1. What is the forward bias of the diode? A positive voltage is applied to the p-segment and a negative voltage is applied to the n-segment. It is a positive bias.
The positive bias is that the diffusion current is greatly increased, and the reverse bias causes the drift current to increase. However, the drift current is formed by the movement of the minority, so there is a reverse saturation current!
2, the general low-frequency signal, the thickness of the resistance wire is to flow how much current, and the size of the resistance caused by the thickness does not count, because the resistance of the copper wire itself is very small, of course, exceptional circumstances!
3. The mos tube is a kind of carrier that is dependent on multi-sub-electrons. It is different from the case where the multi-child and the minority of the transistor participate in conduction. It is a self-isolated device that does not require the isolation islands in the transistor to save the core area and is suitable for very large-scale circuits.
It is characterized by voltage control! That is, the control terminal requires almost no current and is easy to integrate.
4. How to judge the cbe pole of the triode? And how to determine the gds of the mos tube.
a Directly check the data, b use the multimeter diode file, p is connected to positive, n is negative when there is a digital display, so there are several measurements, you can know whether it is pnp type or npn type, b end can be determined. Then use the multimeter's hef file to measure the magnification. If the pair is correct, the result can be judged.
The mos tube is generally connected to the heat sink portion by d. It is simple to determine d, add a power in gs, that is, use the multimeter diode to point to gs, and then the amount of ds will have a digital display. If there is a short circuit ds, then there is no number. Can determine gs.
6. DC feedback is to stabilize the static working point, and AC feedback is to improve the performance of the amplifier.
7. The series-parallel relationship between capacitor and resistor is reversed. The inductor should be the same as the resistor, but there is also the concept of mutual inductance, so there is still a difference? Need to ask for a certificate!
8, many of the oscilloscope digital display is only displayed in the screen for multiple cycles, too much is not displayed!
9. The common-base amplifier is a non-inverting amplifier with a large output resistance and a voltage gain of 1, which is called a freewheel.
The common-amplifier is a non-inverting amplifier with a large input resistance and a current gain of 1, known as a voltage follower.
The common amplifier is an inverting amplifier, and the input and output resistors are between the upper two, and the voltage gain is large and the current gain is also large.
Therefore, the common-source, common-base amplifier knows that the output resistance is large, and the input current is not attenuated to the end of the output resistance.
A common set of common, apparently large input resistance, the input resistance is not attenuated to the end of the output resistance
10. The output average voltage of the sinusoidal voltage is 0.9 times the effective value of the input voltage in the full-Bridge Rectifier circuit, so the average value of the output current (equivalent to a multimeter) is 0.9 times the effective value of the input voltage divided by the load resistance.
11-2011, the two probes of the oscilloscope are common ground, pay attention when double tracking, these two grounds must be connected together, especially when high voltage! Not all oscilloscope probes are common ground, and some are independent.
12, mos tube measurement method, generally gds arrangement. To use a multimeter, first power up the gs at both ends, that is, use the multimeter to point gs and then ds, you can measure the number. These are judged based on its own characteristics. Note that the capacitance at the gs terminal is small, u=q/c. If it is affected by the outside world or electrostatic, it can make u large and burn it off with a small charge.
13, u disk can not test a lot of small files, not then can not drive, stuck. For example, the IE file, these things are best compressed and then admitted.
14. Before measuring with a multimeter, you must understand what signal to measure and what gear to use.
15, tvs tube response speed is generally very fast!
16, the parameters of the triode and the parameters of the op amp need to be reviewed frequently, understand! Because it is very important!
17, a few working states of the triode need to be thoroughly understood!
18, the experimental inspection method summary: First, we should look at the circuit has several parts, each of which can be divided into three parts, power, input, output, one by one check, it must not be wrong. The other is to see if the measuring instrument is set correctly.
19, waveform superposition as long as you master the principle of Uac=Uab=Ubc.
