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# NE5517DG View Datasheet(PDF) - ON Semiconductor

 Part Name Description Manufacturer NE5517DG Dual Operational Transconductance Amplifier ON Semiconductor NE5517DG Datasheet PDF : 14 Pages
 1 2 3 4 5 6 7 8 9 10 Next Last INPUT NE5517, NE5517A, AU5517 APPLICATIONS +15V 0.01mF 10kW 51W 390pF 1.3kW 3, 14 2, 15 4, 13 − 11 62kW NE5517 1, 16 7, 10 5, 12 + 6 0.01mF −15V 10kW 0.001mF Figure 20. Unity Gain Follower 8, 9 OUTPUT 5kW −15V CIRCUIT DESCRIPTION The circuit schematic diagram of one-half of the AU5517/NE5517, a dual operational transconductance amplifier with linearizing diodes and impedance buffers, is shown in Figure 21. Transconductance Amplifier The transistor pair, Q4 and Q5, forms a transconductance stage. The ratio of their collector currents (I4 and I5, respectively) is defined by the differential input voltage, VIN, which is shown in Equation 1. VIN + KT q In I5 I4 (eq. 1) Where VIN is the difference of the two input voltages KT ≅ 26 mV at room temperature (300°k). Transistors Q1, Q2 and diode D1 form a current mirror which focuses the sum of current I4 and I5 to be equal to amplifier bias current IB: I4 ) I5 + IB (eq. 2) If VIN is small, the ratio of I5 and I4 will approach unity and the Taylor series of In function can be approximated as KT q In I5 I4 [ KT q I5 * I4 I4 (eq. 3) and I4 ^ I5 ^ IB KT q In I5 I4 [ KT q I5 * I4 1ń2IB + 2KT q I5 * IB I4 + VIN I5 * I4 + VIN ǒIBqǓ 2KT (eq. 4) The remaining transistors (Q6 to Q11) and diodes (D4 to D6) form three current mirrors that produce an output current equal to I5 minus I4. Thus: ǒ Ǔq VIN IB 2KT + IO (eq. 5) The term ǒI qǓ B 2KT is then the transconductance of the amplifier and is proportional to IB. V+ 11 D4 Q6 Q7 D6 Q10 Q11 Q14 Q12 7,10 Q13 8,9 2,15 D2 −INPUT 4,13 Q4 Q5 D3 +INPUT 3,14 V OUTPUT 5,12 1,16 AMP BIAS Q2 INPUT Q1 Q9 Q8 Q15 Q16 Q3 D7 R1 D8 D1 V− 6 D5 Figure 21. Circuit Diagram of NE5517 http://onsemi.com 7