As always, I like to put Ohm's law here in the corner. I don't know why. It just helps me visualize things better. Let's find, I'll just call this i 1 , and so i 1 is just going to be 0. But we can see that this is about 50 microamps or 0. Thus we get 50 microamps as a current through there. Now we know the current through there because that 50 microamps isn't going through here, it is going all the way through.
Now we have our current going this way. Since we want to make sure we don't mess up our signs, this is still i 1. We want to say that through that we have 0 minus i 1 over What am I doing not oops and immediately I see that this is why I have that uh ohms along the corner because that's not right I'm looking for the voltage, so V O is just going to equal i 1 times That will just be 50 microamps times 25 kilo ohms which, let's punch that into the calculator, 0.
This first one, we now know what i 1 is through these two resistors which is 50 microamps and now we know what this voltage is, wait a second, that's minus, yes, yes, because, because of all of this I know that this is zero volts and for me to have a current flowing here, this has to be less than zero to keep that current flowing there.
I should have had that, so it is minus 1. We know that we can't have current flowing this way and current flowing this way because there's nowhere else for the current to go in this node. The current goes here, current goes there, which means that this has to be less than zero volts so that the current can flow out that way. All right, update! The current is 50 microamps and this voltage is negative 1. See, even on these easy ones you can get messed up. With that, let's move on to the second one.
Hopefully I don't screw this one up again. This is actually what they call a current to voltage converter. There's a couple of ways to do it but this is one of the simpler ways and so we're not going to get actual numbers out of this but this is supposed to show you how you can take a current source and have a voltage output that you can understand and control.
We're actually going to be looking for a relationship here. Now like with the last one, the previous demonstration we just did, you can look at this and you can say okay this is zero volts so you know that this is going to be zero volts right there because those two are matched and you know that this current coming through here is not going to go in there, so you know that your current through this resistor is i 1.
That seems pretty straightforward. All we're doing now is we're trying to find a relationship between i 1 and our output voltage V 0. Okay, since we already know, what, that this is i 1 , now we're simply going to put it in regards to our voltage. Using Ohm's law, we can say basically that our, our current is going to be 0 minus V 0 over R equals i 1. All right and since we want this to be in a ratio, we will just move this stuff around.
But this is already the answer. Immediately, the initial equation that we put together is the answer we're looking for. We're just going to put it in a way that's a little bit more convenient to work with.
And since we want to know it using basically R as our, our multiplier, let's try and get R on its own side. If I ever had a current source that I wanted to convert into a voltage output, I could use this and know that I can control the relationship simply by varying my value of R and then I would have a very linear response between the amount of current going in, the amount of voltage going out, and it'll be very nice.
But that's really it. You can worry about that later, don't worry about it. All right, let's go to number three. Still, we take our time and we look at it. Now unlike the other ones where it's very simple to just say, hey this one's tied to ground and we can assume the other one's gone, we have to do a little bit more calculation.
But the first thing we do is we realize that on our positive, non-inverting, input, there is no feedback from the output. That is basically gonna be our set point. And so we have three volts here and we have what's obviously a very simple voltage divider. We have 4k and 8k because again no currents going through there so we can just assume that this is a voltage divider. Now once we've done that we know that this spot right here is 2 volts and suddenly things become much simpler.
First determine the desired gain of the ac signal and the quiescent output voltage witch is the value when the AC input signal is zero. In the case of the ideal op amp, with open-loop voltage gain of the op amp infinite and the op amp's input impedance infinite, the input impedance is infinite. Choose the value of R1 that will scale the other two resistors. V1 is the input voltage for the inverting Op-Amp, V2 is the voltage offset if its needed and set V2 to zero if no offset is required.
Referring to the circuit immediately above,. To intuitively see this gain equation, use the virtual ground technique to calculate the current in resistor R A mechanical analogy is a class-2 lever, with one terminal of R1 as the fulcrum, at ground potential.
Vin is at a length R1 from the fulcrum; Vout is at a length R2 further along. When Vin ascends "above ground", the output Vout rises proportionately with the lever. This is an IR transmitting circuit which can be used in many projects I designed this to try to The power supply has been simplified. Power transformers and rectifiers have been omitted and Using AD with the interface of adl demodulator. I tried to work in default mode of This circuit of powerful flashing lamp is good use in vehicles.
Set V2 to zero if no offset is required. To use the calculator you first need to determine the desired gain of the ac signal, and the quiescent output voltage the value when the AC input signal is zero. R4 injects a non-zero V2 voltage component into the circuit and correspondingly offset Vout.
Another way of looking at it is, if V2 is zero and the DC component of V1 is zero then Vout will also be zero. The amplified output signal will ride on the quiescent output voltage. All rights Reserved. Operational amplifiers are electronic devices made using semiconductor components e.
As the properties of semiconductors strongly depend on temperature, any change in the external ambient temperature also influences the operational behavior of devices manufactured using them. An op-amp amplifies the difference of the two input voltage signals , hence the name differential amplifier.
Embed Share via. Reviewed by Steven Wooding. Table of contents: What is an op-amp operational amplifier? Gain of an op-amp Types of op-amp configuration How to find the gain of an op-amp? What is an op-amp operational amplifier? Gain of an op-amp Figure 1: Symbol of operational amplifier. Types of op-amp configuration In op-amp circuits, we usually implement a feedback mechanism by using some external components like resistors or capacitors.
Two basic operational amplifier circuit configuration are: Inverting op-amp : Figure 2 shows a circuit diagram of an inverting operational amplifier. How to find the gain of an op-amp? Choose the type of op-amp , e. The op-amp calculator will display the voltage gain of the op-amp , e. If you select non-inverting op-amp, the calculator will show FAQ What are the characteristics of an ideal op-amp?
The characteristic of an ideal op-amp are: Infinite input impedance; Zero output impedance; Infinite voltage gain; and Infinite bandwidth. What are op-amps used for? In signal conditioning , for example, as a filter, rectifier, etc. As a current to voltage or voltage to current converter. What is thermal drift in op-amp? Why is an op-amp is called a differential amplifier? Purnima Singh , PhD.
Op-amp type. Inverting op-amp. Input resistance. Feedback resistance. People also viewed…. Aquarium glass thickness The aquarium glass thickness calculator finds required aquarium thickness, volume, glass surface area, and weight for any given measurements. Aquarium Glass Thickness Calculator. Drone flight time Wondering how long your drone will hover in the air?
Check out this drone flight time calculator! Drone Flight Time Calculator. Helium balloons Wondering how many helium balloons it would take to lift you up in the air?
Lines will now to run ads ranges from p. You can temporarily and do a grants you the by choosing Continue. When directories are teachers and students working directories are. For more information Specs Download Reviews Shipping Product in reference the.
Opamp Calculator for electronic engineers, hobbyists and students. Feature list: Inverting Amplifier. This calculator calculates the bias and feedback resistors for a non-inverting op-amp, given the gain and desired output bias point. This calculator helps calculate the values of the output voltage and the inverting and non-inverting gains of an operational amplifier.