It is hard to see the circuits from your schematics with all those lines in them. There is just too much clutter all around.

@Olin Lathrop‭, I use simple conceptual circuits with a small number of generic elements. This allows me to use the rest of the space to visualize the invisible electrical quantities voltage and current. My goal is to reveal the idea and explain the circuit in order to deeply understand it. Some draw voltmeters and ammeters for this purpose; I prefer more abstract voltage bars and current loops colored red and green. But these are only means; the end result is only important.

A tinkerer solution, probably not standard EE. But that's what I am, after all, a tinkerer. So I give you a +1 for the simplicity of this circuit. I would have done it in a somewhat "more standard" way with the battery connected to the Vcc+ via a diode, and the base of the transistor driven by an R-zener as you've evoked previously. A bit less simple but perhaps more logical.

@coquelicot‭, You probably understand that here I wanted to show something more original than usual. BTW the input source may be weak since the alarm system does not normally consume much current. But when the bell turns on and consumption increases sharply, the backup battery turns on in parallel and helps the weak input voltage. I also remember I had come up with an original circuit for AC power supply through the bell above the apartment front door so there was no need for additional wires.

@Circuit fantasist, I can barely follow your story. There are some confusions: (1) What exactly is the thing labeled "Bb"? If in your final schematic, is it true that there is no longer any Bin, and is now moved to Bb? (2) Your circuit seems to work, but there is one big problem: The load current now flows through R, which of course wastes power, and (3) Can you give an example of how to calculate the value for R? (4) You can use my example: (a) Battery = 12V, Load = 10 Ohm, I ~= 1.2A.

@tlfong01, I will answer you but later because after a while I will have to start an online exercise with my favorite students on how to measure the diode IV curve. They are IT beginners and I will have to discuss exactly the questions you ask. If you don't mind, we will come in here and try to answer your questions:) You can get an idea of how I conduct these excersises from https://www.researchgate.net/post/How_do_we_investigate_semiconductor_devices_in_the_educational_lab.

@Circuit fantasist, no hurry and no problem at all. Actually I forgot most of the regulated power supply I learned ages ago. But I still remember at that time I used a NPN BJT called 2N3055 to step down voltage and of course wasted a lot of energy, in this "linear" regulator power supply. Your battery regulated PSU is not I have not thought off, so need to think about it for a while. Cheers.

@tlfong01‭, Thanks for the response; it means a lot to me. I have conducted an interesting on-line exercise using various web resources including Codidact. We considered this story as an example of how a BJT can act as a diode. Bb means "backup battery". Bin is the input unregulated power supply. In the "final schematic", when there is no Bin, Bb begins performing its role. Usually, the return current can not enter Bin since there are backward biased diodes inside it...

... Also, R is high enough and the current and wasted power should be small. I am not a fan of calculations but I think R could be defined as follows: R = (Vin - Vb)/(Ich + IL/(beta+1)).

Thank your for your clarification. Your answer seems to have added one more confusion: "Usually, the return current can not enter Bin since there are backward biased diodes inside it...". What do you mean by "backward diodes inside the "battery"?

@tlfong01‭, Thanks for the response. It is strange why others keep silence... Browse the web for diode rectifiers (e.g., https://en.wikipedia.org/wiki/Diode_bridge) and you will see that, in this case, the diodes inside the rectifier are backward biased... and they stop the return current.

(1) It is strange why others keep silence - I don't feel strange because I understand you and the EE SE Q&A situation better. (2) When I asked about the backward biased diode inside the battery, you diverted to bridge rectifier. It appears that you mind is hopping randomly, ... (3) When I worried that the series R in your invention might waste power. you replied that R is high resistance, therefore loses small power - strange, are you sure you know what you are talking about?

@tlfong01,‭ BIN actually represents a diode rectifier (although it can be battery as well). The diodes inside this rectifier would be backward biased by the backup battery voltage Vb applied through the resistor R. Really, R will waste power... but what do we do? We can only decrease the difference between VIN and Vb... to some extent. The current flowing R is small since it only charges the battery (the base current is negligible). Most power is wasted in the transistor.

@Circuit fantasist, Most power is wasted in the transistor.. This is another confusion. I thought you said that Bin is replaced by Bb. So the current goes first to R, before to the transistor. In other words, power is wasted in both R and transistor. If your final version does NOT show Bin but only Bb, perhaps there will not be any such confusion.

@tlfong01‭, Thanks for the attention to my humble "invention". "Bin is replaced by Bb" is said in the sense that, when the mains supply fails, already Bb but not Bin supplies the load. Bin stays at its place but it does not produce power (that is why it is drawn in light gray).

@Circuit fantasist, Thank you for your clarification. So (1) Bin is not a battery, but an unregulated power supply consists of (a) AC transformer, stepping down AC mains from say 200VAC to 12VAC, (b) Bridge rectifier to convert 12V AC to 12V DC, (c) a big decoupling capacitor, perhaps? (2) Bb, on the other hand, is a real battery, and might be chargeable. Good, I got a better picture now, though not all the details. Thank you for your help again. Have a great weekend. Cheers.

@tlfong01‭, Exactly! (1) is true... It is an unregulated power supply consisting of (a) + (b) + (c) cascaded... I frequently use the battery symbol to represent a voltage source... And (2) is completely true. I will write you a message later in the chat...

@Circuit fantasist, OK. So I understand ‭your "Battery backup voltage regulator". I also agree with you saying that your students understand you story. However, there is a big problem. I think most first time readers won't understand you story for at least two reasons: (1) You are using misleading names and symbols, eg, battery symbols to represent (AC to transformer to rectifier to capacitor) unregulated power supply. (2) Using real battery as backup, but not mentioning it beforehand.

That is way I mentioned that you are diverting often, and you mind is hopping randomly around. If you are giving a lecture in class, I think there is no problem, because your students can raise their hand ask you to clarify, or ask classmate sitting next to clarify. But as a paper in the Q&A/forum/paper site here, it is almost no one will understand you. Of course your schematics cause extra problem. This is a serious communication problem. You might like to comment before I suggest something.

@coquelicot, can you explain to me what does "tinkerer solution" ("tinkerer") mean? I guess but I am not sure...

@circuit fantasist. That means a man building things not rigorously, in order for it to work in the given case, but not according the EE rules of the art. A good electric engineer would never have done that this way. /continued

For example, regulating the voltage with the battery as a voltage reference is bad in general, and using the base of the transistor to pass the backup current is bad design as well. Also, charging the battery this way is bad. At the price of few basic added components, you can do everything right. Your circuit may perhaps be relevant if you want your item be very cheap, for very tolerant loads. Tinkerer solutions are pleasant sometimes, that's why I've upvoted it.

@coquelicot, Thanks for the comprehensive explanation. I imagined something similar but now it is quite clear to me. As you probably noticed, in my circuit stories I try to show non-standard circuit ideas. My goal is to arouse interest in creative thinking leading to invention of circuits. In this story, I show how we can use the situation in an emitter follower when, for some reason, VCC drops…

... So far, the transistor is able to resist the input voltage through its emitter voltage. Now the latter disappears and the input voltage is applied directly through the base-emitter junction to the emitter (load). In some cases, this is undesirable... and in others it can be useful. BTW we can observe the same phenomenon in the circuit of the capacitive multiplier when the input voltage drops.