- Example 5 5 Find Ic And Vce For A Pnp Transistor Circuit With These Values R 68 Kn R 47 Kn Rc 1 8 Kn Re 2 1 (72.58 KiB) Viewed 46 times
- Example 5 5 Find Ic And Vce For A Pnp Transistor Circuit With These Values R 68 Kn R 47 Kn Rc 1 8 Kn Re 2 2 (22.62 KiB) Viewed 46 times
the answer is 288 please help I posted this question earlier but the answer is so wrong I received.
EXAMPLE 5-5 Find Ic and VCE for a pnp transistor circuit with these values: R₁ = 68 kN, R₂ = 47 kn, Rc = 1.8 kN, RE = 2.2 kn, Vcc= -6 V, and Bpc = 75. Refer to Figure 5-14(a), which shows the schematic with a negative supply voltage. Solution Apply Thevenin's theorem. Related Problem R₂ Vm= (₁, RR,) vec = (a VTH Vcc= + R₂. = (0.409) (-6 V) = -2.45 V R₁R₂ R₁ + R₂ Use Equation 5-7 to determine IE. RTH = IE = = = 47 ΚΩ 68 ΚΩ + 47 ΚΩ (68 ΚΩ)(47 ΚΩ) (68 ΚΩ + 47 ΚΩ) - VTH + VBE RE + RTH/PDC 3.15 V 2.57 ΚΩ From IE, you can determine Ic and VCE as follows: Ic=IE= 1.23 mA Vc = -Vcc + IcRc = -6 V + (1.23 mA)(1.8 k) = -3.79 V VE-IERE = -(1.23 mA) (2.2 k) = -2.71 V VCE VC VE = -3.79 V + 2.71 V = -1.08 V = (-6 V) OTHER BIAS METHODS = 1.23 mA = 27.8 ΚΩ 2.45 V +0.7 V 2.2 ΚΩ + 371 Ω ◆ 229 What value of BDC is required in this example in order to neglect RIN(BASE) in keeping with the basic ten-times rule for a stiff voltage divider?
RELATED PROBLEMS FOR EXAMPLES 5-1 ICQ 5-2 5-3 5-4 RIN(BASE) 5-5 BDC ICQ = 19.8 mA; VCEQ = 4.2 V; Ib(peak) = 42 μA The voltage divider would be loaded, so VB would decrease. 92.3 ΚΩ = = 288 453 ΚΩ