20, the understanding of the choke: the inductance is blocking AC, through the DC signal, this point is basically the opposite of the capacitance.
The low frequency choke is to suppress AC through DC.
The high frequency eclipse is to suppress high frequency through low frequency and direct current.
21-3021, the amplifier circuit has DC coupling and AC coupling, the difference is self-aware!
22, the basic formula of the transformer turns N = V 4 power / 4.44fBmS, the formula derivation is in the learning materials. Fangbo changed 4.44 to 4.
The transformer design of the switching power supply, the volume calculation formula is:
Vcore=4ueP/fBm*Bm ue is the effective permeability and P is the transmission power
f is the switching frequency and Bm is the maximum magnetic flux density (T)
Bm hot-rolled silicon steel sheet, 1.11-1.5t and cold 1.5-1.7t, should have a better thing like iron-silicon-aluminum. The high-frequency iron powder core mpp is about 0.3t. See the learning materials for details.
23. The relationship between triode b and hef, b is the AC amplification factor, hef is the DC amplification factor, and b is frequency dependent. So there is a difference between the two.
24. The load is light and light and the resistance is small, but the cross current source is different. If the resistance is large, the output power is large. The load is the focus!
25, the production of network cable crystal head. The standard for the straight-through line is 586B, and the standard for the cross-line is one 586A and the other 586B. Among them, 1236 four lines are useful, and the other lines are reserved for the telephone line.
1 output data +, 2 output data - 3 input data + 6 input data - 4578 are for telephone lines
26, inductive load: that is, compared with the power supply, when the load current lags the load voltage by one phase difference, the load is inductive (such as the load is the motor, the transformer).
Capacitive load: Compared with the power supply, the load is capacitive when the load current leads the load voltage by a phase difference (for example, the load is the compensation capacitor).
Resistive load: Compared with the power supply, when the load current load voltage has no phase difference, the load is resistive (such as the load is white light, electric furnace). In the hybrid circuit, the capacitive reactance is large, and the circuit is capacitive and vice versa.
Using the formula, when the capacitor inductance is connected in series, X=j(wL-1/wc) if X>0, w is greater than w0 (resonance frequency) is inductive, and vice versa. When paralleling, admittance is used to calculate Y=j(wc-i/wL). If Y>0, w is greater than w0, that is, X is less than 0, capacitive, and vice versa.
27, air conditioning line 16a ordinary 10a, 1 square mm 4a, the general line is 2.5 square.
28, the Internet cat is connected to the router, (the general router has the function of the switch or hub) can then connect to the switch, or hub. The hub needs to use a dual network card to share the Internet.
29, the power supply filter capacitor charging, according to the formula u = q / c, when q is full, indicating that the charge is completed, the specific calculation should use what indefinite integral
Firstly, the charge of the capacitor plate at time t is q, and the voltage between the plates is u. According to the loop voltage equation, Uu=IR (I represents current), and because u=q/C, I=dq /dt (where d is the differential), after substituting, get Uq/C=R*dq/dt, which is Rdq/(Uq/C)=dt, then find the indefinite integral on both sides and use the initial condition: t=0 q = â—‹ [q = CU [1-e^ - t / (RC)] This is the function of the charge on the capacitor plate with time t. Incidentally, electrical engineering often refers to RC as the time constant. Correspondingly, using u=q/C, a function of the plate voltage as a function of time is obtained immediately, u=U[1-e^-t/(RC)]. From the formula obtained, only when the time t tends to infinity, the charge and voltage on the plate are stabilized, and the charging is over. However, in the actual problem, since 1-e ^-t/(RC) tends to 1 very quickly, after a short period of time, the change of charge and voltage between the capacitor plates is very small, even if we use high sensitivity. The electrical instruments are also unaware that q and u are changing slightly, so it can be considered that the balance has been reached and the charging is over. As a practical example, suppose U=10 volts, C=1 picofarad, R=100 ohms, which can be calculated using our deduced formula. After t=4.6*10^(-10) seconds, the plate voltage has reached 9.9 volts.
