If the input resistor (Ri) of an inverting amplifier is open, the amplifier would not receive any input voltage. Consequently, the output would be undefined or could potentially swing to the positive or negative supply voltage, depending on any stray voltages or noise present in the circuit. Essentially, the amplifier would not function as intended, as it relies on a defined input signal to produce an output.
It will depend on the which kind of amplifier it is ,if it is made from Op amp(Non inverting Amplifier) it will simply amplify(multiply)the signal with a factor depend on the configuration of circuit which is also known as gain of the circuit.For instance if gain is 2(dimension less) and input DC voltage is 2 Volt output would be 4 Volts. and if it an Audio or RC coupled amplifier output would be zero Hence i Wrote in the starting "It will depend on the which kind of amplifier it is"
Due to the manufacturing process of op-amps, the differential input transistors may not have exactly the same values, meaning they are not exactly matched. This means that voltage would have to be placed on the non-inverting terminal, with the non-inverting terminal grounded, in order to produce a zero output. The voltage required at the non-inverting terminal in called the input offset voltage.
To design a simple integrator with an op amp, place a resistor and capacitor in series in the feedback loop, between output and inverting input. Place another resistor from circuit input to the inverting input. Ground the non-inverting input. The current through the input resistor will be balanced with the current through the feedback resistor. Since there is a capacitor also, the voltage slope at the output will be proportional to the current. If you want the capacitor to discharge faster in one direction, you can place a diode (and optional resistor) across the feedback resistor. This works because the capacitor resists a change in voltage, proportional to current, and inversely proportional to capacitance. The equation is dv/dt = i/c. This means that dv/dt is linear with constant i and c. In this configuration, a constant current input will be balanced with a linear voltage ramp on the output, limited only by the range of the op amp. Constrast this with a simple RC circuit - with constant voltage, the RC circuit will exhibit a logarithmic output. If, for instance, you were to drive this circuit with a square wave, the output would be triangular. With the diode, the output would be sawtooth.
Hi, Neither is better than the other. The reason for inverting a signal depends on the next device in the chain or for a specific phasing need in the design. Maybe you need something to operate opposite of the incoming signal, for instance, when a voltage goes positive, you need it to go negative (or opposite), inverting would be used. Of course, you can have the option of having both inverting and non-inverting operations at the same time, and can have multiple circuits performing that function. Form follows function and a designer will do what ever's necessary to make the circuit operate as need requires. Hope this helps, Cubby
An op-amp is a device with two inputs, one inverting and one non-inverting. The output goes to whatever value is needed to make the two inputs the same. This means that, typically, the output is connected to a feedback circuit back into the negative input. This is closed loop, negative feedback operation. Due to the high gain of the op-amp, this configuration makes performance, such as gain and bandwidth, easy to predict.
It will depend on the which kind of amplifier it is ,if it is made from Op amp(Non inverting Amplifier) it will simply amplify(multiply)the signal with a factor depend on the configuration of circuit which is also known as gain of the circuit.For instance if gain is 2(dimension less) and input DC voltage is 2 Volt output would be 4 Volts. and if it an Audio or RC coupled amplifier output would be zero Hence i Wrote in the starting "It will depend on the which kind of amplifier it is"
It is a current amplifier. It can be used to boost current in an output stage to a speaker, for example, or for circuit isolation (two of many purposes). If you took a differential amplifier output stage to an 8 ohm speaker, it is very likely you would load the amplifier to the point of making it worthless (it would not amplify).
The capacitor is used to block DC bias from the output, so that only the AC signal is passed. In an audio amplifier, for instance, unwanted DC in the output would cause distortion when fed to a speaker, or could even damage the speaker or amplifier. In the case of interstage capacitors, they block DC so that the output of the first stage does not affect the bias of the second stage.
It may not be the amplifier at all ... it could be that the speaker cone is torn. If it is the amplifier it would mean that some component in the audio output section is faulty and needs repair or replacement.
The instrumentation amplifier provides isolation, and gain to the output of the Wheatstone Bridge. It is placed before filtering because the low output of the gauges would suffer from induced noise in the filter circuit if left unamplified.
Due to the manufacturing process of op-amps, the differential input transistors may not have exactly the same values, meaning they are not exactly matched. This means that voltage would have to be placed on the non-inverting terminal, with the non-inverting terminal grounded, in order to produce a zero output. The voltage required at the non-inverting terminal in called the input offset voltage.
It has inputs and outputs because an amplifier is a power source and has an audio signal going in to it from the head unit using audio cables(input) and then the signal goes out through the speaker wires(output).
If the RF input and output are reversed on a traveling wave tube (TWT), the amplifier will not function correctly. The input signal would not be amplified but instead sent back out through the input port. This would result in no signal being output through the intended output port.
I have one of these graphic equalizers and if your amp has a dedicated output and input for a graphic via a phono connector then it will work. If it doesn't then you could add a phono connector switch. This device would have inputs for all your devices via phono connectors (cd player, computer, satellite receiver etc) and it would have one output which would go into the graphic equalizer then out of it into one input on your amplifier. You would then select the individual device on the phono connector switch which you are going to play and select the solitary input on your amplifier.
It depends on the output impedance of the circuit, and on the value that the amplifier was attempting to produce. If the amplifier was idle, and properly nulled, then the expected output would be zero, so grounding it might not hurt. If it was attempting to produce an output, then you need to look at the possible current produced vs the maximum rated current of the chip. This is where the output impedance comes in. Usually, there is a resistor there to protect against just this sort of thing. Back to the idle state, however. If you ground the output, you remove the feedback loop and destabilize the amplifier. In all probability, then, it attempted to pull to one or the other rail. Again, this is what that resistor is for. The final answer, in all probability, then, is yes, you damaged the chips, unless the chips themselves were internally current limited. Time to check the datasheet.
That would depend on the decibel range output of the amplifier used by it. A very well designed electric guitar itself without an amplifier is barely audible. A small 30W amp will maximize at around 90dB.
To design a simple integrator with an op amp, place a resistor and capacitor in series in the feedback loop, between output and inverting input. Place another resistor from circuit input to the inverting input. Ground the non-inverting input. The current through the input resistor will be balanced with the current through the feedback resistor. Since there is a capacitor also, the voltage slope at the output will be proportional to the current. If you want the capacitor to discharge faster in one direction, you can place a diode (and optional resistor) across the feedback resistor. This works because the capacitor resists a change in voltage, proportional to current, and inversely proportional to capacitance. The equation is dv/dt = i/c. This means that dv/dt is linear with constant i and c. In this configuration, a constant current input will be balanced with a linear voltage ramp on the output, limited only by the range of the op amp. Constrast this with a simple RC circuit - with constant voltage, the RC circuit will exhibit a logarithmic output. If, for instance, you were to drive this circuit with a square wave, the output would be triangular. With the diode, the output would be sawtooth.