â¢ Related QuestionsHow to build VoltMeter for car battery (found circuit need help understanding it)?
If the battery is discharging over a short area of time e.g. in one day then attempt disconnecting it for the evening. in the journey that your motor vehicle starts right here day then the battery is effective, if no longer the battery or battery terminals are defective. If the battery is effective then attempt right here: place an ammeter interior the battery circuit with the ignition off, any present day drain will manifest on the size. the present drawn would be a demonstration of what it is it is draining your battery. Then attempt pulling fuses separately till the present drain stops and this allow you to comprehend which circuit has a topic, from there it is an argument of discover ways to locate the subject. in case you haven't any longer have been given an ammeter then attempt pulling a number of fuses for the evening till you have stumbled on the circuit that drains your battery. maximum present day vehicles have some consistent present day draw whether it fairly is basically an alarm gadget or electronics yet those should not be sufficient to discharge a healthful battery over a short era. fairly generally hidden lights fixtures may be the wrongdoer final on once you think of they're off e. g. boot, below-bonnet or glove-field. one ingredient to submit to in techniques whilst coping with a DC electric gadget is that the situation of electric connections could be needed, examine for corrosion and tightness of battery, floor, and starter motor circuit terminals - it is conceivable for a bounce start to furnish sufficient extra skill to conquer terminal resistance issues. stable success certainly one of those great style of stable solutions and that i'm so sluggish - could discover ways to style with extra suitable than one finger!
For a buffer amplifier using op-amp, what happens if input voltage is greater than supply voltage?
The input voltage range is usually specified by the input common mode range and should be within the power supply range or sometimes considerable less (as much as half the supply in some). It is a good idea to have series current limiting resistors too if the inputs can be widely separated in voltage, so that high current doesn't pass through the clamp diodes. See the internal circuit of a bipolar op-amp like a 741 on its data sheet for an understanding. Some JFET input op-amps can have the inputs outside the power supply within reason, and also have no clamp diodes. The effect of driving the input beyond the specified range depends on the individual device type, considering there are different input circuits, but the sort of effects are latchup (can be catastropic in some older designs), reversal of input polarity, loss of open loop gain, loss of common mode and power supply rejection. The simple point is that a circuit will not behave as expected. For a buffer amplifier the output is driven to saturation when the closed loop gain x input signal exceeds the output voltage limits, which are often specified one or two volts below the supply rails. This is just overdriving the amplifier as long as the inputs remain within their specified range. This is a different specification to the input voltage specification. When the output is saturated the loop cannot operate as a closed feedback loop - the loop is broken. However if the output is just driven into saturation by exceeding the input drive for that gain, the output signal is just clipped. The tops and bottoms are cut off AC signals. As soon as the input comes back in range the loop closes again and all works as expected.
Design a 10 volt power supply with the same specs as problem 2.3(I have this).?
The common way is to have a current reducing resistor connected to the Collector of the series pass transistor, which has the other end of the resistor connected to the base of the same transistor. From the base of the transistor, connect the Zener diode cathode lead to the transistor base, and the series resistor, with the diode anode lead connected to ground. This establishes a constant input from the rectifiers and filter capacitors supply to the diode, and the base of the transistor. If calculated properly, the diode current will be just enough to keep it at it's break over point, and hold a steady voltage value at the base of the transistor. So long as current draw doesn't exceed the capacity of the transformer, or the transistor, it should remain linear over a wide range of loads. Worst case for the transistor would be just at the point of thermal runaway for the transistor. Once that point is reached, however, there is little chance of saving the transistor. The characteristic curves for the transistor should be consulted as these will give a lot of useful information that your problem wants to know. A good resource for such as this is the Radio Amateurs Handbook which is printed by the ARRL. Using the emitter follower design is to make a low impedance voltage output device. If the collector is used, that is the power supply is connected to the emitter, then the collector becomes the output source, and tends to be a high impedance which is a configuration used for fixed current sources.
How to design a switch mode power supply (SMPS) circuit with dual output voltages?
For the best answers, search on this site A switching-mode power supply (SMPS) is a power supply that provides the power supply function through low loss components such as capacitors, inductors, and transformers -- and the use of switches that are in one of two states, on or off. The advantage is that the switch dissipates very little power in either of these two states and power conversion can be accomplished with minimal power loss, which equates to high efficiency.Usually a switching-mode power supply is circuit that operates in a closed loop system to regulate the power supply output. Although the benefits of switched-mode techniques are great, there is a penalty paid in the increased noise present at both input and output of the supply due to the power switching techniques. Also the associated control circuitry is much more complicated than its linear counterpart.The switching mode power supply contains a transformer/coil and to make this as small as possible, the internal switching frequency has to be quite high, something typically in the range between 20KHz and 1MHz. This also makes the device noiseless to human ears. The oscillator noise is often conducted onto the input and output lines with a frequency that varies with the load.There are many different types of withing power supplies.Off Line Switching Mode Power supply is a power supply in which the ac line voltage is rectified and filtered without using a line frequency isolation transformer. After rectifications and filtering the voltage is converted to the needed voltage using a swithing regulator circuit, which usually provides also isolation function (power goes though high frequency transformer). The typical PC power supplies (AT and ATX power supplies for example)are built in this way.
Why there is sudden increase in the small reverse saturation current at the breakdown voltage in diode?
Reverse biased usually refers to how a diode is used in a circuit. If a diode is reverse biased, the voltage at the cathode is higher than that at the anode. Therefore, no current will flow until the diode breaks down. Connecting the P-type region to the negative terminal of the battery and the N-type region to the positive terminal, corresponds to reverse bias. The connections are illustrated in the following diagram: Because the p-type material is now connected to the negative terminal of the power supply, the 'holes' in the P-type material are pulled away from the junction, causing the width of the depletion zone to increase. Similarly, because the N-type region is connected to the positive terminal, the electrons will also be pulled away from the junction. Therefore the depletion region widens, and does so increasingly with increasing reverse-bias voltage. This increases the voltage barrier causing a high resistance to the flow of charge carriers thus allowing minimal electric current to cross the pn junction. The increase in resistance of the p-n junction results in the junction to behave as an insulator. This is important for radiation detection because if current was able to flow, the charged particles would just dissipate into the material. The reverse bias ensures that charged particles are able to make it to the detector system. The strength of the depletion zone electric field increases as the reverse-bias voltage increases. Once the electric field intensity increases beyond a critical level, the pn junction depletion zone breaks-down and current begins to flow, usually by either the Zener or avalanche breakdown processes. Both of these breakdown processes are non-destructive and are reversible, so long as the amount of current flowing does not reach levels that cause the semiconductor material to overheat and cause thermal damage. This effect is used to one's advantage in zener diode regulator circuits. Zener diodes have a certain - low - breakdown voltage. A standard value for breakdown voltage is for instance 5.6V. This means that the voltage at the cathode can never be more than 5.6V higher than the voltage at the anode, because the diode will break down - and therefore conduct - if the voltage gets any higher. This effectively regulates the voltage over the diode.