The theoretical value of the output resistance of the single-tube amplifier circuit
1. **共射放大电路** - 对于单管共射极放大电路，输出电阻\(R_{o}=R_{c}\)。 2. **共集放大电路（射极跟随器）** - 其输出电阻\(R_{o}=R_{e}//\frac{(R_{b} + r_{be})}{(1+\beta)}\)。 3. **共基放大电路** - 输出电阻\(R_{o}=R_{e}//\frac{(R_{b} + r_{be})}{(1+\beta)}\)。 点击前往免费阅读更多精彩小说
Discussion and conclusion of single-tube AC amplifier circuit
The related conclusions and discussion of the single-tube AC amplifier circuit are as follows: * * 1. In terms of static work points ** 1. * * Important ** - The setting and adjustment of the static working point was crucial. A reasonable setting can make the amplifier work stably and reliable. To obtain the maximum undistorted voltage, the static operating point should be located at the middle of the AC load line. In order to stabilize the operating point, certain conditions must be met, such as <BQ>> II <I21>. 2. * * Calculation Method ** - The static operating point can be calculated by a specific formula, such as <R = U21UU> II>, or <<CBBBQE + RR1EBEQBQEQCQR-CQCCQR>-UE= I <EcCQCreERCCEQ + RR-I = E-U-U = EUbeI = ICQBQ>. The calculation involved the parameters of various components in the circuit, such as the base power supply, bias resistance, collector power supply, collector resistance, etc. These components interacted to determine the state of the static operating point. - The static working point can be measured with the Model MT-47 Multimeter. * * 2. Dynamic parameters ** 1. * * Calculation of voltage amplification and input and output resistance ** - The voltage amplification factor is related to the input and output resistance calculation, and the calculation result is usually affected by certain conditions (such as <26> 1>(IEQHR = 0>). - The input resistance, r_{i}, has the following values: r_{i}= R_times beLiouru = A '-_, and because of the two values, we have the following values: LcL//R = RR'_, beBBBi21BBbeR <<Rr_, so we have the following values: beirR =_, and mVMV +_beta += rr' bbbe_, where Omega = r'bb300c_. The input resistance can also be calculated by using [sisiR-uuu]. - The output resistance, r_{o}, can be calculated by the formula, where, u is the output voltage at no-load, and u0 is the output voltage at load. The calculation of the output resistance is related to factors such as the load resistance in the circuit. When all the excitations are assumed to be zero, the controlled source is cut off, and the output resistance can be calculated accordingly. 3. * * Impact on circuit performance ** - The variation of circuit parameters will affect the static operating point, voltage amplification and output wave. For example, when an AC signal was input, the circuit only had a static operating point when the direct current passed through it. The AC signal would interact with the static direct current, affecting the voltage and current at each point in the circuit, which in turn affected the amplification factor and output wave. For example, in a circuit consisting of a mos tube and a semiconductor, the positive and negative half cycles of the AC signal would change the working state of the mos tube and the semiconductor, thus affecting the amplification performance and output characteristics of the entire circuit. - In the experiment, you can change the component parameters in the circuit (such as R_{C}, R_{L}, etc.) to observe and measure the impact on the static operating point, voltage amplification, and output wave. This helps to understand the working principle and characteristics of the single-tube AC amplifier circuit. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Single-sided push-pull amplifier circuit
Single - ended push - pull amplifier circuit。 <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Single-ended input output circuit
Single-ended input and output circuits are a type of differential circuit classified by signal input and output methods. According to the different input and output methods of the signal, the differential circuit was divided into four types: double-ended input and double-ended output, double-ended input and single-ended output, single-ended input and double-ended output, single-ended input and single-ended output. For a single-ended output differential circuit, regardless of the connection method, as long as it is a single-ended output, its amplification factor is half of the basic amplifier circuit; if it is a double-ended output, its amplification factor is the same as the basic amplifier circuit. Single-ended input and output circuits had specific functions and meanings in practical circuit application scenarios. For example, they played a role in circuit designs that required specific amplification or specific circuit connections. At the same time, when the common-mode input (when the two input ends input signals with the same magnitude, it is the common-mode input) and the difference-mode input (when the two input ends input signals with the same magnitude and opposite poles, it is the difference-mode input), the single-ended output will also show different current characteristics. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Is the feedback amplifier circuit a closed loop circuit?
