Electronics Engineering

OK. I have it. What now ?

A: A rectifier is simply a diode or is it? At hi frequency the storage time or the disconnect time becomes significant that is why it looks distorted to the scope

hie is input impedance when signal applied at its base.

A: There is no calculation involved it is specified by the manufacture as a level +/- volts or even current

A: If the signal is analogue then an ADC analogue to converter is required. The output will be in a digital form and that can be decoded and displayed as readout using LEDS or any other visual media

develop the same charge as the charged body

That's the component to use wherever a capacitive reactance equal to -j(159,155/frequency) is required.

A: They both use transistors. In actuality before an op amp becomes an op amp a discrete transistor made is made to provide it capabilities before it gets integrated into a substrate chip we call op amp. OP AMP have two input to work as a differential amplifier the input are bases closely match to provide virtual no voltage or current offset. Single ended amplifier has one input its reference is the signal source return.

To limit the amount of current to the ports

A: The input peak value is the guide for PIV

Normally if a resistor fails they will either increase in value or open up at all (open circuit).

they are same expect error .in adc0809 1/2 lsb unadjusted error is present and in adc 0809 1lsb error is present.

A: Follow me on this the lm317 is a 1.25v regulator period. So to make it look like a 11.25 volts is simple assume the the regulator load is 125 ohms so the current through will be 1.25 volts divided by the 125 ohms or 10 ma. now install a resistor from the adj terminal to ground to IR drop 10v which will be added to the 1.25v so 10v divided by 10 ma =1000 ohms from adj to ground will offset the voltage by 10 plus the 1.25 volt give us 11.25 output from output to ground. since we offset the output voltage by 10 volts the adj pin only needs very little current we could have use 1250 ohms and 10k to achieve the same thing. There is a formula to calculate the output voltages but if you understand this you will have no problems. And by the way this scheme can be used for a 5 v,12v any of the three pins regulators. By changing the 1k you will change the output volts since the 1.25 volts is fixed and the 125 ohms if fixed. Furthermore to make it behave like a current source up to an amp or so all you need is a resistor in series with the load and take the output at the adj terminal and good luck.

There are several factors that affect the heat output of a voltage regulator. A linear regulator such as a 7805 series will pass the same amount of current as the load draws. The regulator will also have a certain voltage across it and can be calculated by subtracting the output voltage (5V in this case) from the source voltage. The source voltage might be anywhere from 7 volts up to as much as 30 volts.

The heat generated in the regulator can be calculated using the power equation, power = current x voltage. As stated above, the current is dependant on the load. As the current increases, so the power dissipated in the regulator increases. Similarly, as the source voltage increases, so does the power dissipated by the regulator.

The regulator will always show a temperature rise but how hot it gets depends on how effectively the heat is drawn away from the regulator. If the regulator is mounted on an effective heat sink, it can dissipate large amounts of heat without becoming hot. Without a heatsink, the same regulator in the same situation will become far hotter.

All regulators have certain limits for power dissipation and temperature. The data sheet will have information on the limits and they should always be noted during a design phase.

when we replace the resistor with a capacitor ,the current will flow until the capacitor charge when capacitor will fully charged there is no current through the circuit because now capacitor will act like an open circuit.

for more info plz E-mailt me at "zaib.zafar@yahoo.com"

A relay is basically a controllable switch that connects a supply voltage to a load. If the relay is switch such that the load is not connected, that could be considered an open load. If it is switched the other way the load will be connected to its supply voltage. A defect in a relay would be something that caused it not to function correctly. It could be in the switch portion of the supply portion of the relay. Hence you are referring to different things and a defect may or may not cause the load to be open.

1 volt is equal to 0.01 hectovolt.

Therefore, there are 100 volts in one hectovolt

A: Three way must have a resistance and voltage source and current source to a very close tolerance,

To isolate the bias circuitry of the transistor from the feedback network with RF currents.

Yes, but you should add a bleed resistor in parallel to a motor run cap if you want to use it for a hid ballast to drain the cap when its not in use, this is to avoid dangerous voltages remaining in the cap, and remove the potential for electrocution

connect a register in series with line and take the voltage at the register terminal if circuit is close.

A: Seven segment display can give digital number visually from a source like a computer that has binary outputs the number 64 in digital would be 100000 and also can display alphanumeric which is what humans can readily see So basically transfer information from a machine language to human language

Mix it with a local oscillator whose frequency is (the IF frequency) away

from the frequency of the FM signal you're interested in.

yes

3MHz