I am really lost on this. The question asks to find the op amp voltage gain of the non ideal op amp pictured below. It give open loop conditions: Rin = 400kΩ Rout = 2kΩ Gain = 500k Actually the question states: Find the voltage gain (vo/vg), the voltage at the negative input terminal (vn) when the input voltage is 50 mV and the input and output resistances Which is a bit vague but I assume it's asking for the closed loop voltage gain. Also I'm assuming the open loop Rin is the op amps internal resistance between Vn and Vp? With Rout being the op amps internal output resistance between -AVin and Vout? I'm not even sure where to start. I reckon the input current is: (Vn - Vp)/Rin + (Vn-Vo)/Rf and the output current: (Vout - Vn)/Rf + (Vout + AVin)/Rout. Am I along the right lines? Or am I way off? EDIT: Also this isn't technically homework as in it won't be marked so I hope it's in the right thread.
Since it is not homework, here is what you can do with LTSpice. Note that I show the effect of changing Open Loop Gain (olg): I use an ideal voltage-controlled voltage-source as a proxy for the opamp. I model the input resistance and output resistance as shown.
Try to use this circuit. But here I will show you a different approach that we can use to find a close loop gain Acl = Vout/Vin First we assume that voltage at Vy = 10V from this we already know the voltage at Vn. Vn = Vy/Aol = 10V/500K = -20μV Rf current is equal to If = (Vy - Vn)/(RF + Rout) = 31.056μA Next we find Rin current I_Rin = Vn/Rin = -50pA And finally we find R1 current. I_R1 = I_RF - I_Rin = 31.056μA + 50pA = 31.05601273μA We get this current for Vin equal to: Vin = - I_R1 * R1 = - 0.248448V and from there we find voltage gain Vout = Vy - IF *Rout = 9.937888V; Acl = 9.937888V/0.248448V = 39.9998712V/V For the ideal op amp we will get Acl = 320/8 = 40V/V
Lam58, I suppose that you are required to find all the answers by calculation rather than by simulaton, right? For this purpose, you need to study (that means: to apply formulas for) the feedback theory. The formulas you have mentioned do not help to find the correct answers. As you know, the closed-loop gain is set by the feedback path as well as the open-loop gain. And the same applies to the input as well as ouput impedances. Both are strongly influenced by feedback. However, in this context it is important if you know the operating frequency and the frequency properties of the opamp gain - or are you allowed to use the mentioned dc gain value? EDIT: As you can see, my answer came only 3 minutes after Jony130´s post. So I didn`t read it before. And - yes- it provides a good eqivalent circuit diagram with the corresponding set of formulas.
Hi guys thanks, I actually got a similar answer to Jony which I posted in another website using a different method, however, I was told I was wrong. My method probably was wrong but but I still got the same answer for the Gain. Here's the link to my original method. What do you think? (hope I'm not breaking any rules posting a link)
I assume I am allowed to use the dc gain value, tbh the question is a bit vague it's merely a tutorial question. I'm studying this for an exam and the module is half the physics behind the transistors and half analysis of op amp circuits. I've spent a lot of time doing the physics and not much on the circuits because I thought I knew most of the content but I am way behind, my exam is in less than 2 weeks and I am stressing out. It doesn't help that the tutorial questions for the circuit analysis aren't exam quality i.e. not as difficult, yet I'm still struggling.
You achieve a similar result accidentally. In you "method" you do not include the influence of a Rin and Rout. But why you get a a similar answer? The answer is very simple. Because your op amp has very large open loop gain and this is why you can ignore Rin and Rout. Negative feedback and large open loop gain will reduces significantly the impact of these resistors.
You are asking if the result as shown in the link is correct? Question: What is beta and what means "12.2k" ? Resistor or gain?
Sorry to bother you again, it was just to ask if you could look over the last part of my working to find Vn, Rin and Rout when Vs = 50mV. I found Vn when the input V is 50mV by the following: Is = I1 + I2 => (50mV/8k) = (Vn-Vo)/320k + Vn/400k after some rearranging: => Vn = 5.278μV Then for Rin: Rin = Vin/Is Is = 50mV/8kΩ Vin = Vs - Vn = 50mV - 5.278μV = 49.995mV => Rin = 7.99913952kΩ And finally for Rout: Ro = Vo/Io Io = (Vy-Vo)/Ri + (Vin - Vout)/Rf If Vy = Vn*Aol = 5.378μ * 500k = 2.689V Plugging in the numbers given Io = 2.35mA And Rout = 850.7Ω
Hi lam58, Regarding r,out: 1.) Rout is the opamps internal output resistor (2k) according to post#1. Thus, for the ouput resistor with feedback you must use a different symbol - for example r,out. 2.) You must not try to calculate r,out based on voltage-current relationships caused by an INPUT signal. That`s false. Instead, you must ground the signal input and apply a test voltage or test current at the output node in order to calculate the resulting ratio v,out/i,out . 3.) This will result in the following expression: r,out=Rout/(1-LG) with LG=loop gain. and LG=-Ao*Rs/(Rs+Rf) with Ao=500k and Rs=R1||Rin. 4.) Result: r,out=0.166 ohms
Yes, I think so - but I didn`t check all the minor places after the decimal. Do you know which error is allowed (accuracy)? EDIT: See my comments regarding r,in (next post).
In case you do not start with voltage-current relationships you can use standard formulas like: * Closed-loop gain: Acl=Ao*Hf/(1-LG) Hf=-(Rin||Rf)/[R1+(Rin||Rf)] and LG=Loop gain (as before). Note that LG is negative! * r,in=R1+(Rin||Rff) with Rff=Rf/(1-LG)=320k/12k=26.666 ohm. That means: r,in must be slightly LARGER than 8k. This sounds logical since it cannot be smaller than R1.
Here are my simulation results based on the data as given in post#1: * Output voltage: 1.9998V for 50mV input. * Voltage at the inv. input terminal: 4.025µV. * Voltage (internal) Vy=2.0123V * Input current: 6.25µA. Remark: There is a small error in your last calculation: The expression for the input current (50mV/8k) assumes that Vn=0 (which is not the case).