LT6210/LT6211 - Single/Dual Programmable Supply Current, R-R Output, Current Feedback Amplifiers
... a minimum output current drive of 75mA. ...
... a minimum output current drive of 75mA. ...
MAX9130 Single 500Mbps LVDS Line Receiver in SC70 General Description Features
... offset, coupled with the receiver’s 0 to +2.4V input voltage range, allows an approximate ±1V shift in the signal (as seen by the receiver). This allows for a difference in ground references of the driver and the receiver, the common-mode effects of coupled noise, or both. The LVDS standards specify ...
... offset, coupled with the receiver’s 0 to +2.4V input voltage range, allows an approximate ±1V shift in the signal (as seen by the receiver). This allows for a difference in ground references of the driver and the receiver, the common-mode effects of coupled noise, or both. The LVDS standards specify ...
EL2257 Datasheet
... capacitive drive. However, other applications may have high capacitive loads without a back-termination resistor. In these applications, a small series resistor (usually between 5Ω and 50Ω) can be placed in series with the output to eliminate most peaking. The gain resistor (RG) can then be chosen t ...
... capacitive drive. However, other applications may have high capacitive loads without a back-termination resistor. In these applications, a small series resistor (usually between 5Ω and 50Ω) can be placed in series with the output to eliminate most peaking. The gain resistor (RG) can then be chosen t ...
R 12 - Courses
... Resistors R2 and R5 were in series. We have replaced them with an equivalent resistor, which we call R11. Next, we recognize that R4 and R11 are in parallel, and that that parallel combination is in series with R3. Performing both of these two steps at once, we can move to the next equivalent, on th ...
... Resistors R2 and R5 were in series. We have replaced them with an equivalent resistor, which we call R11. Next, we recognize that R4 and R11 are in parallel, and that that parallel combination is in series with R3. Performing both of these two steps at once, we can move to the next equivalent, on th ...
Gr X Physics Practicals Instructions: Please write these experiments
... 2. Least count of ammeter and voltmeter are noted and checked for zero error. 3. First resistor R1 is connected parallel to voltmeter and 3 readings are noted from ammeter and voltmeter by adjusting the rheostat. 4. The resistor R1 is removed and R2 is connected in its place. 5. The procedure is rep ...
... 2. Least count of ammeter and voltmeter are noted and checked for zero error. 3. First resistor R1 is connected parallel to voltmeter and 3 readings are noted from ammeter and voltmeter by adjusting the rheostat. 4. The resistor R1 is removed and R2 is connected in its place. 5. The procedure is rep ...
Solution
... I- (S.2.3) : Diode Characteristics: Power diode is a two-terminal pn-junction device ……….); The equivalent circuit and i-v curve are displayed as follows, where the diode conducts when a battery is connected across its terminals . ...
... I- (S.2.3) : Diode Characteristics: Power diode is a two-terminal pn-junction device ……….); The equivalent circuit and i-v curve are displayed as follows, where the diode conducts when a battery is connected across its terminals . ...
LMH6551 Differential, High Speed Op Amp
... The circuit shown in Figure 1 is a typical fully differential application as might be used to drive an ADC. In this circuit closed loop gain, (AV) = VOUT/ VIN = RF/RG. For all the applications in this data sheet VINis presumed to be the voltage presented to the circuit by the signal source. For diff ...
... The circuit shown in Figure 1 is a typical fully differential application as might be used to drive an ADC. In this circuit closed loop gain, (AV) = VOUT/ VIN = RF/RG. For all the applications in this data sheet VINis presumed to be the voltage presented to the circuit by the signal source. For diff ...
Student Study Aids
... Problem: Express the resistance value of 2,500,000 Ω in engineering notation. Next, substitute the correct metric prefix for the power of 10 used with engineering notation. Answer: In engineering notation 2,500,000 Ω = 2.5 106 Ω. Because the metric prefix mega (M) corresponds to 106, the value of ...
... Problem: Express the resistance value of 2,500,000 Ω in engineering notation. Next, substitute the correct metric prefix for the power of 10 used with engineering notation. Answer: In engineering notation 2,500,000 Ω = 2.5 106 Ω. Because the metric prefix mega (M) corresponds to 106, the value of ...
Document
... upon the components resistance. Let us take a look at the schematic diagrams for the circuits below. In the lower left picture, the two resistors are series. Note that we can move one of the resistors any where in the circuit while maintaining the same current through each. In the lower right pi ...
... upon the components resistance. Let us take a look at the schematic diagrams for the circuits below. In the lower left picture, the two resistors are series. Note that we can move one of the resistors any where in the circuit while maintaining the same current through each. In the lower right pi ...
AD623 Data Sheet
... The input signal is applied to PNP transistors acting as voltage buffers and providing a common-mode signal to the input amplifiers (Figure 40). An absolute value 50 kΩ resistor in each of the amplifiers’ feedback assures gain programmability. ...
... The input signal is applied to PNP transistors acting as voltage buffers and providing a common-mode signal to the input amplifiers (Figure 40). An absolute value 50 kΩ resistor in each of the amplifiers’ feedback assures gain programmability. ...
