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Op Amp Bandwidth
OP_BANDWIDTH1.CIR Download the SPICE file
Its all about speed - at least that's what this topic is about. In an ideal world, an op amp responds accurately and instantly to an audio or video signal. But in the real world, there's a limit on the highest frequency (bandwidth) and fasted edge your op amp can process. A few simple concepts provide insight into an amplifier's bandwidth. And knowing this can help you make better op amp and circuit choices.
THE REAL STORY OF GAIN
To understand bandwidth, we must understand the real gain equation. You've probably seen the ideal Closed-Loop voltage gain equation for a non-inverting amplifier.
But, what's the real story including the op amp's internal gain? It actually looks like this
where A - open-loop gain - internal
gain of the op amp
Here's the beauty of this equation. Check out what happens to GCL if A is made large.
The bottom line? The gain is set by R1 and R2, not the op amp gain A! ( This fact certainly simplifies op amp circuit design.) And consequently, A can vary due to initial tolerances or temperature drift, but the voltage gain holds rock solid set by the resistor values!
So what's the problem with the real gain equation? Although, A is large (+100,000) at lower frequencies, it falls at higher frequencies to well below unity (<<1). And when A drops near (R2+R1) / R1, GCL drops too. The frequency where GCL falls below the ideal gain is called the closed-loop bandwidth fc.
CIRCUIT INSIGHT Run a simulation of OP_BANDWIDTH1.CIR. The closed-loop gain for this circuit is GCL = (10k+10k)/10k = 2 V/ V. Plot the AC Response for the output at V(4) and open loop gain A using the equation V(4)/(V(2)-V(1)). To get a clearer view, select log for the Y-Axis. For this particular op amp, A has a DC gain of 100,000 V/V, then falls off above 100 Hz. What about GCL? You can see that all is well as long as A >> 2. But when A drops close to GCL, the closed-loop gain takes a dive.
Note where GCL begins to drop. The frequency where the voltage falls to 0.707 of its intended value is the cutoff or -3 dB frequency, fc. ( Gain in decibels = 20∙log(0.707) = -3dB. )
HANDS-ON DESIGN Pick a higher gain for this amplifier. Choose R2 somewhere in the range of 10 k to 10,000 kΩ. Rerun the simulation. Yes, GCL looks great at low frequencies, but what happened to the bandwidth? Because GCL is essentially bounded by A, the bandwidth fc gets smaller for a higher gain! Life and circuit design are full of compromise, and gain versus bandwidth is a fine example.
Can you extend the bandwidth GCL? Sure, make A larger by increasing its gain or bandwidth. The op amp model simulates the DC gain with EGAIN 3 0 1 2 100K. Increase the 100k by a factor of 10 or so. Or, you can increase the bandwidth by decreasing RP1 or CP1 by a factor of 10. Run a new simulation. Did your new op amp extend the bandwidth at V(4)?
How can you predict the bandwidth at any gain? A simple equation gets you the answer.
fu (Unity Gain Frequency) - the frequency where the open-loop gain A
falls to unity (1V/V or 0dB).
Why the name "noise gain"? Typically, noise is modeled as a voltage source at the positive input of the amplifier. So the non-inverting gain is used to calculate the resulting output. Does the noise gain equation look familiar? It should! The noise gain same as the signal gain for the non-inverting amplifier GN = GCL.
For example, a non-inverting amplifier having a fu = 10 MHz and R1 = R2 = 10k gives a closed-loop gain GCL = 2 and a noise gain of GN = 2. Calculating its bandwidth fc, we get
HANDS-ON DESIGN Run a few simulations with various voltage gains of OP_BANDWIDTH1.CIR. Plot the AC Response at the output at V(4) and A via the equation V(4)/(V(2)-V(1)). Adjust the gain by varying R2 and R1. You should be able to predict the bandwidth at V(4) for any of your chosen gains.
The term fu is also called the the Gain-Bandwidth-Product (GBP). Why call it this? If we rearrange the above equation, we get
Notice, that the product of gain GN and bandwidth fc is constant and bounded by GBP (fu)!
What does this mean? You can't arbitrarily set the gain and bandwidth for a given op amp. Increase the gain GN, and the bandwidth fc will drop to keep GBP constant. Alternatively, if you need a higher bandwidth, then you must choose a lower gain. If you need both higher gain and bandwidth, you're out of luck with this device. You need to pick an op amp with a higher GBP (fu) on its data sheet.
