This instructable will teach you how to make a great voltage source using a few resistors, a voltage source, and a transistor. Only a minimal knowledge of electronics is required!

## Step 1: Motivation

In simple terms, we can think of a

Say you have a voltage source, but it isn’t adjustable and your load requires a smaller voltage. The simplest way to remedy this is with a set of two resistors, which will make up a

**voltage**(or more specifically, potential difference) as the source of energy in a circuit. Whatever your load may be (iPhone, speakers, etc.), you will need to provide voltage for it to function.Say you have a voltage source, but it isn’t adjustable and your load requires a smaller voltage. The simplest way to remedy this is with a set of two resistors, which will make up a

**voltage divider**. If you know how voltage dividers work, you will know how a large part of electronic circuits work!## Step 2: How a Voltage Divider Works - an Example

The voltage divider works by outputting a fraction of input voltage. This fraction is determined by the relationship between two resistors.

If we hook up our load in parallel with R2, we may give it whatever voltage (less than Vin) with good choice of R1 and R2 values.

For example, if Vin is 15 Volts and R1 and R2 are both 100 ohms (as in the attached file

**Kirchoff’s Laws**tell us that, in a circuit consisting of some input voltage Vin and two resistors R1 and R2, the voltage dissipated across R2 will be*Vin*R2/(R1+R2)*.If we hook up our load in parallel with R2, we may give it whatever voltage (less than Vin) with good choice of R1 and R2 values.

For example, if Vin is 15 Volts and R1 and R2 are both 100 ohms (as in the attached file

*voltage-divider.pdf*), Vout=15*(100)/(200)=7.5 V. Thus we can get an output of 7.5 Volts from a fixed 15 Volt source!## Step 3: Problems With Voltage Dividers As Voltage Sources (or an Introduction to Sag)

The

Although it would be easy enough to just build a voltage divider and use it as a voltage source, we run into one big problem. The actual voltage across the load turns out to be rather dependent on the load's resistance.

This dependence of voltage on load resistance results in

where 1/R3 is the resistance of the load. This allows us to add the resistance of the two together, since it is the equivalent resistance of those two resistors that make up the actual voltage divider. With the two in mind, lets see an example of how much a voltage divider can sag with a small load.

Let’s say we have the same resistor as before. However, this time we will add in a 10 ohm load. Instead of the second resistor in the voltage divider equaling 100 ohms, we have to factor in the parallel resistor and use the Req as our resistance.

With a 10 ohm and a 100 ohm resistor in parallel, the equivalent resistance is then 9.09 ohms (1/10+1/100 = .11, 1/.11 = 9.09). When this is used as the second resistor in the voltage divider, we get a voltage divider that puts out 9.09/109.09* 15= 1.25 V, significantly less than the 7.5 volts that we wanted!

What we ultimately desire is a

**thevenin resistance**(which we can think of as the internal resistance of a voltage source) of the voltage divider is*R1R2/(R1+R2)*.Although it would be easy enough to just build a voltage divider and use it as a voltage source, we run into one big problem. The actual voltage across the load turns out to be rather dependent on the load's resistance.

This dependence of voltage on load resistance results in

**sag**, which is not desirable for a voltage source. Ideally, we'd have a constant voltage across the load, no matter its resistance. However, when we hook up a load, we must consider the load resistance and R2 in parallel. To add these resistances, you simply follow the equation*1/Req= 1/R2 + 1/R3*,where 1/R3 is the resistance of the load. This allows us to add the resistance of the two together, since it is the equivalent resistance of those two resistors that make up the actual voltage divider. With the two in mind, lets see an example of how much a voltage divider can sag with a small load.

Let’s say we have the same resistor as before. However, this time we will add in a 10 ohm load. Instead of the second resistor in the voltage divider equaling 100 ohms, we have to factor in the parallel resistor and use the Req as our resistance.

With a 10 ohm and a 100 ohm resistor in parallel, the equivalent resistance is then 9.09 ohms (1/10+1/100 = .11, 1/.11 = 9.09). When this is used as the second resistor in the voltage divider, we get a voltage divider that puts out 9.09/109.09* 15= 1.25 V, significantly less than the 7.5 volts that we wanted!

What we ultimately desire is a

**stiff**voltage source, or one that doesn’t change voltage output no matter what load resistance.## Step 4: Transistors Solve Our Problem - the Emitter Follower

It turns out that a good solution to this problem is a special circuit called an

There are two main rules of thumb to know when working with transistors.

1.

2.

At first glance, the emitter follower seems like a useless circuit. Our output voltage is simply our input voltage, minus the 0.6 Volts we lose going through the transistor.

However, the emitter follower can be very useful in terms of “stiffening” our voltage source (i.e., reducing sag). Ideally, a voltage source’s internal resistance is minimal, and our load resistance is maximal. We can think of this as voltage sources “liking” loads with a big resistance and loads “liking” voltage sources with low internal resistance.

The factor of ~100 difference in current between the emitter and the base means that the resistance of our voltage source (which in our case is something called the Thevenin resistance of our voltage divider) looks ~100 times smaller to our load, which helps out with our sag issue!

Let’s revisit our previous example, but now using our emitter follower voltage source. Then Vout=Vin*(Rload)/(Rload+Rth/100)=15*(10)/(10+50/100)=15*(10)/(10.5)=14.28 V.

**emitter followe**r. The emitter follower consists of input voltages (which may or may not come from the same source) at the*base*and*collector*of what we call a**transistor**, with output voltage (and our load, eventually) at the transistor’s*emitter*.There are two main rules of thumb to know when working with transistors.

1.

*The emitter voltage will always be the base voltage minus a 0.6 V drop (which is for the diode that connects the base to the emitter.*2.

*The current from the emitter is always equal to the current from the collector, which is ABOUT 100 times bigger than the current from the base. (*There are certain limitations to this: if the collector source can’t put out enough voltage to keep the current at that level, your load won't get the voltage you're trying to give it. Also, the voltage from the collector must always be about 0.2 V higher than the voltage from the base. Otherwise, the transistor will break.)At first glance, the emitter follower seems like a useless circuit. Our output voltage is simply our input voltage, minus the 0.6 Volts we lose going through the transistor.

However, the emitter follower can be very useful in terms of “stiffening” our voltage source (i.e., reducing sag). Ideally, a voltage source’s internal resistance is minimal, and our load resistance is maximal. We can think of this as voltage sources “liking” loads with a big resistance and loads “liking” voltage sources with low internal resistance.

The factor of ~100 difference in current between the emitter and the base means that the resistance of our voltage source (which in our case is something called the Thevenin resistance of our voltage divider) looks ~100 times smaller to our load, which helps out with our sag issue!

Let’s revisit our previous example, but now using our emitter follower voltage source. Then Vout=Vin*(Rload)/(Rload+Rth/100)=15*(10)/(10+50/100)=15*(10)/(10.5)=14.28 V.

## Step 5: A Darn Good Voltage Source (or at Least a Heck of a Lot Better)

This circuit displayed here is one that will deliver a stiff 5V current that will sag only 5% at the maximum current going through the load, which is 25 mA. These are generally good numbers for most circuits that you will be powering and the numbers can be changed accordingly to fit your needs. The second resistor out of the emitter will keep the load from blowing up. To keep that second resistor from affecting your design, you want to keep that resistance significantly higher than the resistance of the load (see the parallel resistance equations if this does not make sense).

This is some great info for a DIYer such as my self. Thanks for the ible as i didn't know about this. <br>You have more to follow?

This was for our electronics class, if you want more check out our group page! <br>https://www.instructables.com/group/pomonaphysics/