I am relatively new to electronics and circuits. I'm modifying an alarm clock to have a battery backup. I've found a few circuits that show how to build a battery backup circuit and it makes a lot of sense to me. A few of these circuits allow the battery to be trickle charged though and I'm not clear on how these work. Here's one:
V3 is the main power. V2 is the battery. If V3 is on, power goes through the resister and back into the battery (the resister value is arbitrary, I just haven't figured out what it should be). The power doesn't go through the D2 diode because it's reverse biased. But why doesn't the power go from the battery, V2, through the resister and to the load as well? And when V3 is out and the battery is powering the load, which path does it take? Does it go through the D2 diode or the resister? Apologies if this seems like a stupid question. I've been doing my best to learn this, but I can't find answers to these particular questions. Thanks for the help!
Best Answer
Correct. With the values shown the V2 battery will be charged with a current given by \$ I = \frac {V_3 - V_2} R_1 = \frac {5 - 4.5} {100} = 5 \ \text {mA} \$.Now that I've run the simulation I can see that for a 100 Ω load BAT2 will never charge. Due to the voltage drop of D1 V(B) is lower than that of BAT2.
Correct. Actually from the simulation we can see that D2 is always forward biased (for R2 = 100 Ω).It would if the diodes didn't need several hundred millivolts to "crack open". Since the top of R1 and D2 are connected there should be no voltage between the two so there is no voltage across D2 and it won't conduct. I've explained this in an article What is an LED? which you might find helpful.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Circuit for simulation.
Figure 2. Results of simulation. Battery V3 has been replaced by a 5 V square-wave source to show what happens with V1 connected and disconnected.
Note from the simulation:
If you hit the "Simulate this circuit" link you should be able to play with the battery voltages and load resistance to get a better idea of what's going on.