Transistor electronic technologies

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1 Transistor electronic technologies Bipolar Junction Transistor discrete or integrated circuit discrete = individual component MOS (Metal-Oxide-Silicon) Field Effect Transistor mainly used in integrated circuits driven by digital applications but analogue Junction Field Effect Transistor similar in many ways to MOS FET discrete, not easily implemented in ICs so - is this a course on circuits? No but it is necessary to understand some basics to be able to deal with more complex elements, including some features of op-amps g.hall@ic.ac.uk 1

2 Bipolar transistor pnp or npn semiconductor, usually Si, but also Ge heavily doped emitter, lightly doped base base emitter collector n++ p+ n Operation - npn base is biased more positive than emitter so a forward biased diode collector more positive than base = reverse biased diode majority carriers from emitter diffuse across base to collector small fraction combine with majority carriers in base current reaching collector is = α I B = (1-α) I B arrows show direction of current flow npn I B = βi B = [α/(1- α)]i B β = α/(1- α) = d.c current gain = h fe eg α = 0.99 β = 99 g.hall@ic.ac.uk 2

3 pnp transistor Works like npn transistor bias arrangements different most positive but if emitter is positioned at top pnp npn I B I B easy to remember both pnp and npn most negative g.hall@ic.ac.uk 3

4 Slightly more precise picture to turn npn transistor on V B -V E = V BE > V (invert for pnp) 2n3906 npn transistor so we can use it as a switch by controlling V BE V BE 0 = = 0 however, if transistor is ON V BE 0.7V this is a consequence of the diode behaviour - discuss in a moment in contrast, β is not a reliable parameter for design β NB log scale for NB both log scales (ma) g.hall@ic.ac.uk 4

5 Ebers-Moll model Transistor can be modelled as two back-to-back diodes I-V behaviour of diode I I 0 [exp(qv BE /kt)-1)] Base-emitter diode is forward biased = 0.[exp(qV BE /kt)-1] 0 exp(qv BE /kt) ie V BE (kt/q)log e Base-collector diode is reverse biased I BC = O [exp(qv BC /kt)-1] 0 this explains why V BE varies so little with I also basis of band-gap T reference - which is small so current arriving at collector is dominated by current from emitter, which has diffused across base How does current vary with small change in V BE? d /dv BE = i e /v be = (q/kt)0 exp(qv BE /kt) = (q/kt) i e r e = v be with r e = kt/qi = 25Ω/ (ma) NB we don t usually need to distinguish between and - consider them equal ie to ac current signals transistor looks like dynamic resistance g.hall@ic.ac.uk 5

6 Emitter-follower DC conditions ±6V are example values, but results don t depend on them at all apply our rule that V BE 0.7V = [V E - (-6V)]/ (V B + 5.3V)/ +6V Now ac behaviour V C v in = v b = V B = (V E +0.7V) = v e = v out v in amplifier with gain = 1 - not very interesting!!?? V B V E vout Input impedance -6V R in = v in /i in i in = i b = i c /β i c = i e = v e / R in = β high, eg β ~ 100, ~ 1kΩ (more careful treatment => R in = β( + r e ) this is promising for a voltage buffer - what is the output impedance? g.hall@ic.ac.uk 6

7 Emitter-follower output impedance How to find it? Consider the black box vary v out and see what happens to i out keep other conditions fixed R O i out v out Use Ebers-Moll result V BE = (kt/q)log e dv BE /d = v be /i e =(kt/q ) V C +6V v in If V B is constant V B V E vout v out = v e i out = i e Z out = (kt/q ) = r e = 25Ω/ (ma) -6V small, as required for buffer g.hall@ic.ac.uk 7

8 Short footnotes In analysing circuits for small signal (AC) behaviour all fixed DC levels are equivalent to ground ie ac current does not need to distinguish voltage at other end of path This is often useful in looking at circuits to tell if routes are in parallel Calculations keep simple try to make approximations - 1% answers are almost never required if so better tools exist eg parallel resistances transistor β - assume β unless better value known or is critical 47Ω 51Ω 50Ω g.hall@ic.ac.uk 8

9 Common-emitter amplifier DC conditions ±6V are example values again +6V V C = 6V - R C V E = -6V + C R 1 R C v out Since V E = V B - 0.7V, & defined by bias network v in small signal, AC behaviour v e = v b = v in = i e R 2-6V v out = -i c R C so v out /v in = -R C / what s the purpose of C? amplifier with gain Input impedance: signal sees bias network in parallel with transistor so R in = R 1 R 2 β( +r e ) - typically a high value g.hall@ic.ac.uk 9

10 Common-emitter output impedance Play same trick as emitter-follower but this time, from output terminal, the two paths for i out are collector-base junction reverse biased diode = high impedance R C v out R C usually much lower than r cb no need to worry about any source impedance driving amplifier so Z out R C usually relatively high g.hall@ic.ac.uk 10

11 Reading circuits look for the building blocks usually blocks are "vertical columns" look for feedback paths horizontal paths,which are not DC bias, or output-input 5kΩ 18nF 25Ω 350Ω 18nF V CC Out In 10kΩ 750Ω 50Ω 18nF Ground 300Ω V EE g.hall@ic.ac.uk 11

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