Generally, when t = rc, the capacitor discharges to 0.36u, or charges to 0.64u.
30, 78.79 series of pin arrangement is 132, the voltage drop is arranged in order, 2 is always the output.
31-4031, if you want 7824 voltage regulator, the front requires 28v DC signal, 28/1.2 = 23.3 AC voltage, these are the experience.
32. For the requirements of power supply or amplifier, the general output resistance is small, and the load capacity is good. The internal resistance is small!
33, lm3886.gif look at the file in the electronic diagram, the description: the upper left 22u capacitor, is to make the circuit's DC working state with 100% negative feedback. That is, the DC gain is 1, the operating point is very stable, and the change of the power supply voltage can be tracked. The DC signal is the same as the voltage follower.
47p capacitor 18k, resistance, phase adjustment (pid proportional integral derivative control). Here, the change of the frequency of the signal changes the magnitude of the gain. The lower the frequency, the larger the impedance of the 47p capacitor is, and the gain is still 18 times. However, there is a calculation for the high-frequency signal, which has an attenuation effect on the high-frequency signal.
2.2 ohms, 100nf is a high frequency noise bypass to improve the output.
220p is estimated to be a high-frequency signal to the 2.2 ohm resistor to pav a road.
The 0.7uh inductor is useless for low frequency signals and can be used completely. The input DC blocking capacitor is recommended to be non-polar.
To analyze such an op amp circuit, you should first understand which circuit type it is, whether it is AC in-phase or DC reversal.
34. How to do the problem: Understand the background; list the technical indicators; perform system partitioning, estimate the time, list the required instruments, and raise the difficulties; do hardware block diagram; perform software design; make circuit board debugging, and finally Conduct a finished product test.
35, the multimeter, especially the current file measurement current is, if the battery is insufficient, it will cause waveform distortion.
36. Use the secant method to check the short circuit problem of pcb and gradually narrow the scope.
37. Instrument amplifier, when measuring floating signal, there must be a connection between the input and output signals. If there is no electrical connection, the common mode voltage is floating, which will generate a lot of interference. For example, the DC signal amplification will come out one way. Wave, of course, is consistent with the frequency of the power supply. Figure instrumentation amplifier ad620.jpg. You can look at it and have a detailed discussion of this issue on 21ic. If the input is not electrically connected, at least 2, 3 feet should be added to the ground of the op amp to add a capacitance of about 1u to eliminate the voltage between the two grounds, that is, to solve the effect of the common mode voltage. Generally, the input and output grounds are connected, and the data of the ad620 is connected.
The common mode signal is generated because there is a distributed capacitance between the primary and secondary of the transformer, and this capacitance is much larger than the capacitance of the signal ground. Therefore, the two grounds are unequal, and the maximum voltage of 220v can be generated. Measured with the digital multimeter's AC file, this figure produces a floating common mode voltage of 10V. In the 2, 3 feet connected to the ground 1uf capacitor, the capacitance of the op amp ground to the signal input ground is greatly increased, the voltage of 220v will not pass, basically two grounds are equipotential. Maybe there is still a little deviation.
38, part of the use of protel:
Solve the problem that the LAN can't be used. It's very simple, just let dxp not access the network.
Set dxp to an inaccessible network in the properties of the network connection - Advanced - Settings - Exceptions. Of course you have to turn on the firewall. Keep open to prevent dxp from accessing the network
The method can be used in many places, such as prohibiting certain software from automatically upgrading.
On the copper, in order to fit 99, you need to choose mode when using ad6.3, for hatched, you can not choose solid
Align ega, coordinate eos, measure rm unit transform vu, dr design rule, tp online drc remove, (preferential option) pt interactive line, pl normal line, dl+ enter, open layer, 99 is dr, shift+s is per Layers can be viewed separately. You can use the +-number to change the layer. Shift+c often uses this to cancel the selection when the group is changed. Otherwise you will not be able to point to anything. Do set the board, visual network 0.508 25.4mm, the component grid is the smallest, the electrical network 0.2mm can be, this is more important, the wiring can be positioned at the time.
