JS-1200-545/DT – Digitally controlled charger
... The JS-1200-545/DT is a microprocessor controlled battery charger. Digital communication line allows remote adjusting of the charging current or voltage and alarms and live report of system values and alarms. It can be switched on/off by an external TTL signal.The charger has more than 90% efficienc ...
... The JS-1200-545/DT is a microprocessor controlled battery charger. Digital communication line allows remote adjusting of the charging current or voltage and alarms and live report of system values and alarms. It can be switched on/off by an external TTL signal.The charger has more than 90% efficienc ...
Small-Signal Equivalent Circuit
... Also shown in Figure 6.2 are the sinusoidal variations in the gate-to-source voltage, drain current, and drain-to-source voltage, as a result of the sinusoidal source vi. The total gate-to-source voltage is the sum of VGSQ and vi. As vi increases, the instantaneous value of vGS increases, and the bi ...
... Also shown in Figure 6.2 are the sinusoidal variations in the gate-to-source voltage, drain current, and drain-to-source voltage, as a result of the sinusoidal source vi. The total gate-to-source voltage is the sum of VGSQ and vi. As vi increases, the instantaneous value of vGS increases, and the bi ...
lab4a - inst.eecs.berkeley.edu
... In a BJT, shorting the collector and base forms a two-terminal device that behaves just like a diode. In a MOSFET, shorting the drain and gate gives a two-terminal device with behavior that is qualitatively diode-like, but has a (roughly) quadratic current-voltage relationship. Even though it’s deci ...
... In a BJT, shorting the collector and base forms a two-terminal device that behaves just like a diode. In a MOSFET, shorting the drain and gate gives a two-terminal device with behavior that is qualitatively diode-like, but has a (roughly) quadratic current-voltage relationship. Even though it’s deci ...
sot-23 bipolar transistors transistor(npn)
... Transition frequency (VCE= 10V, IC= 1mA, f= 30MHZ) CLASSIFICATION OF hFE(1) RANK ...
... Transition frequency (VCE= 10V, IC= 1mA, f= 30MHZ) CLASSIFICATION OF hFE(1) RANK ...
Pass Transistors Beef Up Voltage Regulators Last month we
... Last month we learned how to make any fixed voltage regulator adjustable. However, many fixed regulators such as the 7805 limit their output current to 1.5 amps. While this may be adequate for many QRP transceivers, amateurs would often like to run a low power mobile 2 meter transceiver on their hom ...
... Last month we learned how to make any fixed voltage regulator adjustable. However, many fixed regulators such as the 7805 limit their output current to 1.5 amps. While this may be adequate for many QRP transceivers, amateurs would often like to run a low power mobile 2 meter transceiver on their hom ...
Lecture-3: Transistors - Dr. Imtiaz Hussain
... • A transistor is basically a Si or Ge crystal containing three separate regions. ...
... • A transistor is basically a Si or Ge crystal containing three separate regions. ...
The Rating of Power Transistor
... • Typically VCBO ~ VCES > VCER > VCEO (VCER resister value less than 1K), After test, VCES is close to VCBO. • If VCER resister value less than 1K, VCES ~ VCER VCER is close to VCES, After test, VCER is close to VCES. • The voltage rating is tested according to the leakage current – No IC current in ...
... • Typically VCBO ~ VCES > VCER > VCEO (VCER resister value less than 1K), After test, VCES is close to VCBO. • If VCER resister value less than 1K, VCES ~ VCER VCER is close to VCES, After test, VCER is close to VCES. • The voltage rating is tested according to the leakage current – No IC current in ...
Transistor Switch and Emitter Follower Phys 3610/6610 Lab 18 Student: TA:
... 1.) Apply the input and examine the output waveform using a scope. Document the phase shift. What is the output impedance when Vout = 5 V? 2.) Let us load this transistor switch with a 10 kΩ resistor to simulate the more realistic situation where your circuit drives some other input. How does the lo ...
... 1.) Apply the input and examine the output waveform using a scope. Document the phase shift. What is the output impedance when Vout = 5 V? 2.) Let us load this transistor switch with a 10 kΩ resistor to simulate the more realistic situation where your circuit drives some other input. How does the lo ...
Electrical Engineering Department EENG351
... 1. The conductivity between the plates of a transistor changes according to their connections. 2. A npn transistor is almost considered as a diode characteristics. 3. The Base-Emitter represents in a npn transistor a diode in a Forward-Bias state. 4. The input characteristics with an open or closed ...
... 1. The conductivity between the plates of a transistor changes according to their connections. 2. A npn transistor is almost considered as a diode characteristics. 3. The Base-Emitter represents in a npn transistor a diode in a Forward-Bias state. 4. The input characteristics with an open or closed ...
VEGEtek - 003 - Instructables
... High-side and low-side sensing Direct sensing has 2 methods: High-side and low-side sensing. It depends on the position of the shunt resistor with respect to the load. This op-amp configuration is called “differential amplifier” which it amplifies the voltage difference between its inputs. ...
... High-side and low-side sensing Direct sensing has 2 methods: High-side and low-side sensing. It depends on the position of the shunt resistor with respect to the load. This op-amp configuration is called “differential amplifier” which it amplifies the voltage difference between its inputs. ...
Homework 5
... Label the voltage V = 25.0 V and the resistances (clockwise from b) R1 = 20.0Ω, R2 = 5.00Ω, R3 = 10.0Ω, R4 = 10.0Ω, and R5 = 5.00Ω. Computing some equivalent resistance of R1 and R2 in series we have Rs = R1 + R2 = 25.0Ω Computing the equivalent resistance of Rs , R4 , and R5 in parallel we have ...
... Label the voltage V = 25.0 V and the resistances (clockwise from b) R1 = 20.0Ω, R2 = 5.00Ω, R3 = 10.0Ω, R4 = 10.0Ω, and R5 = 5.00Ω. Computing some equivalent resistance of R1 and R2 in series we have Rs = R1 + R2 = 25.0Ω Computing the equivalent resistance of Rs , R4 , and R5 in parallel we have ...
ch 16 Electricity Essential Questions
... The following questions are to be reviewed at the beginning if the chapter to familiarize yourself with what is to be expected. At the end of the chapter you will be answering the questions as a study guide tool to prepare for the test. Name: __________________________Date: __________ Bl: ________ ...
... The following questions are to be reviewed at the beginning if the chapter to familiarize yourself with what is to be expected. At the end of the chapter you will be answering the questions as a study guide tool to prepare for the test. Name: __________________________Date: __________ Bl: ________ ...
