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25:[["$","$L34",null,{"feature":"flashcards"}],"$L35",["$","$L36",null,{"href":"/a-level/physics/aqa/non-inverting-amplifier-configuration-a-level-only/flashcards","label":"non-inverting amplifier configuration (A-level only)","subjectCode":"a_level_physics__aqa","surface":"flashcards"}],["$","$L37",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L38",null,{"decks":[{"id":"4fa2e411-0778-45e1-b525-e645e62146fd","title":"non-inverting amplifier configuration (A-level only) Flashcards","topicId":"a79ca4b4-c1a3-5e82-945b-07a3436b669a","topicSlug":"non-inverting-amplifier-configuration-a-level-only","cards":[{"id":"c5b4e81b-3944-4c0d-9983-00c70dc7a891","cardType":"basic","points":1,"front":"To which terminal is the input voltage $V_{in}$ connected in a non-inverting amplifier?","back":"The non-inverting ($+$) terminal.","mcqOptions":[{"text":"The non-inverting ($+$) terminal.","correct":true,"explanation":"Applying $V_{in}$ to the non-inverting terminal produces an output with the same polarity as the input, which defines the non-inverting amplifier configuration."},{"text":"The inverting ($-$) terminal.","correct":false,"explanation":"Connecting $V_{in}$ to the inverting terminal creates an inverting amplifier instead. In a non-inverting amplifier, the input is connected to the non-inverting ($+$) terminal."},{"text":"The output terminal.","correct":false,"explanation":"The output terminal provides the amplified signal rather than receiving the input signal. $V_{in}$ must be applied to the non-inverting ($+$) terminal."},{"text":"The 0 V (ground) terminal.","correct":false,"explanation":"Connecting the input signal to ground would fix it at $0\\text{ V}$. In the non-inverting configuration, $V_{in}$ is connected to the non-inverting ($+$) terminal."}]},{"id":"fcca5b9d-ef0a-42fb-bd2c-14408de9118b","cardType":"basic","points":1,"front":"What is the phase relationship between the input and output signals in a non-inverting amplifier?","back":"They are in phase (0° phase shift).","mcqOptions":[{"text":"They are in phase (0° phase shift).","correct":true,"explanation":"A positive input produces a positive output in a non-inverting amplifier, so corresponding points on both signals occur together and their phase difference is 0°."},{"text":"They are in antiphase (180° phase shift).","correct":false,"explanation":"A 180° phase shift occurs in an inverting amplifier. A non-inverting amplifier preserves the input signal’s polarity, so the input and output are in phase."},{"text":"They are out of phase (90° phase shift).","correct":false,"explanation":"A 90° shift can arise in reactive circuits, but the ideal resistive non-inverting amplifier introduces no phase reversal or delay, giving a 0° phase shift."},{"text":"They are out of phase (270° phase shift).","correct":false,"explanation":"A 270° shift is not produced by the ideal resistive non-inverting configuration. Its output follows the input’s polarity, so the phase shift is 0°."}]},{"id":"7692e269-5566-4b34-b3bb-26594a78f54c","cardType":"basic","points":2,"front":"Describe the connections for the two resistors in a non-inverting amplifier circuit.","back":"

$R_f$ (feedback resistor): connects output to the inverting terminal ($-$).
$R_1$: connects the inverting terminal ($-$) to 0 V (ground).

","mcqOptions":[{"text":"

$R_f$ (feedback resistor): connects output to the inverting terminal ($-$).
$R_1$: connects the inverting terminal ($-$) to 0 V (ground).

","correct":true,"explanation":"These resistors form a negative-feedback potential divider. $R_f$ feeds back part of the output, while $R_1$ completes the divider between the inverting terminal and ground."},{"text":"

$R_f$ (feedback resistor): connects output to the non-inverting terminal ($+$).
$R_1$: connects the non-inverting terminal ($+$) to 0 V (ground).

","correct":false,"explanation":"Feeding the output to the non-inverting terminal gives positive rather than negative feedback. Both resistors must meet at the inverting terminal, with $R_1$ continuing to ground."},{"text":"

$R_f$ (feedback resistor): connects 0 V (ground) to the inverting terminal ($-$).
$R_1$: connects the inverting terminal ($-$) to the output.

