**Description: **A review of how counting works with normal (base 10) numbers. Then it is relate to the new concept of binary numbers; a system which is implemented in computers to represent numbers.

**Script:**

First Let's look at how we're used to counting.

We have 10 digits in our number system - 0 through 9.

There's no number bigger than 9 that we can fit in the 1's place. Or any place for that matter.

Suppose we have the number 327.

7 in the ones place

2 in the tens place

and 3 in the hundreds place.

Now let's count up from here.

328

329

Hmm, we can't fit any more value in the 1's place so we have to "carry over" - Like this.

This system can't be directly transferred to digital circuits, since we can only do true or false comparisons. On or off, 1 or 0.

But even with that, we still have two digits to work with.

Let's explore this system by counting upwards using the same carry over rules as before.

Start with zero

1

Now we can't fit any more value into this place, so we have to carry over.

Add 1 to this place.

And set this place to 0.

The value shown here is 2, so we can conclude that this is the 2's place.

3

Now we're faced with a carry again.

Add 1 to this place

It can't hold any more value so we have to carry again.

Add 1 to this place

set the 2's place to 0.

and set the 1's place to zero.

We know this number is 4, so this must be the 4's place.

We can continue this trend onward

5

6

7

The system that we are exploring is called binary, and each place is called a "bit"

This is a 3 bit number

Binary enables us to execute highly accurate and repeatable arithmetic and comparison using digital logic circuits.

Using what we learned -- see if you can try and count up to the number 12 using binary.

That's it for today's Short Circuit. Thanks you for Watching!

Any articles concerning power electronics

Articles discussing the process of converting DC to another level of DC

V_{SAT} |
Saturation voltage of the output transistor | |

V_{F} |
Forward voltage drop of the diode | |

V_{IN} |
Typical input voltage | |

V_{MIN} |
The minimum voltage of the input | |

V_{OUT} |
Desired output voltage | |

I_{OUT} |
Desired output current | |

f_{MIN} |
Minimum desired output switching frequency | |

V_{RIP} |
Desired peak to peak output ripple voltage | |

R_{1} |
Resistor for setting V_{OUT} |

R_{1} |
Resistor for setting VOUT | |

R_{2} |
Resistor for setting VOUT | |

R_{3} |
Switch Biasing (constant) | |

R_{PD} |
Pull down resistor for transistor (constant) | |

R_{SC} |
Current sense resistor | |

C_{T} |
Timing capacitor | |

C_{O} |
Output capacitor | |

L_{MIN} |
Minimum inductance for output |

V_{SAT} |
Saturation voltage of the output transistor | |

V_{F} |
Forward voltage drop of the diode | |

V_{IN} |
Typical input voltage | |

V_{MIN} |
The minimum voltage of the input | |

V_{OUT} |
Desired output voltage | |

I_{OUT} |
Desired output current | |

f_{MIN} |
Minimum desired output switching frequency | |

V_{RIP} |
Desired peak to peak output ripple voltage | |

R_{1} |
Resistor for setting V_{OUT} |

R_{1} |
Resistor for setting VOUT | |

R_{2} |
Resistor for setting VOUT | |

R_{PD} |
Pull down resistor for transistor (constant) | |

R_{SC} |
Current sense resistor | |

C_{T} |
Timing capacitor | |

C_{O} |
Output capacitor | |

L_{MIN} |
Minimum inductance for output |

In this video the energy efficiencies of Low Drop Out (LDO) regulators and Buck converters is discussed. A quick graph is generated in Sage Mathematics to compare the two circuits power consumption

**LDO Efficiency:**

$$P_{loss}=I*(V_{in}-V_{out})$$

Where $P_{loss}$ is heat dissipation, $I$ is the current drawn on the output, $V_{in}$ is the input voltage, and $V_{out}$ is the output voltage.

**Buck Efficiency:**

$$P_{loss}=(1-E)*V_{out}*I$$

Where $P_{loss}$ is heat dissipation, $E$ is the calculated efficiency, $I$ is the current drawn on the output, and $V_{out}$ is the output voltage.

**Sage Simulation Script:**

```
var("Vin",latex_name="V_{in}")
var("Vout",latex_name="V_{out}")
var("I")
var("E")
Vin=17
Vout=12
E=.87
PlossLDO=I*(Vin-Vout)
PlossBuck=Vout*I*(1-E)
plot(PlossLDO,(I,0,1.5))+plot(PlossBuck,(1,0,1.5),color="red")
```

**References:**