What controls battery discharge? Can you use a transistor to lower the amount of battery capacity used?

Question

What controls battery discharge? I want to be able to save battery capacity by using a transistor to amplify current but don't know how current even would flow from the battery in the first place.

Is it the component it is powering that controls discharge (made up example: if it needs 10ma the battery discharges 10ma) or does the battery need something to control discharge? If so what would that be? My idea was that I could use a transistor to boost the current so the battery wouldn't have to discharge as much of its capacity and the transistor could be used as an emergency digital off switch because the gate will be controlled by my main IC.

This would not work,however, if the battery gives what is normally needed without amplification to power the component. Would the battery discharge current be lowered for amplification and therefore save battery capacity or does it have to be controlled by an IC? Either is fine and will save capacity I just need to know if it would work out automatically or if I need to get an IC or something else to control the current.

This is a simplified version of what I am trying to do V2 is the battery and V3 is the IC. The transistor is just a random one I found:

Answer 1

My idea was that I could use a transistor to boost the current so the battery wouldn't have to discharge as much of it's capacity

This doesn't work because your idea, in effect is trying to magic current from an invisible energy source. A transistor does amplify current but, it doesn't create current; the current has to come from somewhere and, that current comes from your battery. I mean if you think about it, you'd save all the world's energy problems if this idea worked.

You can say that an electric car amplifies the movement of your foot to create a massive power gain but, that power (voltage or current amplification or both) has to be sourced from somewhere and, that somewhere is the battery.

the transistor could be used as an emergency digital off switch because the gate will be controlled by my main IC.

This is common practice.

Would the battery discharge current be lowered for amplification and therefore save battery capacity

This is impossible.

Answer 2

A transistor does not really "amplify" current but lets a small current control a large current. You still need to supply that large current in order to control it, so you cannot save energy using transistors. What you can do is use a transistor for turning a battery off early to save it from deep discharge. But that will lower the total amount of energy you can draw, not increase it.

Answer 3

You seem to be rather confused about basic electric concepts, so I advice you to get an entry-level textbook and study it (there's tons of freely downloadable material online too), otherwise you risk to confuse yourself even more.

Electrical concepts are not intuitive at all, so you shouldn't expect yourself to grasp them simply performing some hobby experiment. Building electronics circuits is fun, but you rarely can get much theory out of it without some basic theoretical knowledge (unless you are a real genius!).

Keep in mind that after the laws of basic mechanics (those still used today to design, say, automobiles and buildings) were published by Newton's (1687) it still took centuries to understand electromagnetic phenomena with some detail and precision. Ohm stated its law around 1825, but his work wasn't widely accepted until 1840. Kirchhoff stated its laws in 1845 and Maxwell's published his comprehensive theory on electromagnetism in 1861. And still, the microscopic nature of electricity and related phenomena had to wait yet more decades to be uncovered by quantum mechanics (early 20th century).

So what you should learn is at minimum Ohm's law (and to which components it is applicable), Kirchhoff's laws and, for good measure, a bunch of easy formulas derived from those, like voltage divider and current divider formulas. To understand those you need to understand some basic concepts, like current and voltage (and relative sources).

Extremely important, as in all fields of physics, are also energetic concepts like power and energy. I will focus on these here, which is tangentially related to your problem,

You should understand that in a circuit, as in every physical situation, energy is conserved (this is an universal law). In particular power (which is the energy transferred/converted in unit time is conserved. This means that in every instant of time the power that goes out a part of the circuit must flow in the other parts it is connected to.

This means that whatever your battery does, the energy stored in it cannot be "amplified" by a transistor. The power P (in watts, W) that that battery can provide is its voltage times the current it provides to the circuit ("P = V x I"). This is the energy lost by the battery in 1 second. The energy E stored in the battery is some fixed amount (expressed for example in watt-hours, Wh), so to get how many hours of power you can get from your battery you simply calculate \$T = E/P = E/(V \times I)\$.

So you can see that if your current increases, the time decreases. A transistor cannot change this basic statement. A transistor can limit the current delivered to the load (and the voltage seen by the load), but the very presence of the transistor means that a part of the energy delivered to the load will be wasted in heat in the transistor itself.

There are circuits (that comprise transistors and other components) that can increase or decrease the voltage seen by the load or the current delivered to it:

^{simulate this circuit – Schematic created using CircuitLab}

but whatever the circuit interposed between the battery and the load, the conservation of energy will still enforce the following relation:

$$ V_{out} \cdot I_{out} = V_{in} \cdot I_{in} - P_{losses} $$

Or also

$$ V_{out} \cdot I_{out} = \eta \cdot V_{in} \cdot I_{in} $$

Where \$\eta\$ is the efficiency of the converter.

Therefore inverting that equation gives: $$ I_{in} = \frac {V_{out}} {\eta \cdot V_{in}} \cdot I_{out} $$

So you can use a circuit to change the current "taken out" of your battery, but this won't increase the capacity of your battery. It can, of course, prolong the battery life, i.e. the time during which the load is powered, but not the total energy delivered to the load.

---

BTW, what is called battery capacity and expressed in ampere-hours, Ah, (or its submultiple milliampere-hours, mAh) is essentially a simplification used in industry in substitution for the actual energy content (also called capacity) expressed in Wh. This simplification stems from the fact that batteries are essentially voltage sources, i.e. devices that maintain the voltage at their terminal essentially constant until they are discharged.

This is an approximation (see below), but as long as we assume that the voltage across a battery is a constant Vbatt, then its capacity Cbatt in Ah is related to the energy content Ebatt by the simple formula: $$ E_{batt} = C_{batt} \times V_{batt} $$

However that approximation is quite rough. It's common knowledge that the voltage of a battery changes with time and decreases as the battery is discharged.

For example, an AA alkaline battery starts fresh with a voltage of about 1.6V and can be used until it reaches 1.2V-1.1V before it can be considered "empty" (some low power circuits can even work with alkaline batteries down to 0.9V). A fully-charged common Lithium cell has a 4.2V voltage, that decreases to 3.5V until a recharge is warranted.

Battery manufacturers actually provide what are known as discharge curves that show how exactly the voltage of the battery changes depending on its state of charge (SOC) and other parameters.

Here is an excerpt from the datasheet of a Duracell AA alkaline battery:

In particular this is a graphics with discharge curves: