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NE5517A View Datasheet(PDF) - ON Semiconductor

Part Name
Description
Manufacturer
NE5517A 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
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