HINT: voltage regulate = maintain voltage

In electronics, voltage regulators maintain a constant output voltage.

So, whether the question mentions Linear or Switching voltage regulators, choose the answer that mentions maintaining a constant output voltage.

The resistance of the regulator varies in accordance with the load resulting in a constant output voltage.

https://en.wikipedia.org/wiki/Linear_regulator

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A Zener diode is typically used as a stable reference voltage in a linear voltage regulator. When reverse biased, the diode breaks down and allows current to flow in the reverse direction when the voltage is above the avalanche point. The Zener voltage is the voltage necessary to cause the avalanche to occur and allow current to flow. As long as the reverse bias voltage remains above the avalanche point, the voltage drop across the Zener diode remains constant to provide a stable reference for regulating power supply output voltage.

Hint: this is a stretch but it worked for me...
stables are for horses
Zorro had a horse
Z is for Zener

A Zener Diode is the simplest type of voltage regulator.

Hint 2: a ZEE-ner is LEEnyer (linear). Weird, but helps me remember.

Remember seeing Varactor somewhere as an answer? Both hint 1 and 2 work as 'stable' and 'linear' are not a part of a different, similar question:

What type of semiconductor device is designed for use as a voltage-controlled capacitor? Varactor diode

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An example of a shunt regulator is using a Zener diode in series with a resistor between supply and ground. The Zener diode establishes a constant voltage drop and the resistor sets a constant current.

Note: You can immediately dismiss two of the possible answers. "Regulator" appears in both the question and the answer.

HINT: While performing a carotid endarterectomy, (removing the blockage from the carotid artery), the surgeon must SHUNT the blood around the blockage to have CONSTANT blood flow to the brain.

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Battery capacity is measured in amp-hours; how many amps a battery can deliver in one hour.

Intuitively, if we are given a value in amp-hours and we need to solve for hours, we divide by amps:

(amps*hours)/amps = hours

Example: a 12 amp-hour battery will operate a 4 amp load for 3 hours (12 Ah / 4 A = 3 hours).

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Hint: smaller transformer and components = lighter and more often less expensive.

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Linear voltage regulators maintain a constant output voltage in a circuit by using a control element that responds to current. If the "pass element" in a series regulator doesn't receive enough input voltage it can't perform that control function. Consequently output voltage begins to drop below the regulated value.

The minimum voltage a regulator requires before drop out occurs is called the "drop-out voltage."

Hint: A voltage regulator maintains regulation. Only one answer choice has this word.

Another hint: Many students who only do the minimum amount of required work to maintain their grade may eventually dropout.

Hint: dropout + output = dropOUTput

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Electrical power is measured in Watts and calculated by multiplying the voltage by the current. In an electrical circuit, every device will have either a voltage "drop", i.e. consumes power, such as a resistor, or a "voltage gain", i.e. produces power such as a battery.

A linear voltage regulator is no different in this respect. A linear voltage regulator is explicitly designed to do this in a reliable fashion, by taking a higher (usually unregulated) voltage and producing a lower, regulated voltage.

A power calculation across a regulator can be difficult if the input is highly unregulated, or the current demand of the regulated circuit changes frequently but that's not the point of the question. Power consumption, and in most cases "dissipation", meaning wasted power, is always the voltage drop multiplied by the current through the device.

An example could be a 9-volt battery connected to a 5 volt linear regulator, powering a circuit that consumes 2 amps. Since current is always conserved, the 2 amps at 5 volts going into the powered circuit comes from 2 amps at 9 volts coming from the battery. While the powered circuit would consume 10 Watts

\[P=IV=(2 \text{ A})(5\text{ V})=10 \text{ W}\]

the regulator has to dissipate the excess energy from the 18 Watts

\[P=IV=(9\text{ V})(2 \text{ A})=18 \text{ W}\]

coming from the battery. In this case, you would see 2 amps of current moving through the regulator, with 9 V as its input and 5 V as its output. The regulator would be dissipating 8 Watts.

\[(9 \text{ V} - 5 \text{ V}) \times 2 \text{ A} = 4 \text{ V} \times 2 \text{ A} = 8\text{ W}\]

Linear regulators are specifically designed to do this, but care has to be taken that the heat generated by the regulator is properly dissipated and within the design limits of the regulator.

• Hint * the correct answer is the only one that has "Voltage" as the first word.

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Answer: All of these choices are correct.

Think of a simple capacitor: it is two plates separated by an insulator. When you connect capacitors in series, you are charging up all of the capacitors with the same number of electrons (the same current flows through each because they are in series).

If the capacitors have any imperfections, they will each end up with a different voltage, and resistors can help balance the capacitors because the fuller ones will have a higher voltage, and therefore get discharged faster by the resistor.

The value of the resistors chosen is high such that the resistance is at least 10 times the highest value of \(X_C\), where

\[X_C = \frac{1}{2πfC}\]

Hint: Equal-value -> Equally important (All these choices are correct)

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A "step start" circuit gradually increases the voltage applied to filter capacitors. The reason we use this circuit, is to avoid putting the full voltage on the capacitor all at once. If we were to put the full voltage on the capacitor all at once, it might fail (literally blow up), or at least result in a decrease in the capacitor's useful life.

Hint: Gradually

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