To counter it then do my resistors need to be in parallel with the amp?If it had zero input impedance it wouldn't be an amplifier it would be a short circuit, and there would be no way around that. It is true that 10^12 is a large number, but typical of a FET input amplifier. If you like assume something smaller like 10^5 Ω or 10^6 Ω if that will make you happy.
im confusedThe unity gain buffer amplifier may have extremely high input resistance....but the input divider defines low input resistance.
Non-inverting amp cannot have gain less than - hence the divider.
Looks like you are using "design-by-happening". Instead of trying to understand and apply circuit analysis and design principles, you are simply trying nearly random circuits in the hopes that, at some point, one of them will happen to work.trying to figure this out, trying different things
Actually the last circuit was the answer the book says so I know it works but yall are saying it doesnt so im confused. Thanks for thatLooks like you are using "design-by-happening". Instead of trying to understand and apply circuit analysis and design principles, you are simply trying nearly random circuits in the hopes that, at some point, one of them will happen to work.
How does V2 come into play?Zinput = R * (V2/(V1-V2))?
For the second design, the input impedance of the amplifier is in parallel with 6kΩ. That pretty well counteracts it.No you don't. The input impedance is not a function of only the external components. The amplifier itself In the case of the TL081) has an input impedance of 10^12 Ω. That's huge. How do you plan to counteract that?
It would be considered to have infinite impedance (the ideal opamp model).Ok but that information wasnt given in the question so how would I know that about the amplifier? I would assume it is considering it to have zero impedance.
If you follow a unity gain buffer by a voltage divider with a gain of 3/4 .....Let's start with some basics. The non-inverting configuration cannot have a gain of less than 1. So to get a gain of 3/4 you must use the inverting configuration. The gain is then the feedback resistor divided by the input resistor. The non-inverting input is connected to GND or to a virtual ground derivied from the + and - power supplies. Starting with that configuration we can proceed to calculate the input and output impedance.
The last circuit will work. But you are still using "design-by-happening". That is further evidenced by your claim here -- you know it will work not because you understand the design or the analysis of it, but only because "the book says so".Actually the last circuit was the answer the book says so I know it works but yall are saying it doesnt so im confused. Thanks for that
It's debatable whether you have helped or not. The goal of the Homework Help forum is to guide students into solving their problems for themselves so that they learn the concepts. Just presenting them with a full-up solution seldom accomplishes this. Consider that, in most cases, the student has already seen the material developed in the text, has already seen examples worked in class, and has already seen examples worked in the text. If that has not been sufficient for them to learn the concepts, then there is no reason to suspect that seeing their homework worked for them by someone else is going to change that. At that point, in most cases, they need to fight and struggle with the material in order to learn it. We are here to help guide that struggle, not to do away with it.The last circuit is correct because of the following reasons.
- The transfer function of the resistive voltage divider would be multiplied by the transfer function of the unity gain buffer. So, the overall gain would be 3/4 as you have expected.
- The input impedance of the overall circuit would really be 8 kΩ. This is because theoretically the ideal operational amplifier would be replaced with an open circuit.
I hope that I have helped.
It looked to me like you were telling him that he MUST use an inverting configuration:I was trying to get the TS/OP to show how he arrived at the configuration shown. Especially to show how the configuration counteracts the high impedance of the input. Oh well, I like the phrase "design-by-happening". It's sorta like stumbling into a building and discovering you live there.
I was concerned that that would just lead to lots of confusion, especially since he would need to add another inverting stage to meet spec. Having him start with the inverting configuration just seems, to me, like an unnecessary rabbit hole.Let's start with some basics. The non-inverting configuration cannot have a gain of less than 1. So to get a gain of 3/4 you must use the inverting configuration. The gain is then the feedback resistor divided by the input resistor. The non-inverting input is connected to GND or to a virtual ground derivied from the + and - power supplies. Starting with that configuration we can proceed to calculate the input and output impedance.
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by Jake Hertz
by Luke James