PROBLEM 065 · Easy · The virtual ground
Minus Ten
Inverting amplifier: gain of −10
The problem
Build an inverting amplifier. The non-inverting input is tied to a 0 V Input source named “inp”. The signal source named “in” drives a resistor Rin to the inverting input “inn”, and a feedback resistor Rf runs from “out” to “inn”. Choose them for a gain of −10 (20 dB in magnitude).
The feedback holds “inn” at the same voltage as “inp”, so it is a virtual ground: the input current is simply Vin/Rin and it all flows through Rf, giving Vout = −(Rf/Rin)·Vin. The input resistance seen by the source is Rin (not infinite as for the non-inverting amplifier).
The test bench
- The test bench supplies the op amp as a ready-made block with terminals on the nets named “inp”, “inn” and “out”: DC gain 80 dB, one pole at 1 kHz (unity-gain frequency 10 MHz), output resistance 100 Ω.
- The test bench applies the AC stimulus to “in” (the source “inp” stays at 0 V).
Numbers to hit
- Gain magnitude between 19.5 dB and 20.2 dB
- Bandwidth (−3 dB) at least 0.7 MHz
- Peaking at most 0.5 dB
Design it yourself
Draw the circuit on a schematic canvas, size every transistor, and see it simulated in your browser with ngspice and the SkyWater SKY130 models. Every transistor lights up green, gold or red by operating region as you move the sliders.
Sign in or create a free account →More on feedback and stability
- Times TenWrap two resistors around an op amp: the gain is set by their ratio, not by the op amp.
- Short on GainWhen the op amp’s gain is not huge, the textbook formula is off by several percent. Correct for it.
- Ramp MakerReplace the feedback resistor with a capacitor: the output is the integral of the input, with a gain that falls at 20 dB per decade.
- Light CatcherTurn a current into a voltage with one resistor, then stop the input capacitance from making it ring.
- No Ringing ZoneA capacitive load and a weak output stage make a second pole. Cancel its effect with a lead capacitor across the feedback resistor.