Help with some circuit analysis. - Page 1
Forum section: electronics / beginners

author: Anonmis
author_rank: Contributor
timestamp: July 29, 2012, 02:58:27 am
content: My brother sent me a circuit that he saw from one of his instructors. It's part of a pre-amplifier circuit connected to a guitar (usually the guitar would input at the non-inverting pin of first op-amp). I took only this portion of the circuit to try to understand what is going on at certain nodes. I have done the following....1) Tried nodal analysis etc to find voltages, currents etc.2) Simulated in LTSpice3) Simulated in Circuit Wizard (the screenshot)4) Built the circuit on a breadboardAfter doing all those, ALL of the results are different so I don't know which one is the correct (if any).Can someone please help me figure out currents, voltages etc. or maybe even tell me if the circuit doesn't make sence.Thanks in advance.

author: IanB
author_rank: Super Contributor
timestamp: July 29, 2012, 04:10:51 am
content: Well I'm not an expert, so take my advice with caution.But it looks like a straightforward two stage amplifier.If the input signal is at TP1, then you need this voltage to rest around halfway between 9 V and GND, so that the op amp is operating in the middle region between the power rails. I don't see a biasing arrangement to set this up. Once the biasing is in place you would normally have a capacitor to couple the input signal and probably the output signal too. I don't see those capacitors.Have you removed some important parts of the circuit?

author: Anonmis
author_rank: Contributor
timestamp: July 29, 2012, 04:23:40 am
content: Yes I have. As you mentioned, the original circuit does have capacitors coupled at the inputs and outputs (and includes another amplifying stage also). Is there away to analyze current flow and voltages at nodes with the circuit given? (I set the probe tips on the nodes for personal reference) I want to figure out the current flow and voltages in the circuit though.

author: David_AVD
author_rank: Super Contributor
timestamp: July 29, 2012, 04:59:00 am
content: It looks like the LED is being used an a power indicator and also to provide biasing for the 2 amplifier stages.

author: amspire
author_rank: Super Contributor
timestamp: July 29, 2012, 05:09:33 am
content: Lets start with the non inverting input. There is a 60mV drop across the 47K, so that means in input bias current to the opamp of .06/47000 = 1.3uA. Now it should be less then 0.1uA, so something is wrong here. I think the Circuit Wizard is way out of spec for the input bias current for the LM358.Anyway, lets accept that the drop is 0.06mV. you have to compare all the voltages in the circuit to the DC reference voltage (1.21V), so that makes the input about -0.06V. The first stage has a gain of about 6, so you expect an output of -0.36V which is about right. The second stage has a gain of 6, so the second stage should have an output of -2.16V which it cannot do, so instead you have a saturated low output. So after the input, the rest of the circuit is behaving about right.Now for a guitar amplifier, you really do not want the same gain for DC as for AC. A 1Hz wave going through to the speakers will not give any useful sound output from the speakers, but it can certainly do a brilliant job in burning out the speakers.If you add a capacitor in series with the 2K resistors, the DC gain will reduce to unity, so even if there is this odd -0.06V on the input, it will just give -0.06V on the output, while you will still get a total of x36 AC gain for the audio.Next thing I do not like is that the LED is not a great way to regulate the 1.21V reference level. The current drain from the second stage 2K resistor will modulate the 1.21V across the diode and this will feed back into the first stage causing distortion, given the non-linear nature of the LED. An electrolytic cap across the LED will greatly improve the circuit performance.In conclusion, the only mystery in the circuit is "why you are getting the massive 0.06V drop across the 47K resistor to the first opamp input". I would expect the output offset on the breadboard to be a lot less than 300mV from 1.21V, but the exact amount depends on the values of input bias current and offset voltage for your opamp.Richard.

author: IanB
author_rank: Super Contributor
timestamp: July 29, 2012, 05:13:21 am
content: Quote from: Anonmis on July 29, 2012, 04:23:40 amYes, there is.Starting from the left, then as David said the voltage at TP2 will be the forward bias voltage across D1 when subjected to the current flowing through it through R6. You will have to solve R6 in series with D1 to find out what this current and intermediate voltage is. The result will depend heavily on the exact characteristics of D1.Next, you know that the voltage at TP1 must equal the voltage at TP2 since no current can flow through R5. (An op amp has high impedance inputs.)It follows if the feedback arrangement is working then the voltage at TP4 equals the voltage at TP1, which is also the voltage at TP2. Therefore no current flows through R1. If no current flows through R1 then no current flows through R3 and TP3 must also be at the same voltage as TP1, TP2 & TP4.By a similar process you can discover that TP6 is also at the same voltage as TP3.

author: IanB
author_rank: Super Contributor
timestamp: July 29, 2012, 05:15:24 am
content: Quote from: amspire on July 29, 2012, 05:09:33 amSince it's a guitar amp, perhaps this distortion is intended?

author: Anonmis
author_rank: Contributor
timestamp: July 29, 2012, 05:52:48 am
content: Thank you very much for your quick response guys. I believe it is starting to click now.Amspire: I was confusing myself by trying to figure out what the voltage at TP1 should be. One would assume that it would be the Diode voltage if the op-amp was ideal (no current flowing into it's inputs) right? Since no component is actual ideal in the real world, is TP1 calculated by using the "Input Bias Current" from the data sheet and calculating the voltage drop across 47k. The typ. value is 20nA.... so TP1 should be 1.21 - (47k*20nA) = 1.209 right?IanB: Thanks for your help. At one point I had gone through the circuit and ended up with the same results as you did. But I guess I was confusing myself by asking "then what is the purpose of the amplifier stages or this circuit at all?". Even though it should have been obvious since i'm dealing with Audio, but this circuit is only useful (and makes sense) only when there is an AC input.Thanks again for your guys help. Any more input is welcome. I am really noobish when it comes to actual circuit practical experience. I finally purchased some equipment recently to help me practice more at home.

author: amspire
author_rank: Super Contributor
timestamp: July 29, 2012, 06:06:22 am
content: Quote from: Anonmis on July 29, 2012, 05:52:48 amYes, but in understanding the circuit, you will find it easier if you make the 1.21V point the 0V rail for the circuit, which will make the two supply rails -1.21V and 7.79V.This way, the TP1 input is 47K*(-20nA) = -0.94mV.This would make the output of the first opamp (if you could ignore offset voltage) would be -0.94 X 6 = -5.64mV.

author: efron
author_rank: Contributor
timestamp: July 30, 2012, 04:18:49 pm
content: Quote from: amspire on July 29, 2012, 05:09:33 amThis is not really exact. The output of the first OA is not proportional to the entry but has an equation as follows:Vout(OA1) = 1.2 + (VTP1 - 1.2)*6Therefore, if VTP1 = 1.15V (for any reason different from 1.2V), the Vout = VTP3 will be 0.9V.The second OA configuration behaves like the first one, so Vout (OA2) = 1.2 + (VTP3 - 1.2)*6 = -0.6V --> this will force the output to 0.Obviously, under these conditions, it is not possible to amplify any ac signal.@AnonmisYou said that you tested in breadboard. This is almost real life; what values you measured?