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Q. Explain common collector configuration?
It is called the common-collector configuration because (ignoring the power supply battery) both the signal source and the load share the collector lead as a common connection point:
Since the emitter lead of a transistor is the one handling the most current (the sum of base and collector currents, since base and collector currents always mesh together to form the emitter current), it would be reasonable to presume that this amplifier will have a very large current gain .the current gain for a common-collector amplifier is quite large, larger than any other transistor amplifier configuration.
conditions where the transistor is conducting. Cutoff occurs at input voltages below 0.7 volts, and saturation at input voltages in excess of battery (supply) voltage plus 0.7 volts. Because of this behavior, the common-collector amplifier circuit is also known as the voltage-follower or emitter-follower amplifier. This amplifier configuration has a voltage gain of slightly less than 1In the common-collector configuration, though, the load is situated in series with the emitter, and thus its current is equal to the emitter current. With the emitter carrying collector current and base current, the load in this type of amplifier has all the current of the collector running through it plus the input current of the base. This yields a current gain of ? plus 1:
Redraw the Wheat stone bridge circuit of Figure and show that Equation holds good for the null condition when the meter A reads zero current.
Q. Parallel - flash converter? Parallel / flash converter. Also known as parallel A/D converter, this circuit is the simplest to understand. It's formed of a series of co
its defination and mathematical expression if any
Starting Methods for Polyphase Induction Motors When high starting torques are required, a wound-rotor induction motor, with external resistances inserted in its rotor circuits
Thermal considerations: At continuous current, the voltage across the emitter-base junction V BE of a bipolar transistor get decreases 2 mV (silicon) and 1.8mV (germanium) fo
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Q. Consider a 1-bit version of the digital comparator shown in Figure. Note that the operation of this circuit is such that whichever output is 1 gives the desired magnitude compar
Dry type 10KVA-3ph-440/230v Dyn11. If we decide to wound above such specification, what current to be taken i.e. phase current or line current 7.57A or 13.12A
what is zeenar doide and explain its working
Determine the total energy loss: Two capacitors C 1 = 50 μF and C 2 = 100 μF are connected in parallel across 250 V supply. Determine the total energy loss. Figure
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