In an amplifier circuit, a circuit with feedback was called a closed-loop circuit. The feedback amplifier circuit had a feedback path from the output to the input. This feedback circuit was in a closed-loop state, so the feedback amplifier circuit was a closed-loop circuit. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
R2 of the inverse proportional amplifier circuit
In an inverted proportional amplifier circuit, R2 usually referred to the balancing resistance or compensation resistance. Its resistance is equal to the value of R1 and Rf in parallel (R2 = R1//Rf). When R2 = R1//Rf, the output voltage caused by the input bias current of the op amp can be zero, thereby eliminating the effect of the input bias current on the output voltage. However, due to the difference in bias currents between the non-inverted and inverted ends of the op amp, the input stage devices are not exactly the same.(The difference between the two is the input offset current Ios). Even if a balancing resistance is introduced, the bias current will still produce a certain output voltage. However, under normal circumstances, the error caused by the very small Ios (usually nA) can be ignored. In the case of amplifying a very weak signal (such as a uV level signal), the error caused by the input bias current may not be ignored. In this case, a precision op amp with a smaller input bias current and offset voltage should be selected. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The Principle of the Single-Chip Computer's Automatic Voltage-output Circuit
The following are some principles of voltage output circuits related to single-chip processors: ** 1. Power supply self-locking to open the circuit (Take a circuit as an example)** 1. ** Initial state ** - Before the circuit is powered up, the switch "STAR" is turned off. When the single-chip computer is not powered up, the base of T1 is grounded through R9(100k) and is in the cut-off state. Test and T1, which were connected to the base resistance R7 of T3, were both in the off state, so T3 was also in the off state. At this time, the power supply +9V was isolated by T3 and was not loaded on the voltage regulator chip IC2. The output of IC2 was maintained at a low level. 2. ** Starting Circuit ** - Press the button "Test" to start the circuit. The base of T3 is grounded through R7, Test, and b-e of T2, making T3 turn on. At this time,+9V was added to the voltage stabilizing chip of IC2 through T3, and the output of IC2, VRCC, was added to the single-chip computer. 3. ** Self-locking open ** - After the single-chip computer was working, it would output a high voltage through IO2 and turn on T1 through R8. At this time, even if Test is released, the base of T3 can be grounded through R7, LED1, and T1 to achieve power self-locking. After that, the single-chip software could make the IO2 port turn low again, causing T1 to be turned off, and then T3 to be turned off. ** II. Principle of the direct current step-down circuit (part related to the voltage output of the single-chip computer)** 1. ** Resistance voltage division ** - The simplest way to step down the voltage is to divide the voltage with a resistance, such as the formula [V out = 12> times5> div7 + 5]=5V. However, this method cannot be used with a load. 2. ** Zero-Stabilizer Diode Step-Down ** - The principle was to use the characteristics of the voltage stabilizing circuit, which was cheap and practical. For example, in a series voltage stabilizing circuit, the base of BG was clamped at 13V by the voltage stabilizing device D, and its transmitter output a constant 12V operating voltage. 3. ** Low-voltage-difference linear regulator ** - The circuit is simple, easy to learn and use, and can achieve stable voltage output. 4. ** Switch power supply step-down circuit ** - This kind of circuit had a large voltage drop and a large output power. ** 3. Limiting circuit principle (related to voltage output)** 1. ** Basic limiting circuit ** - In the positive half cycle, when the input voltage is greater than or equal to 0.7V, the positive direction of the circuit is turned on. At this time, the output voltage is clamped at 0.7V. When <VIN> is less than 0.7V, it is in the cut-off state. During the negative half-cycle, it is equivalent to the current reversing, and it is also in the cut-off state. At this time,<VOUT>= VIN>, the <VOUT> wave follows the change of <VIN>>. 