Electronics Flashcards

1
Q

saturation (define, transistor)

A

transistor acts like a short circuit (current flow freely from collector to emitter). Being driven so hard that there is no longer any amplification

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2
Q

cut-off (define, transistor)

A

transistor acts like an OPEN CIRCUIT (no current flows from collector to emitter)

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3
Q

why would you use a transistor as a switch?

A
  • switches faster
  • less mechanical wear
  • less “noise” (switch bounce) - better at higher freq.
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4
Q

what are some design factors when using a transistor as a switch?

A

make sure it’s at saturation (e.g. set correct operating point)

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5
Q

astable (define, 555)

A

555 acts as a rectangular-wave generator

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6
Q

monostable (define, 555)

A

555 acts as a “one-shot timer” - when an input voltage is applied, it goes from low to high

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7
Q

Basic op-amp operation

A

Switches from one maximum output to another when there’s a voltage difference between it’s inputs.

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8
Q

Negative feedback

A

When voltage is fed back to inverting terminal, gain can be controlled.

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9
Q

Basic op-amp rules

A

1) v(out) = A0(V+ - V-)
2) A0 is infinity (open loop)
3) R(in) is infinity (therefore no current can flow into op-amp)

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10
Q

Advantages and disadvantages of centre-tapping transforming/rectifier set-up.

A

Advantages: need only 2 diodes instead of full wave bridge
Disadvantages: centre-tapping means half of the secondary voltage as output

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11
Q

What is ‘duty cycle’?

A

Fraction of the time output is high

= t(high)

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12
Q

mark-space ratio concept

A

Literally mark (time high, or active) divided by space (time low)

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13
Q

t(high) for a 555 astable

A

0.693R2C2

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14
Q

t(low) for a 555 astable

A

0.693(R1 + R2)C1

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15
Q

duty cycle

A

fraction of time the output is high.

duty cycle = t(high) / t(high) + t(low)

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16
Q

frequency of astable waveform

A

f = 1 / period
period = t(high) + t(low)
therefore f = 1 / t(high) + t(low)

17
Q

period of monostable wave

A

1.10R1C

18
Q

Why

A

Because