Due to the encapsulation of the PCB, the VALUE value of the SCH graph will not be automatically led to the PCB. It must be entered in the COMMENT command by pressing the command "=value". How important it is to dig haha, especially for the production staff. A very important thing, that is, according to the pcb to generate the library file. Ad6.3 in the design menu, choose to generate pcb library. Can also be achieved in 99.
In addition, AD6.3 has a cross selection mode in the tool menu, which is convenient for selecting components.
The date should be written after the pcb is drawn, and the detection point should be set when drawing the pcb.
The shortcut key (document option) color selection in DXP is 214, which is better for the eyes.
When soldering, please use the following method. First, you must use a lot of equipment to install the chip resistors, capacitors, etc. This is convenient for finding.
Put together things of the same value, such as the resistance and capacitance of 104. And indicate the package.
Secondly, when soldering, open the pcb, click the pcb on the lower right foot, and there is a pcb option in the pop-up menu. When you select it, a window will appear. In that window, there is a drop-down menu. The default is the nets point. Components, where you select the components inside will jump directly to that component. Convenient for welding to find.
Again, solder the chip first, then solder the chip resistors and capacitors in order, according to the selection method in the components, the same value is soldered together. Then solder other things.
40, protel draws the date and the version number.
41-5041. Ordinary potentiometer, if the 1 and 2 feet are connected clockwise, the resistance of the 1, 3 feet can be reduced, and the counterclockwise resistance is increased.
42. If you write the program in assembly language, the header file needs to be added by yourself. If you use c programming, then the header file is written by yourself, such as #include, and the assembly, for example, plus file startup.a51.
43. The function of the in-circuit emulator is to replace the mcu on the target board, so that it does not need to be written to the real mcu microcontroller.
44, protel component list export method shortcut rri and then export can become xls file can be opened with execl haha.
45, liquid crystal display module, such as 160 * 128 for 16 * 16 Chinese character lattice is 10 columns and 8 rows.
46. ​​The current measured by the multimeter DC gear is the average value, while the AC gear measures the effective value.
47, the difference between self-excited shock parasitic oscillation needs to be clarified!
48, single-chip learning experience
Use c language to use c compiler, compile c program into machine code.
Keil51 usage
New project is named test.uv2
Select single chip at89c51
New program, c is test.c compiled as .a or .asm
Add the c program to the project, point source group
Compile can point target to generate hex file Enter debug mode, press d button, d with a circle of the thing Run the program Press Stop, then press the d, turn off the debug mode.
To understand the hardware structure with c or assembly, especially assembly.
The minimum system of the 89c51 should include a crystal oscillator circuit, a reset circuit, and a led display.
The c language is case-sensitive. It is also possible to use the programmer to write the target program code to the microcontroller or use the emulator instead of the mcu.
In the 51 MCU, the external memory ram can only be the A accumulator, so the data operation of the external ram is sent through A. In addition, the internal ram can directly exchange data, but the external is not, such as the external 0100H. Data is sent to the 0200H unit and must be implemented by accumulator A. And to read or write the external ram, of course, you must also know the address of the ram. In the last two instructions, the address is placed directly in the DPTR. The address to be read or written should first be sent to the DPTR or Ri, and then read and write commands.
50. Method of dissecting the circuit board: It is necessary to arrange the components one by one in the sch diagram, connect the lines like the pcb, and then slowly analyze each chip in the circuit.
51, too small signal try not to use switching power supply to deal with, such as 10a constant current source through 0.001 ohm resistor.
51-6051, 50hz of the work frequency signal can be used to solve the pull with a band-stop filter. (The effect may not be good)
52. Several main indicators of switching power supply Current regulation rate: also called voltage stability, which indicates the relative change of output voltage when the current is from zero to maximum.
