem to be normal since I have two identical chargers) and in a moment of boredom, not seriousness, I grabbed a small 12v computer fan off my shelf and connected in parallel with the charger, all was well, the fan operated seemingly at its intended speed, and the wattage shown on the power supply went up a fraction, fine, as expected. It got my wonderingbelow, I've numbered a few statements, perhaps you can let me know if I'm correct with my thinking ?sorry again that this is real school boy stuffas I think I understand it, in parallel circuits, each load receives the same voltage, (1) yet the resistance apparently goes down since there are more paths for the current to flow.I've just been and taken some measurements (from the display on the power supply)with just the charger connected (likely just discharging) ... I have 12.1v / 0.09amps, so 134.44ohms ?with two small computer fans also connected ... I have 12.1v / 0.18amps, so 67.33ohms ? (2) and a lower ohms value shows that the resistance in the circuit has decreased (3) thus allowing more current to flow around the circuit (4) and more power to be consumed(5) what are the practical limitations of this setup (I don't mean I'm interested in adding fans to a battery charger, simply understanding the principal of adding loads) ?(6) what would happen if I reach and attempt to exceed the 2amp limit which I've set on the power supply ?and now for more silly questions :(7) increasing the amps of the power supply will affect nothing other than the allowable maximum(would I be correct in assuming that this type of load, charger and fan alike, simply 'take' what they need in terms of power, or recieve what they us in terms of energy ? or in terms of power ? or just have a resistance themselves ? and that allows them to do a particular amount of work ? they are 'accepting energy' based on what's available rather than it being forced through them ?(9) however, if I turn the voltage up, I can burn out my load, how is this explained ?thank you for readingPaul

author: KT88
author_rank: Frequent Contributor
timestamp: October 23, 2022, 12:47:12 pm
content: Such loads don't respond like a (Ohmic) resistor as they - like you have a hunch already - take current based on their demand.If you block the fan as an example the load would change -it could go up first and then drop if the fan detects an overload condition.The Battery charger, if it is a swithched-mode type would draw current based on the charge power, which varies over time with the SoC of the batteries.Also swithched-mode regulators draw (almost) constant power - that means if you increase the supply voltage the current would drop - this appears like a negative resistance.Of course you only can vary the supply voltage within the allowed limits of the load.In case of a switched-mode charger and a linear supply it would be benficial to increase the supply voltage. This would reduce the voltage drop inside the power supply thus reducing the overall losses...CheersAndreas

author: Brumby
author_rank: Unknown
timestamp: October 23, 2022, 12:55:43 pm
content: Quote from: carpetpaul2129 on October 23, 2022, 10:55:01 amYou have a pretty good understanding of the basics - so don't be too hard on yourself.QuoteNot a problem. It's always good to get the basics straight.QuoteExactly right.QuoteExperimental demonstration of what you outlined above.QuoteEach additional load connected to a constant voltage supply (the most common type) will draw its own current - as you have already worked out. Adding more loads will increase the current drawn up the the stated limit of the supply. Once you exceed that, there are a few things that can happen, depending on the supply and the settings.1. For a supply with current limiting circuitry, the voltage will be reduced in order to keep the current from exceeding the selected maximum current. The supply should survive this.2. For a supply without current limiting circuitry, the supply will do its utmost to provide the current being demanded. While most supplies will have a little extra "headroom" and can handle a bit more current for a time, there is a point where the demand is just too much - which usually results in two things:a) Voltage sag. While a little bit like current limiting, this is purely a factor of the capability of the components being pushed further than they are meant to be, which usually results in:b) Overheating. Heat is the number on enemy of electronics. Things get too hot and the magic smoke escapes. That's usually when things die.QuoteCorrect.QuoteAgain, correct. Voltage is what is applied to a load and the current which flows is dependent on that load. A 2 amp (capacity) supply set at a given voltage will not force 2 amps into a load. The load will take whatever current it "needs".QuoteThe answer to this can get a bit complicated as it depends on the load - however if we stick to a simple resistive load, you will get the idea.Say we have a supply set at 10V and connect a 100 ohm resistor. A current of 100mA will flow through that resistor giving it 1W of power which will get turned into heat. Assuming the resistor is rated for 1W operation, then all is well.Now let's look at bumping the voltage up to 20V with that same resistor connected. The current is now 200mA which means there is now 4W of power being handled by the resistor. This resistor is now going to heat up WAY more than it was designed to handle and it will fail. Note that it won't fail immediately - but over a period of time ... which, in this case, is likely to be measured in seconds.QuoteHope this helps a bit.

author: rstofer
author_rank: Super Contributor
timestamp: October 23, 2022, 04:03:58 pm
content: All of the above follow from Ohm's Law, Kirchhoff's Laws, Thevenin's Theorem and Norton's Theorem. There are videos on each of these topics. It is definitely worth the time to nail down the basics. There are also complete DC Circuits courses on the Internet.I'm not sure this particular course discusses Thevenin and Norton:https://www.engineer4free.com/circuits.htmlThere are many other links directed at those specific topics.

author: carpetpaul2129
author_rank: Contributor
timestamp: October 23, 2022, 05:12:59 pm
content: wow, thank you so much for the comments, really great, I'm loving this topic, it's both seemingly insanely complicated, yet as I slowly start to understand it, seems straightforward, what a subjectindeed, I will check the link provided and continue learningMany thanks for your replies, as always,Paul

author: carpetpaul2129
author_rank: Contributor
timestamp: October 23, 2022, 05:20:45 pm
content: https://www.engineer4free.com/circuits.htmlthis link is really useful to me, thank you so much, I've got much watching to do !Paul