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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: enter image description here

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!

Transistor
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namarino41
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2 Answers2

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If V3 is on, power goes through the resistor and back into the battery (the resister value is arbitrary, I just haven't figured out what it should be).

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.

The power doesn't go through the D2 diode because it's reverse biased.

Correct. Actually from the simulation we can see that D2 is always forward biased (for R2 = 100 Ω).

But why doesn't the power go from the battery, V2, through the resistor and to the load as well?

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.

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?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Circuit for simulation.

enter image description here

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:

  • V(A) alternates between 5 V and 0 V. When high the current through D1 is 40 mA.
  • The simulator has added a 2 Ω internal equivalent series resistor (ESR) so we can see a little variation on V(B) depending on whether it's being charged or discharged.
  • Notice that I(R1) [blue] is always positive (for the point I'm measuring at). This means that BAT2 is always discharging even when V1 is high.
  • When V1 is switched off V(B) drops to about 3.8 V and so D2 conducts and the resistor current increases to about 8 mA.

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.

Transistor
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  • I didn't quite understand the second to last paragraph. Would you mind explaining that again? Sorry – namarino41 Nov 10 '20 at 19:14
  • Did you read the linked article? – Transistor Nov 10 '20 at 19:16
  • Yep I did. And it made sense. But when the main power is on, why doesn't the battery the load also draw power from the battery? I get that it doesn't draw power though the diode, but why not the resister? – namarino41 Nov 10 '20 at 19:36
  • I may have simplified things a bit much in that I have kind of treated D1 as ideal. Let me edit later. – Transistor Nov 10 '20 at 20:27
  • @namarino: See the update. – Transistor Nov 10 '20 at 22:31
  • Oh very cool. So, correct me if I'm wrong, but basically when the main power source is on, no current is flowing through the resister to the battery. The battery is just discharging at all times. What do I do to prevent that? – namarino41 Nov 11 '20 at 00:42
  • Try replacing the diodes with Schottky diodes which have a lower voltage drop. Run the simulation. Change R2 to something similar to your expected load. – Transistor Nov 11 '20 at 07:38
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Not a dumb question.

R1: R1's value should be chosen so that current into your battery(V2) does not exceed the limits of battery(v2) For instance V2 has a min voltage of 4.1V with a max charging current of 100mA. R1 should be chosen such that ((V3-V(D1))-V2)/R1 <= 100mA. (Make sure R1 has an appropriate wattage rating)

On to where the current flows.

In the circuit below I have set an arbitrary load = 10 ohms. (Also from here on out we will use the circuit belows part numbers.)

Using Kirchhoff's current law we can show where the current is going.

Lets assume no current flows through D1. All current should flow through the resistors or 4.5V/(110 Ohms) = .041 Amps

Lets check if this is right by finding the voltage drop across R1. .041 Amps * 100 Ohms = 4.1 V drop.

If D1 was off then the drop across R1 would be less than Vf of the diode.

Its not which means that the diode is on.

Now we can resolve the circuit with the diode on.

4.5V-Vf(.7V) = 3.8V at the node between the two resistors.

Total Current through the diode and the resistor R1 is given from 3.8V/10 = .38 A.

Current through R1 then is (4.5-3.8)/100 = .007 A

Finally current through the Diode is Total Current - I(R1) = .373 A.

This circuit analysis should be all you need to design a good charging circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Parker
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