You first have to look at your circuit and understand it. You know already it is a rectifier. This is not a linear function, where does it happen? There is no diode. Now you have a look at your opamps. Normally Vcc is on pin7 and GND/Vee on pin4. If you connect it the wrong way, you would see the blue smoke, what opamp is this supposed to be? Now to your homework. Analyze the circuit with two different cases: Vin>0 and Vin
Of course the pin connections for a quad op amp are different than those that Ban gave, typical for a single op amp in an 8-pin package...but why would one not use two sections in the same quad package, instead of apparently two separate packages?
Answers to Ban's questions should lead you in the right direction: there is a very good reason that Vee is indeed tied to ground.
Yet another question: Does the circuit present the same resistance load to the input capacitor through both the positive and negative halves of an input waveform cycle? If it does not, what happens to the DC level on the right side of the capacitor when you put in a signal? Part of the answer to these questions has to do with what protection there is on the op amp input pins: what does the input pin current do if you try to drive the input negative?
Actually, this is not a homework problem. I got this design from EDN.com (search for active full-wave rectifier). The SPICE simulation works, so I just wanted to see if I can understand its operations more by writing the transfer function.
So far, my initial calculations indicate that VOB (output) will always be positive because it relies on the relationship between the resistors used in the circuit. But then again I'm not positively sure if I did the analysis the right way. I just used the two golden rules of opamps and also some nodal analysis.
I'll try to plugin some values in my equation once I can get access to the university computer labs for Matlab.
I've only got the macromodel of one opamp from National Semiconductors. That's all they have. I really don't know much about SPICE modeling. I haven't delve much into it.
Not really sure, I had the impression that the opamp inputs are high-impedance. I don't know how to go about answering this question. Any tips?
I'm kinda new to electronic design, so please excuse my lack of knowledge.
Well, most ICs have either intentional or parasitic diodes from inputs to at least the negative supply, so if you try to drag the input more negative than ground in this case, they start to conduct, and will only go "one diode drop" below ground at modest current.
Presumably those op amps are ones that work when the inputs are at the same potential as the negative supply. The data sheet might say something like "input common mode range includes the negative supply voltage." And presumably the outputs will behave reasonably all the way to the negative supply voltage. Typically, they can't actively pull down quite all the way, but if you have a resistive load to the negative supply, the output just stops sourcing any current, and the resistive load drags the voltage down to zero. It's the inability of the op amp output to go negative that gives you the "diode" effect that makes the circuit a full wave rectifier. As Ban suggested, break the analysis into two pieces, one when the input is positive and one when it's negative. For a first pass, ignore any effects like current in op amp inputs when they are dragged negative, but if you build the circuit, such an effect would explain why you got a DC offset...
Imagine, for example, +1V at the junction of R1 and R3 (the input, ignoring the capacitor). By inspection, what's the voltage at VA1, and therefore at VB1? (Recognize R1 and R2 as a 2:1 voltage divider.) If feedback is working around the second op amp, VB2 must equal (very nearly) VB1, and you can calculate the currents in R3 and R4, knowing the voltage at each end of each of them. The current in R5 must make the net current in the VB2 node zero. That tells you the voltage drop across R5. Be sure to get the direction right, and that drop added to the voltage at VB2 tells you the output voltage.
The analysis for -1V at the input is different, because the output of the first op amp cannot go negative...it will just go to zero. And it should be clear how to do the rest of the analysis for that case.
You should be able to find LOTS and LOTS of op amp macromodels. If you download LTSpice for free from Linear Technology, you'll get a whole bunch of macromodels included with it. And you should be able to find Spice macromodels from the TI, Analog Devices, and National Semiconductor web sites, to name just three of many, for a great many parts. You DO have to be careful about WHICH version of Spice the model was written for. Most seem to be Spice2, and often use a POLY keyword in some of the dependent sources. That's bad news for use in Spice3, and you need to translate them, if you use a Spice3-based simulator such as LTSpice. Well, as far as I know, LTSpice doesn't automatically do the translation for you, but it's surprised me with its capability before, so maybe it does. Anyway, a time-domain simulation over one cycle should show you just what's going on in the circuit, and make it clear how the "rectification" happens.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.