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Flashcards in Electronics Deck (17):
1

saturation (define, transistor)

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

2

cut-off (define, transistor)

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

3

why would you use a transistor as a switch?

- switches faster
- less mechanical wear
- less "noise" (switch bounce) - better at higher freq.

4

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

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

5

astable (define, 555)

555 acts as a rectangular-wave generator

6

monostable (define, 555)

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

7

Basic op-amp operation

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

8

Negative feedback

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

9

Basic op-amp rules

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)

10

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

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

11

What is 'duty cycle'?

Fraction of the time output is high
= t(high)

12

mark-space ratio concept

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

13

t(high) for a 555 astable

0.693*R2*C2

14

t(low) for a 555 astable

0.693*(R1 + R2)*C1

15

duty cycle

fraction of time the output is high.
duty cycle = t(high) / t(high) + t(low)

16

frequency of astable waveform

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

17

period of monostable wave

1.10*R1*C