Okay, before we get started, I want to differentiate between purchased followers and real followers on Facebook. You can do any type of Google search and find hundreds of websites that will have their robot and spam account followers go and like your facebook page in just a few hours. And if you do that, you will ruin your Facebook page. You see, Facebook has an algorithm that looks for these types of activities. And if they find that your page is indulging in a quick fix, they will penalize you. Going forward your organic reach will be stymied and your page will never have the potential it could have had if you had built it the correct way.

In electronics, a common collector amplifier (also known as an emitter follower) is one of three basic single-stage bipolar junction transistor (BJT) amplifier topologies, typically used as a voltage buffer.


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In this circuit the base terminal of the transistor serves as the input, the emitter is the output, and the collector is common to both (for example, it may be tied to ground reference or a power supply rail), hence its name. The analogous field-effect transistor circuit is the common drain amplifier and the analogous tube circuit is the cathode follower.

The common collector amplifier's low output impedance allows a source with a large output impedance to drive a small load impedance without changing its voltage. Thus this circuit finds applications as a voltage buffer. In other words, the circuit has current gain (which depends largely on the hFE of the transistor) instead of voltage gain. A small change to the input current results in much larger change in the output current supplied to the output load.

From this result, the gain approaches unity (as expected for a buffer amplifier) if the resistance ratio in the denominator is small. This ratio decreases with larger values of current gainĀ  and with larger values of R E {\displaystyle R_{\text{E}}} .The input resistance is found as

where || denotes a parallel connection, and R {\displaystyle R} is defined above. Because R {\displaystyle R} generally is a small resistance when the current gainĀ  {\displaystyle \beta } is large, R {\displaystyle R} dominates the output impedance, which therefore also is small. A small output impedance means that the series combination of the original voltage source and the voltage follower presents a Thvenin voltage source with a lower Thvenin resistance at its output node; that is, the combination of voltage source with voltage follower makes a more ideal voltage source than the original one.

The resistor before an op-amp follower serves as a voltage divider, which helps to reduce the input voltage to a level that the op-amp can handle. This prevents the op-amp from being overloaded, and also ensures that the input voltage is within the op-amp's input range.

The resistor before an op-amp follower does not affect the output voltage, as the op-amp follower has a high input impedance and low output impedance. This means that the output voltage will be the same as the input voltage, regardless of the value of the resistor.

The value of the resistor before an op-amp follower depends on the input voltage and the op-amp's input range. It is typically chosen to be large enough to prevent overloading the op-amp, but small enough to maintain a high enough input voltage for accurate amplification.

Yes, a potentiometer can be used as the resistor before an op-amp follower. This can be useful for adjusting the input voltage to the desired level. However, care should be taken to ensure that the potentiometer's maximum resistance is high enough to prevent overloading the op-amp.

In most cases, a resistor is necessary before an op-amp follower to prevent overloading the op-amp and to ensure accurate amplification. However, in some cases where the input voltage is already within the op-amp's input range, a resistor may not be needed.

We report the first detailed accounts of bites by the Balkan adder, Vipera berus bosniensis from Bulgaria. Documentation of bites by this subspecies is very rare in the literature and most available accounts are from the northern limit of its distribution. V. berus bosniensis is considered to possess neurotoxic venom but little evidence has hitherto been available to support this supposition. In this case series symptoms typical of adder bites developed including oedema, nausea, dizziness, lymphangitis, vomiting, and diarrhoea together with aberrant symptoms such as diplopia and ptosis that confirm the presence of neurotoxic venom in Balkan adders. In addition, unusual and atypical symptoms of adder bites such as painless bites and muscle cramps appeared. The inadequate treatment in hospital and the remote habitats in which this species is encountered are potential sources of complication.

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If you are adding the first two bits, wouldn't the initial carry-in always be zero considering that nothing has been added yet? Couldn't the first full adder be substituted with a half adder and still work?

$\bullet$ This is a 4-bit adder. Maybe you want to take two or more of these and hook them together to make an 8-bit adder (or a 16-bit adder or whatever). In that case the carry-out of each 4-bit adder would be connected to the carry-in of the next 4-bit adder.

$\bullet$ You might want to use this to implement subtraction. To subtract $A-B,$ where the numbers $A$ and $B$ are in two's-complement form, you'll complement all the bits of $B$ (by using four NOT gates in parallel) to get the four values $B',$ and you'll add $A+B'+1$ (because that's how two's-complement arithmetic works). The easiest way of doing this is to add $A$ and $B'$ with an adder like you have, but with the carry-in tied to $1$ instead of $0.$

$\bullet$ You could build a similar device that could either add or subtract, based on an extra selector input $Q$ which would be $0$ for addition and $1$ for subtraction. You would then use four XOR gates to XOR each of the 4 bits of $B$ with the selector input $Q$ to produce either the bits of $B$ (for addition) or the complement of the bits of $B$ (for subtraction). You would also connect $Q$ to the carry-in so that you would add an extra $1$ for subtraction but not for addition. 589ccfa754

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