2. ** Bias limiting circuit ** - In order to generate different magnitude of the limiting voltage, a bias voltage, VBIAS, was added to the circuit. Only when the voltage was greater than or equal to the voltage, the voltage could be turned on. At this time, VOUT was clamped, and its value was 0.7V+VBIAS. ** 4. Principle of clamp circuit (related to voltage output)** 1. ** Basic clamp circuit (Diode-in-Combination with a Condenser)** - Assuming that the time constant is large enough so that the output wave will not be distorted. When the input is negative in the negative half cycle, the voltage is turned on and the voltage is gradually charged to the voltage. During this process, the voltage is zero. When the input voltage is positive in the positive half cycle, the voltage is cut off, and the voltage is equal to the voltage on the voltage plus the positive half cycle voltage. At this time, the voltage is equal to 2V. 2. ** Bias clamp circuit ** - Similar to the limiting circuit, in order to obtain the required clamp value, a bias voltage was added to the circuit. When the applied bias is in the same direction as the conducting direction of the LED, the clamp value will increase by <<V1>>,<<OUT = 2V +V1>>. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Single-phase heating tube power and resistance
For single-phase heating tubes, there was a certain relationship between power and resistance. According to the formula, resistance = voltage * voltage/power. For example, for a 220V, 1500W heating tube, its resistance value is 220 * 220 / 1500 = 32.27 Ohms. There were also some common power resistance conversion relationships, such as 100w = 484 Ohms, 200w = 242 Ohms, 300w = 161.33 Ohms, 400w = 121 Ohms, 500w = 96.8 Ohms, 600w = 80.67 Ohms, 700w = 69.14 Ohms, 800w = 6…Wait. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Why is the common-emit amplifier circuit inverted?
In the common-emitting amplifying circuit, when the input signal is positive for half a cycle, the base potential of the NPM triode is raised by the input signal, the base voltage to the ground increases, and the degree of continuity increases, that is, the resistance of the collector and the transmitter decreases, the current increases, the voltage at both ends of Rc increases, the voltage of the C pole of the triode to the ground decreases, and the output capacity enters a discharge state when the potential of the output is higher than that of the C pole. At this time, the lower end of the load Ri is positive, the upper end is negative, and the output signal is negative for half a cycle. When the input signal is in the negative half cycle, the base potential is pulled down by the input signal, the voltage of the B pole to the ground is reduced, the degree of continuity is reduced, the current is reduced, the voltage at both ends of Rc is reduced, the potential of the C pole is raised, and the output voltage is lower than that of the C pole. At this time, the lower end of the load Ri is negative, the upper end is positive, and the output signal is in the positive half cycle. Therefore, the input and output of the common-emit amplifier circuit were reversed. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What is the resistance of the single-phase steamer heating tube?
If it was a single-phase steamer heating tube, the resistance would be about 16 Ohms, but this was relatively rare. Usually, two to three were connected in parallel. If no answer is given, it can be calculated according to the formula R = U*U/P (R is resistance, U is voltage, and P is power). The result calculated above was the resistance of the heating tube when it was working. The cold resistance might be smaller. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
How big is the resistance of the single-phase steamer heating tube?
For the single-phase steamer heating tube resistance, it can be calculated according to the formula R = U*U/P (R is the resistance, U is the voltage, and P is the power). If the power and voltage of the steamer were known, the resistance could be calculated. It was also mentioned that the resistance of a single-phase rice steamer (similar to a steamer) was about 16 Ohms, but this was relatively rare, usually 2 to 3 in parallel. The resistance of the heating tube of the steamer was usually between tens to hundreds of Ohms, but the specific value was also affected by the power, voltage, material and other factors of the steamer. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>