ngt Posted November 20, 2004 Report Posted November 20, 2004 Can anyone explain to me the function of an op-amp in a simple manner ? Especially this IC > LM358Nhttp://rocky.digikey.com/WebLib/ST%20Micro/Web%20Data/LM158_258_358_A.pdf Quote
audioguru Posted November 21, 2004 Report Posted November 21, 2004 Hi NGT,A good course about opamps is in our Articles section at the top of this page and is here:http://www.electronics-lab.com/articles/files/6_Op_Amps.zipCheck the Articles index, it has all kinds of information about electronic devices. Quote
lisa08 Posted November 26, 2004 Report Posted November 26, 2004 Can anyone explain to me the function of an op-amp in a simple manner ? Especially this IC > LM358Nhttp://rocky.digikey.com/WebLib/ST%20Micro/Web%20Data/LM158_258_358_A.pdf :D what exactly do you mean by coupling?in practice where is an op amp used? what is meant by virtual ground? Quote
Guest Yevgenip Posted November 26, 2004 Report Posted November 26, 2004 OpAmp = Operational Amplifier, As in amplifing....Another use is as a comparator.Plenty more... Quote
Kevin Weddle Posted December 5, 2004 Report Posted December 5, 2004 An opamp uses transitors in the pushpull configuration to achieve high gain. One interesting aspect is that the difference between the inputs get amplified while a zero difference produces a zero ouput. Let's say you apply 5 volts to each input. Then you apply -5 volts to each input. The result is the same, 0 volts. This is because the pushpull is biased by the two inputs and the two inputs are biased by a constant current which gives you a constant voltage applied to the pushpull. It is a little difficult to explain, but consider that the inputs directly affect the pushpull, while a bias of equal magnitude produces the same amount of bias. Quote
Guest Yevgenip Posted December 6, 2004 Report Posted December 6, 2004 Wow Kevin, Talk about expanding an explenation... 8)Congradulations on becoming an electronics god!!! ;D Quote
trigger Posted December 6, 2004 Report Posted December 6, 2004 can we say virtual ground = opamp dc offset voltage? Quote
Kevin Weddle Posted December 6, 2004 Report Posted December 6, 2004 Thanks Yevgeni. I am really proud to help out those in need of electronics advice. I have 10 years experience and feel that I am still learning. Thank you again. Quote
audioguru Posted December 6, 2004 Report Posted December 6, 2004 Hi Kevin,Congratulations for becoming a god! ;DHi Trigger,A virtual ground is the inverting input of an opamp that has negative feedback. The feedback resistor will supply exactly the same current as the input resistor, except with the opposite polarity so the currents cancel. If the non-inverting input of an opamp with negative feedback is connected to ground, the input resistor to the inverting input will have the same current with both polarities if it was connected directly to ground.Offset voltage is caused by a slight mismatch in the opamp's input transistors. It usually doesn't cause any problems unless the opamp's DC gain is very high or you need an exact DC output. On cheap opamps it can be adjusted away with a pot, or you can purchase an expensive "laser-trimmed" opamp with an extremely low offset voltage. Quote
trigger Posted December 7, 2004 Report Posted December 7, 2004 Thanks audioguru. That refreshes my old time memeory.... ;DOne more question: if I got a sigal, say 1KHz, 2Vp-p with DC offset at 2Vdc; how can I use an opamp to adjust the DC offset to 1Vdc? Quote
audioguru Posted December 7, 2004 Report Posted December 7, 2004 Hi Trigger,If an inverting opamp has a positive DC output voltage with or without a signal, you can reduce it by applying a positive current (from a positive voltage through a resistor) to its inverting input. Quote
steven Posted December 7, 2004 Report Posted December 7, 2004 :) hi since i became an electronics god im still waiting for me magic wand, sorrey merlin. Quote
Kevin Weddle Posted December 7, 2004 Report Posted December 7, 2004 I have luck with setting the output bias current. You do that with a resistor to the negative supply. This will give you the offset. Without the bias current, you are dealing with a dead opamp. But of course you will still get the output voltage by virtue of the total current going around the opamp. Be mindful of the impedance. I have found that in order to get the ouput current right and the impedance you need this resistor to the negative supply. If you don't operate the opamp the way it was intended, you won't get the change in input current. The inputs will sit there at ground and all the current still goes around the opamp. I think what happens is that the beta of the output current transistor, the top one, becomes too high and the feedback resistor looks to be too large. The feedback resistor, after all, is really a collector resistor as seen by the interior of the opamp circuit. The base resistor is seen an emitter element from the viewpoint of the other input that is divided by beta. Quote
audioguru Posted December 7, 2004 Report Posted December 7, 2004 Kevin,The bias current for the output transistors of most opamps is set internally to avoid crossover distortion. An external resistor from the opamp's output to the negative supply won't change the bias current and it won't change the output voltage of the opamp, unless the resistor's current exceeds the output current rating of the opamp and therefore the opamp is overloaded.An opamp without your "overload resistor" is not dead but is very much alive and works fine.The offset voltage of an opamp is changed by applying a very small current to its input, not its output.Because of its extremely high open-loop gain, the output impedance of an opamp that has negative feedback is extremely low, much lower than its load. That is why its output voltage doesn't change with or without a reasonable load. The beta of the output transistors of an opamp is not used and the feedback resistor is not a collector resistor, because the output transistors of most opamps are emitter-followers. Quote
Kevin Weddle Posted December 10, 2004 Report Posted December 10, 2004 Audioguru, I knew you would disagree with me on this. Haven't we had this discussion elsewhere? The collector resistor I am talking about is the effect the feedback resistor has on the internal stages of the opamp. Now, If you look at the input stage, the input resistor beomes an emitter element. That is, looking into one of the inputs, the input resistor is divided by beta and is an emitter element in parallel with the other resistor. Quote
audioguru Posted December 10, 2004 Report Posted December 10, 2004 Hi Kevin,Please forget about collector resistors, beta and "emitter elements" when discussing opamps.An inverting opamp circuit simply consists of an opamp, a feedback resistor and an input resistor. The charactristics of an opamp are that it has an enormous voltage gain, differential inputs with very high impedances and a very low output impedance. The opamp's characteristics cause the current in the feedback resistor to be equal to the current in the input resistor. Since the circuit is wired to be inverting, the currents in those resistors cancel at the inverting input of the opamp. Since the currents in those resistors are equal, the voltage gain of the inverting opamp circuit is directly set by the ratio of the value of those resistors. Quote
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.