THE INVERTING AMPLIFIER
What about the inverting amplifier? The results are similar with a slight twist. Let's start with it's closed-loop gain equation - significantly different than the non-inverting gain.
where A is the internal gain and the feedback factor is
β = R1 / (R1+R2).
Note, the feedback factor, β = R1 / (R1+R2), is the same for inverting or non-inverting amplifier. That's because β is simply the gain of vo to the neg input (β = V- / Vo), just like the non-inverting amp.
Similar to the non-inverting amplifier, when A is large, the ideal inverting gain is achieved.
However, as frequency increases and A drops close to the ideal gain, GCL begins to drop. How do we predict this frequency where the gain falls off? Similar to the non-inverting amplifier we calculate
The fc calculation also uses the noise gain. But here's the twist, the noise gain for the inverting amp is the same as the non-inverting amp!
For example, an op amp having a fu = 10 MHz and R1 = R2 = 10k gives an inverting gain of GCL = -1. However, the bandwidth is reduced by the noise gain GN = (R1 + R2) / R1 = 2 giving
Here's the disadvantage of the inverting amplifier: signals are amplified by the R2 / R1 ratio, but the bandwidth is knocked down by the larger ( R1+R2 ) / R1 ratio.
CIRCUIT INSIGHT Try out the inverting amplifier in the OP_BANDWIDTH1.CIR. Plot the AC Response for the output V(14) and the op amp's internal gain V(14)/V(12). A log plot on the Y-Axis can give a better view. The gain should match the ideal -R2 / R1 as long as A is larger than (R1+R2)/R1.
HANDS-ON DESIGN Crank up the gain by choosing R2 somewhere in the range of 10 k to 10,000 kΩ. Rerun the simulation. For any gain, GCL should be bounded by A. Does the bandwidth get smaller as you increase gain?
NON-INVERTING VS. INVERTING AMPLIFIER
Here's a showdown between the two classic amplifiers. The challenge is this: for the same gain, which amplifier has the greater bandwidth? We'll use a voltage gain of 2 for both circuits.
The bandwidth champion is the non-inverting amplifier! However, this bandwidth difference is less for larger gains. Still, all other things being equal, choose the non-inverting amplifier to maximize your bandwidth.
CIRCUIT INSIGHT Run a simulation of OP_BANDWIDTH1.CIR. Set the resistors in the non-inverting and inverting amplifiers to values in the table above. Plot the AC output at V(4) and V(14). Did the non-inverting gain live up to expectations?
Another way to measure circuit speed is how fast the amplifier responds to a step input. In the real world, the step input represents a quick brightness change in a video signal or the rising / falling edge of a clock signal. Run a simulation and plot the Transient Response at V(4) and V(14). How much faster does the non-inverting output reach 90% of its final value compared to the inverting output?
MORE TO BANDWIDTH
Although fu (GBP) and gain play a major role in determining bandwidth, there's more to it. Slew rate and heavy loading can cut into your bandwidth. Slew rate would not only make an interesting design topic, but a challenging simulation topic as well. Its on the list of future topics.
For a more detailed description of the op amp, see the Basic Op
Download the file or copy this netlist into a text file with the *.cir extension.
OP_BANDWIDTH1.CIR - OPAMP BANDWIDTH * VS 1 0 AC 1V PWL(0US 0V 0.01US 1V 100US 1V) * * * NON-INVERTING AMPLIFIER R1 2 0 10K R2 2 4 10K XOP1 1 2 4 OPAMP1 * * INVERTING AMPLIFIER R_1 1 12 10K R_2 12 14 10K XOP2 0 12 14 OPAMP1 * * * OPAMP MACRO MODEL, SINGLE-POLE * connections: non-inverting input * | inverting input * | | output * | | | .SUBCKT OPAMP1 1 2 6 * INPUT IMPEDANCE RIN 1 2 10MEG * DCGAIN =100K AND POLE1=1/(2*PI*RP1*CP1)=100HZ * GBP = DCGAIN X POLE1 = 10MHZ EGAIN 3 0 1 2 100K RP1 3 4 1000 CP1 4 0 1.5915UF * OUTPUT BUFFER AND RESISTANCE EBUFFER 5 0 4 0 1 ROUT 5 6 10 .ENDS * * * ANALYSIS .AC DEC 5 10 100MEG .TRAN 0.01US 0.5US * * VIEW RESULTS .PRINT AC VM(4) .PRINT TRAN V(4) V(14) .PROBE .END
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