Loading ...The Operational Amplifier: How it works
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The operational amplifier is probably the most widely used component in ICs (integrated circuits). The name is self-explanatory: it is an amplifier with a high gain (40dB or more) and with additional circuitry it can be used to do a lot of arithmetic operations such as sum, difference, integrals and so on.
In this article I will try to explain you how it works.
The architecture
Most op-amps available on the market make use of a two-stage configuration. In fact, only the differential stage and the high-gain stage contribute to the total amplification.
This configuration provides a differential stage, a gain stage, a level translator and a buffer. They are joined in DC, because of the diffuculty to obtain large capacities in ICs.
The first stage, the diff-amp, is used to amplify the difference between two signals, and to furnish a hign input impedance. The gain stage provides a darlington cuple to raise gain. Being joined in DC, there is the possibility that you have to shift the working point (read: change voltage) between two stages. In fact, the eventual difference is treated as a signal, and it is amplified! The level translator provides for that. Finally, the buffer gives low output impedance, that is necessary if you want to be able to treat low loads (consider the series of two resistances).
The ideal caractheristics are:
- Infinite gain
- infinite badwidth
- Infinite CMRR (Common Mode Rejection Ratio)
- Infinite Rin
- Rout = 0
- Iin = 0
- Perfect balance (V1=V2=0 => Vout=0)
Let’s see in detail the various stages.
The Differential Amplifier
You can say that this is the heart of the op-amp. It provides the two inputs, the basic amplification between their differential part (see later) with high CMRR and a high input resistance, necessary to treat low signals.
Any generic signal can be divided in a differential part and a common part, as follows:
V1 = Vc + Vd/2 and V2 = Vc - Vd/2
then
Vc = (V1 + V2)/2 and Vd = V1 - V2
The diff-amp will amplify both Vc and Vd, but you ideally would Vc not be treated: you are interested only in the amplification of the differential signal! So, you have to require a high CMRR:
CMRR = Vdm/Vcm
where Vdm is the differential gain and Vcm the common gain. If the common gain is 0, Vc will not be amplified at all and CMRR -> infinite.
How can you obtain such a high CMRR? Let’s consider the structure of a simple diff-amp:
As you can see, there are two BJT npn transistors (if you don’t know what they are, consider them like current ampilifiers. The current in the collector is the base current amplified) in common emitter configuration. The circuit has to be as simmetric as possible.
Consider the following signals: V1 = -V2 = Vx/2.
Because of the circuit simmetry, the two emitter currents (Ie = Ib + Ic = Ib(1 + Av)) are opposite, so the tension Ve at the strands of the resistor Re in zero. You can so consider, thanks to the general simmetry, the simple common-emitter BJT.
Making some calculus, you obtain
Adm = -gm*Rc
Make attention to the minus sign! The gain is negative, so you will obtain, in output, a negative tension, if you apply a positive Vd. That’s why there is an ap-amp terminal called inverting. If you apply to this terminal a positive voltage, you will obtain a negative output.
Consider, now, the following signals: V1 = V2 = Vx‘. The two emitter currents will now be equal (Ie1 = Ie2), so Ve is not zero! You have to consider the followind circuit, an emitter-degenerated common emitter:
In fact, you can consider Re as the parallel of two resistor with 2Re resistance. Again, you obtain:
Acm = -gm*Rc / (1 + 2gm*Re)
So that
CMRR = Vdm/Vcm = 1 + 2gm*Re ~ 2gm*Re
It’s very important to have a high Re! As you can’t obtain big resistances in ICs, you will have to turn to current generators, characterized from high output resistances.
As regards Rin, it is Rin = 2*hie, where hie is the a11 parameter or the hybrid BJT model (little signals) we have used in the analysis.
Gain stage
There’s no much to say here. Simply, this stage takes the diff-amp output and amplifies it. It is realized with a darlington couple, that is two BJT in CC - CC or CC - CB configuration (-> Google to know more). The total gain is the product of the two separate gm gains.
Level translator
Sometimes, you can have to shift tensions to line up working points. If they are different, the op-amp can treat this difference as differential signal! You can realize a level translator with a simple tension divider, but you will attenuate not only bias, but also signal. So, it’s better to use a current generator or a Zener diode (read: voltage generator) in appropriate configurations.
This way, you will not absorbe power. Ideally, you could simply use a trasformer, but you can’t to because of his narrow bandwidth.
Buffer
There are many ways to realize a buffer (high Rin - low Rout), but you can simply use an emitter follower. Of course, there are better solutions, but they are not so important to understand how ther op-amp works IMHO.
Real vs Ideal
Let’s take into consideration the ideal characteristics described above:
- Infinite gain
- infinite badwidth
- Infinite CMRR (Common Mode Rejection Ratio)
- Infinite Rin
- Rout = 0
- Iin = 0
- Perfect balance (V1=V2=0 => Vout=0)
Of course, gain will not be infinite. It will be 10^5, or more. It could be raised using another darlington, but you’ll have to fight against noise.
Banwidth will be limitated, too. It is usually raised with a Cascode configuration in substitution of the only BJT in the diff-amp. Thanks to the reduced Miller effect, Bw is a lot higher (~80 MHz).
Using a current mirror instead of Re in the diff amp, CMRR reaches ~90dB. Really high.
Rin is about 2MOhm, Rout ~ 50 Ohm. Input currents are ~ 50nA, perfect balance is never reached: Short circuit output current ~ 20 mA.
The above characteristics can be found in the constructor’s datasheets, for example: uA751 TEXAS INSTRUMENTS
Conclusions
The operational amplifier is probably the most important IC in use nowadays. It allow us to realize almost every arithmetic operation in hardware. To give an idea, this simple configuration it is an integrator:
In fact, Rin infinite means that the op-amp input current is zero, so V+ = V- = 0.
The only current is in the resistor and the capacitor. Ir = Ic = C(dVout/dt) but Ir = Vin/R. So, Vo = -(1/RC)*integral(Vin dt)
I hope this will be useful to understand its enormous potential.
Bibliography
Jacob Millman - Arvin Grabel, Microelectronics, 2nd ed.. McGraw-Hill, Inc, New York, 1987.
uA741 General-Purpose Operational Amplifiers datasheet by TEXAS INSTUMENTS
License
This article is under the Gpl License.
Hope you have enjoyed it. ©2004 The Cicoandcico company.
Tags: electronics, hardware, science