Voltage regulation rate: The input voltage changes and the output voltage changes relative to the load. Also known as the range in which the input voltage is allowed to vary.
Ripple Voltage Power Supply Efficiency Temperature characteristics.
53. The phenomenon of magnetic saturation in electromagnetic induction: the magnetic flux of the iron core in the winding has a certain amount, and does not increase indefinitely with the increase of current. When the current is large and a certain value, the magnetic flux does not change to reach magnetic saturation. , can not produce self-inductance voltage, it loses the ability to resist the applied voltage, that is to say, there is no inductive reactance, there is no inductance L. Basically it is close to zero.
54. Saturation of the triode: When Ib increases to a certain extent, and Vce is small, such as 0.3v, then Ib is continuously increased, and Ic is basically unchanged. This is the saturated working state of the triode. It will be saturated because c has internal resistance, or a series resistor, or the power supply voltage is not high enough. If the voltage is high enough, the resistance will not be saturated.
55, the so-called accuracy of 0.1 and 0.1%, followed by a percentage.
56. The energy storage state of the capacitor is that the open circuit voltage of the capacitor is constant, the current is zero, and the energy is 1/2 (CU*U).
The energy storage state of the inductor is short-circuited, the current is constant, the voltage is zero, and the energy formula is 1/2 (LI*I)
57, common mode voltage, understanding of differential mode voltage
For example, the inverse proportional operation circuit, if the input signal is +/-3v to ground AC signal, the input signal is equal to Un at any time and equal to 0 (in-phase grounding), instead of the end of the op amp to withstand 3v Differential mode voltage. Note that the differential mode voltage is the voltage between + and - of the cloud. As long as the amplitude of the input signal does not exceed the power supply voltage, it does not matter with the differential mode voltage of the op amp. This is only reflected when the comparator is connected. The two terminals are connected to different voltages, and the open-loop amplification factor is large. Not a positive supply voltage or a negative supply voltage.
58. What is the excitation current, which can be simply understood as the secondary no-load, flowing through the primary current, so the primary inductance is large, so the excitation loss will be small. For AC, the larger the inductance, the greater the inductive reactance, just like the concept of larger resistance.
59. How does the emitter amplifier not be distorted?
Mainly depends on RC, RB1, RB2, VCC, and also need to add a Re, play a negative feedback role, stabilize the working point.
IB must have a certain current value, IC can play a significant role in its control, the most important thing is that the input signal can not be too large, too large will cause distortion, on the contrary, if the signal is small enough, even if the static working point is not in the load line The center point is also not distorted. For example, the input signal amplitude is small, and distortion may not occur even if the operating point is higher or lower. So, to be exact, the generation of waveform distortion is caused by improper matching of the signal amplitude with the static operating point setting. To meet the requirements of large signal amplitudes, the static working point should preferably be as close as possible to the midpoint of the AC load line.
Whether the static operating point is appropriate or not has a great impact on the performance and output waveform of the amplifier. If the working point is too high, the amplifier will easily produce saturation distortion after adding the AC signal. At this time, the negative half cycle of uO will be cut off; if the operating point is low, the cutoff distortion is easy to occur, that is, the positive half cycle of uO is shrunk (generally cut off) Distortion is not as obvious as saturation distortion). None of these conditions meet the requirements for undistorted amplification. Therefore, dynamic debugging must be performed after the selected working point, that is, a certain input voltage ui is added to the input end of the amplifier to check whether the size and waveform of the output voltage uO meet the requirements. If not, adjust the position of the static working point.
When the current flowing through the bias resistors RB1 and RB2 is much larger than the base current IB of the transistor T (generally 5 to 10 times), the RB1, RB2 can be calculated according to the load current.
60, rail to rail The difference between an op amp and a normal op amp is not the same as their output. The former is the c-pole d-pole output, while the latter is the emitter output or the source output.
