1. For the circuit shown in Fig.1 assume that R11 = 1.8 kW, TR2 transistor b = 100, and the resistance of the speaker element equals to 100 W and answer the following questions:
a) (4 points) What is the function of TR2? Is it PNP or NPN transistor?
The transistor TR2 acts as a switch. When the signal is high, the transistor turns on, and current is pulled through the speaker. For a low signal no current flows. The signal bounces up and down with the frequency of the oscillator, and this causes speaker to beep. This is NPN transistor.
b) (4 points) What is the voltage at the speaker element, when VO (input to the circuit) equals to 5V?
The transistor will be in SATURATION region with VCE = 0.2V, hence voltage on the speaker will be 4.8V. Remember, that in saturation state the transistor can be modeled as a pair of batteries with VBE-ON = 0.7V, VCE-SAT = 0.2V. (We can easily verify that the transistor is indeed in saturation. Remember conditions required for saturation: IB ³ 0, IC £ bIB. In this case IB = (VO – VBE-ON)/ R11 = 2.4 mA, and IC = (5V - 0.2V)/100 = 48 mA £ bIB = 240 mA).
.
Figure 1.
2. For the speaker circuit shown in Fig. 2,
a) (3 points) Why do we connect the input to the speaker to Vout from the second inverter (as shown) instead of the third inverter?
The 2nd inverter has a zero output when the oscillator is turned off and we do not want to use battery power to drive DC current trough the speaker.
b) (3 points) How will the frequency of the oscillator theoretically be changed if we double the value of R12 (R12 = 136 kW)?
If we double the value of R12 the frequency will decrease 2 times.
Figure 2.
3. For the op-amp circuit with a resistive divider in the input network shown in Fig. 3 answer the following questions (Rs = 5kW, Rf = 1MW, R1 = R2 = 10kW):
a) (3 points) Is this op-amp inverting or not inverting?
It is inverting op-amp.
b) (4 points) Write out the equation for Vout as a function of Vin.
In order to write this equation you should use the assumption of ideal op-amp and virtual short condition
V- = V+ = 0, (1)
It means, that V- is effectively grounded.
Now we will use Ohm’s law to find a voltage in node A
VA = Vin - Iin Rs
And KCL in nodes B and A
I1 = V1/R1 = (Vin - Iin Rs)/R1 = - If = - Vout/Rf , (2)
Iin = I1 + I2 = (Vin - Iin Rs)/R1 + (Vin - Iin Rs)/R2, (3)
From (3) Iin is
Iin = Vin (1/R1 + 1 /R2)/(1 + Rs/(1/R1 + 1/R2)), (4)
substituting (4) in (2) and solving for Vout/Vin, we obtain the answer
Vout/Vin = - R2 Rf / (R1 R2 + R1 Rs + R2 Rs). (5)
c) (4 points) What is the input impedance (Vin/Iin) of this circuit?
The equation (4) in part (b) is actually the solution for this problem too. We can write that
Vin/Iin = Rs + 1/(1/R1 + 1/R2), or
Vin/Iin = Rs + R1 R2 /(R1 + R2), or
Vin/Iin = (R1 R2 + R1 Rs + R2 Rs)/ (R1 + R2). ]
Figure 3.