The approximate voltage across the forward-biased base-emitter junction is 0.7 volts.
The voltage across a semiconductor diode (and across the base/emitter junction of a transistor) decreases as temperature increases: the actual figure is -2mV/°C.
The voltage drop across the emitter-collector junction develops the output signal with the help of a resistor or two in series. The output is 'seen' at the collector.
In the common emitter configuration, gain is hFe or collector resistance divided by emitter resistance, whichever is less. Placing a capacitor across the emitter resistor effectively makes the emitter resistor less, for higher frequencies, so the gain is higher for higher frequencies. This creates a high pass filter, or a low cut filter, depending on what you want to call it.
It can be used as a feedback and to ground unwanted signals and frequencies
The voltage across a battery in a parallel circuit is equal to the voltage across each bulb because Kirchoff's Voltage Law (KVL) states that the signed sum of the voltages going around a series circuit adds up to zero. Each section of the parallel circuit, i.e. the battery and one bulb, constitutes a series circuit. By KVL, the voltage across the battery must be equal and opposite to the voltage across the bulb. Another way of thinking about this is to consider that the conductors joining the battery and bulbs effectively have zero ohms resistance. By Ohm's law, this means the voltage across the conductor is zero, which means the voltage across the bulb must be equal to the voltage across the battery and, of course, the same applies for all of the bulbs.
Collector-emitter saturation voltage refers to the voltage drop across the collector-emitter junction of a transistor when the transistor is in saturation mode. It is the minimum voltage required to keep the transistor in saturation, where the transistor is fully turned on and conducting maximum current.
In the cut-off region of a transistor, the base-emitter junction is not forward-biased, meaning the transistor is effectively turned off. As a result, no current flows from collector to emitter, and the voltage across the collector to emitter (V_CE) is approximately equal to the supply voltage (V_CC) connected to the collector. Therefore, V_CE is at its maximum value, close to V_CC, indicating that the transistor is not conducting.
A: Because it is a voltage amplifier the current will inversely reflect the voltage across a resistor
To calculate the output voltage (V_out) of a common emitter amplifier, you first determine the voltage drop across the load resistor (R_C) using Ohm's Law: V_out = I_C × R_C, where I_C is the collector current. The input voltage (V_in) is then related to the output voltage through the transistor's current gain (β) and the base-emitter voltage (V_BE). The final output voltage can be expressed as V_out = V_CC - V_CE, where V_CC is the supply voltage and V_CE is the voltage across the collector-emitter junction. Ensure to account for the biasing and any AC signals superimposed on the DC operating point.
The voltage across a semiconductor diode (and across the base/emitter junction of a transistor) decreases as temperature increases: the actual figure is -2mV/°C.
If a bypass capacitor is used the voltage drop across emitter resistance is reduced which in turn increases the gain.....
Emitter biasing is when you add a resistor between the emitter of a transistor and the 0v rail so that any voltage developed across the emitter will subtract from the voltage on the base and effectively turn the transistor OFF. We are talking about an NPN transistor and the transistor is an "ordinary transistor" or BJT (bi-polar Junction Transistor). For more information on transistor biasing see: Talking Electronics website.
It depends on the transistor. Minimum base-emitter junction voltage can be as low as 0.6 volts for a silicon transistor, and as low as 0.2 volts for a germanium transistor.
The voltage drop across the emitter-collector junction develops the output signal with the help of a resistor or two in series. The output is 'seen' at the collector.
Generally about 0.6V, but it could be a bit higher, 0.7 or so, if the base/emitter current is high.
In a ce amplifier, an increase of base voltage causes the collector current to rise. This causes an increased voltage drop through the collector load resistor, so the collector voltage drops. With a cc amplifier the increase in current causes more voltage across the emitter load resistor, therefore the emitter voltage rises.
A: Vce is the voltage across the transistor . Ie is the emitter current. Ico is the collector current with the base open. Or really the leakage.