61-6961. The discharge of the capacitor to the resistor or the charging of the power source E to C is the concept of direct current. Do not engage with the RC oscillator circuit. The resonant frequency of the RC oscillator circuit is f=1/2piRC.
62. Why do you want high input impedance and low output impedance?
One possible reason is the high input impedance, the load is very light on the input signal, does not affect its waveform, (equivalent to making the input signal an ideal current source) and the low output impedance can carry a relatively heavy load (equivalent To make the op amp itself an ideal voltage source). This is more in line with the actual situation, the signal is better, and the loading capacity is very strong.
63. The induced electromotive force in the magnetic field is related to the amount of change in the magnetic flux, that is, the magnetic field strength B and the v of the loop motion, both vector and directional.
Ampere's law has a cloud, and the magnetic field strength B established around the long straight wire with current is proportional to the current on the wire, and inversely proportional to the cube of the distance of the wire.
Ampere formula F=Bvq=BIL=BLq/t, I=q/t
64, whether it is an integral circuit or a differential circuit, should meet RC
Integral circuit, in order to prevent the low-frequency signal gain from being too large, the resistor is connected in parallel with the feedback capacitor (such as 30k for 5k signal). In the differential circuit, a 100 ohm resistor is connected in series with the input capacitor to limit the current, and two Zener diodes connected to the k pole are connected in parallel with the feedback resistor R, and a small capacitor is used to compensate the phase, so that the circuit does not oscillate and the stability is improved. Sex.
65. Inductive excitation is the accumulation of energy, and the degaussing of the inductor is to release energy.
66, the inductance value will not change, but the inductive reactance will change according to the signal frequency. The capacitance value will not change, but the capacitive reactance will change according to the signal frequency.
67. Capacitance formula, inductance formula V=(di/dt)*LI=(du/dt)*C
68, inductor open circuit and capacitor short circuit
Since the inductor has the characteristic of preventing the current from changing, when the inductor is disconnected, the inductor keeps the current unchanged, but since the loop impedance is greatly increased, according to Ohm's law: U=IR, or U=IZ, it can be seen that the inductance is generated. The voltage will be very large, although the inductance also has distributed capacitance, but the distributed capacitance is very small, then the impedance will be very large, so the instantaneous high voltage is very high.
If the instantaneous energy generated by the inductor is not released by its own distributed capacitance, it will release to the air, and the air will form a loop. The resistance of the air is very large, which will generate a large high voltage, spark, and even explosion. Electrical energy is converted to heat and light energy.
The open circuit of the inductor is actually the same as the short circuit of the capacitor.
Capacitance open circuit generates a large current, and the capacitor has the characteristic of keeping the voltage constant. The current is so large that the voltage is constant, and the instantaneous power is very large. Therefore, since the capacitor itself contains a certain resistance, the capacitor will be overheated and burned. Even the explosion.
69. The current of the motor changes with the speed; when the load is overloaded, the motor speed decreases, the current is greater than the rated current, and the time is long. The motor overheating causes insulation breakdown and burns.
The motor is not a purely resistive load. The external reflection of the motor is impedance. The inductive reactance of the motor is not fixed. The induced current in the rotor is different with different speeds. The external reactance is different. The motor is the motor, not the resistor. It is not working under the rated voltage. It cannot be calculated with a simple ohmic rate.
Interpretation and self-understanding: When the load is overloaded, it means that the torque of the motor is not enough, and it can't work at the best efficiency point. If the torque is not enough, the inductance of the motor will decrease, then the current will be larger than normal, and the torque will not be enough to cause the motor to turn. Do not move (such as with a pliers can not turn), the number of revolutions decreased. According to the formula 9550*P=T*n, the rated speed of the motor is relatively reduced or the point power is increased (refer to the input active power * efficiency = P, that is, the load power, which refers to the mechanical output power of the motor), The button moment of the motor will be suitable for the original overload situation.
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