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When switch \$S\$ is at position \$1\$:

$$V_B = R_2 \mathbin{/} (R_1 + R_2)*V_{CC} = 30 \mathbin{/} 130*-20=-4.62\mathrm{V}$$ $$R_{TH}=100*30 \mathbin{/} 100+30=23.1\mathrm{kOhm}$$ $$V_{TH}-I_ER_E-V_{BE}-I_BR_{TH}=0$$

where \$I_B= I_E \mathbin{/} \beta\$ and \$V_{BE} =0.7\mathrm{V}\$.

So,

\$I_E= V_{TH}-V_{BE}\mathbin{/} (R_E + R_{TH} \mathbin{/} \beta)\$

\$I_E= (-4.62-0.7) \mathbin{/} (500+23100 \mathbin{/} 110)\$

\$I_E=-7.49\mathrm{mA}\$

$$V_E = V_B + 0.7 = -4.62+0.7 = -3.92\mathrm{V}$$

\$V_C = -V_{CC} + I_CR_C\$

\$V_C = -20 + 7.49*2000 \mathbin{/} 1000\$

$$V_C = -5.02\mathrm{V}$$

$$V_{CE} = V_E -V_C = -5.02 - (-3.92) = -1.1\mathrm{V}$$

$$P = IV = -1.1 * -7.49 \mathbin{/} 1000 = 8.24\mathrm{mW}$$

Am I right?

fuzzyhair2
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Timoo
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  • The bias resistors are not connected to the base at all in the schematic. (Where is the junction dot?) – Kaz Dec 28 '13 at 16:57
  • With a supply of -20V on the divider, current will flow from ground through the bias resistors, right? That would make Vbase = -20*(100/130) = -15.4V – sherrellbc Jun 26 '14 at 19:56

1 Answers1

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Yes. (But I should type at least 30 characters, where 3 are enough.)

johnfound
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  • i don't know how to do with it when it is in position 2. i know the two 30k ohm resistor are parallel. but how about the resistor R?? – Timoo Nov 28 '13 at 10:05
  • @Timoo U(r)=Ib*R and it is actually very small value that can be ignored. (Well, if R is very big it can't be ignored, but then you need to know the value of R then) – johnfound Nov 28 '13 at 10:09
  • The circuit of fig.Q3 shows a voltage-divider biasing scheme for a pnp transistor with β = 110. Switch S in position 1 shorts out resistor R, but when in position 2, it connects R1 in parallel with R2 and also maintains R in circuit. Find the value of R such that the power dissipated by the transistor with the switch in position 1 is half the value when the switch is in position 2
    • – Timoo Nov 28 '13 at 10:10
  • Here's the questions that it said i have to calculate the value of R. – Timoo Nov 28 '13 at 10:10
  • @Timoo - according to what conditions you have to calculate the value of R??? If there is no conditions defined, then R can be random. – johnfound Nov 28 '13 at 10:13
  • Find the value of R such that the power dissipated by the transistor with the switch in position 1 is half the value when the switch is in position 2 – – Timoo Nov 28 '13 at 10:17