Analogue Circuit Design

[fm4743]

Design an amplifier circuit which uses an OpAmp and is powered by dual 15 V power
supplies, and which has the following characteristic at low frequencies:

v(out) = 3v(1) - 5v(2)

first of all how can an op amp be powered by a dual 15 v supply i don't know.. in our class we haven't covered it but here is a question..

i want to know how to analyse a circuit and extract characteristics at low frequency or even high frequency when need to ..

 

[chuckey]

+- 15V is THE standard for supply line voltages, others tend to be exceptional. Your equation makes no sense, do you mean Vo = (3 X Vin) - 5V ?
Frank

 

[KerimF]

Perhaps he meant:
Vout = 3*V1 - 5*V2
where V1 and V2 are two input voltages.

If this is the case:
R1 is between V2 and the inverting input of the opamp.
R2 is between Vout and the inverting input of the opamp.
R3 is between V1 and the non-inverting input.
R4 is between ground and the non-inverting input.

Vout = V1 * R3 / ( R3 + R4) * ( 1 + R2 / R1) - V2 * R2 / R1
But
Vout = V1 * 3 - V2 * 5

I guess you know how to find R1, R2, R3 and R4.
Hint: the values of R1 and R3 could be assumed as 22K for example.

 

[Borber]

This is diferential amplifier with gain 3 on noninverting input and gain 5 on inverting input.

KerimF changed his post in meantime.

Just to add quick calculation about above problem.
Gain at inverting input is calculted as usual it is R1/R2 and it is 5. Gain at noninverting input is 1+inverting gain and it is equal 6. Because gain at noninverting input must be 3 attenuator 0.5 R3 R4 was added.

 

[KerimF]

Vout = V1 * R3 / ( R3 + R4) * ( 1 + R2 / R1) - V2 * R2 / R1 [equation_1]
But
Vout = V1 * 3 - V2 * 5 [equation_2]

From comparing [equation_1] and [equation_2] we get:
R2 / R1 = 5 [result_1]

R3 / ( R3 + R4) * ( 1 + R2 / R1 ) = 3
R3 / ( R3 + R4) * ( 1 + 5 ) = 3
R3 / ( R3 + R4) = 0.5
R4 = R3 [result_2]

 

[LvW]

Design an amplifier circuit which uses an OpAmp and is powered by dual 15 V power
supplies, and which has the following characteristic at low frequencies:
v(out) = 3v(1) - 5v(2)
first of all how can an op amp be powered by a dual 15 v supply i don't know.. in our class we haven't covered it but here is a question..
i want to know how to analyse a circuit and extract characteristics at low frequency or even high frequency when need to ..

Hi fm4743,
I`ve got the impression that your knowledge of operational amplifiers is limited.
Thus, some basic information:
* It is quite normal to use a dual power supply (each opamp has two input nodes for this purpose), because in this case the dc operational point is at Vdc,out=0V (good enough for this purpose).
Thus, no specific bias circuitry at the input and no coupling capacitors at the input and at the output are necessary.
* The task has to be solved for "low frequencies" because for very high frequencies the formulas as given in previous answers must not be used.
These formulas apply for a very large gain of the opamp unit only (approximated to be infinity).