","correct":false,"explanation":"This reverses the resistor labels and therefore the gain ratio. $R_f$ must connect output to the inverting terminal, while $R_1$ connects that terminal to ground."},{"text":"

$R_f$ (feedback resistor): connects output to the inverting terminal ($-$).
$R_1$: connects the inverting terminal ($-$) to the positive supply.

","correct":false,"explanation":"Connecting $R_1$ to the positive supply biases the feedback divider and changes the circuit’s reference voltage. In the standard configuration, $R_1$ connects to 0 V."}]},{"id":"c6322b2d-a5ec-4fb4-99d5-60615b2c59ee","cardType":"cloze","points":null,"front":"The voltage gain equation for a non-inverting amplifier is $$\\text{[...]}$$.","back":"The voltage gain equation for a non-inverting amplifier is $$\\boxed{\\frac{V_{out}}{V_{in}} = 1 + \\frac{R_f}{R_1}}$$.","mcqOptions":[{"text":"$$$\\frac{V_{out}}{V_{in}} = 1 + \\frac{R_f}{R_1}$$","correct":true,"explanation":"Negative feedback makes the input terminal voltages approximately equal. Applying the potential-divider relationship then gives the non-inverting voltage gain as $1 + R_f/R_1$."},{"text":"$$$\\frac{V_{out}}{V_{in}} = -\\frac{R_f}{R_1}$$","correct":false,"explanation":"The negative ratio is the gain equation for an inverting amplifier, where the input is applied through a resistor to the inverting terminal. Here the gain is $1 + R_f/R_1$."},{"text":"$$$\\frac{V_{out}}{V_{in}} = 1 - \\frac{R_f}{R_1}$$","correct":false,"explanation":"Using subtraction incorrectly treats negative feedback as a negative contribution to the closed-loop gain. The non-inverting gain is positive and equals $1 + R_f/R_1$."},{"text":"$$$\\frac{V_{out}}{V_{in}} = \\frac{R_f}{R_1}$$","correct":false,"explanation":"This omits the unity term arising from the non-inverting configuration. The complete closed-loop voltage gain is $1 + R_f/R_1$, not merely the resistor ratio."}]},{"id":"b9a7afca-0b5f-402e-aa62-c573bdeade41","cardType":"basic","points":2,"front":"Why is the voltage gain of a non-inverting amplifier always 1 or greater?","back":"Because the formula is $1 + \\frac{R_f}{R_1}$ and resistance values cannot be negative.","mcqOptions":[{"text":"Because the formula is $1 + \\frac{R_f}{R_1}$ and resistance values cannot be negative.","correct":true,"explanation":"Both resistance values are non-negative, so their ratio cannot reduce the unity term. Consequently, the ideal non-inverting voltage gain cannot be less than 1."},{"text":"Because the formula is $1 - \\frac{R_f}{R_1}$ and resistance values cannot be positive.","correct":false,"explanation":"This uses the wrong sign and incorrectly claims resistances cannot be positive. The gain is $1 + R_f/R_1$, and positive resistance values make it at least 1."},{"text":"Because the formula is $\\frac{R_f}{R_1}$ and resistance values must always be equal.","correct":false,"explanation":"The two resistors do not have to be equal, and their ratio alone is not the gain. The unity term in $1 + R_f/R_1$ ensures the gain is at least 1."},{"text":"Because the formula is $1 + R_fR_1$ and resistance values cannot be fractional.","correct":false,"explanation":"Multiplying the resistances is dimensionally unsuitable for voltage gain, and resistance values may be fractional in a chosen unit. The dimensionless formula is $1 + R_f/R_1$."}]},{"id":"52957095-6825-4089-a66e-821edadf7b21","cardType":"basic","points":2,"front":"How does negative feedback affect the voltages at the op-amp input terminals?","back":"The output adjusts so the voltage at the inverting terminal ($-$) equals the voltage at the non-inverting terminal ($+$).","mcqOptions":[{"text":"The output adjusts so the voltage at the inverting terminal ($-$) equals the voltage at the non-inverting terminal ($+$).","correct":true,"explanation":"With a very large open-loop gain and negative feedback, even a tiny input-voltage difference changes the output strongly. The stable result is approximately equal input-terminal voltages."},{"text":"The output adjusts so the voltage at the inverting terminal ($-$) equals the voltage at the output terminal.","correct":false,"explanation":"The feedback divider means the inverting-terminal voltage is generally only a fraction of the output. Negative feedback instead makes it approximately equal to the non-inverting-terminal voltage."},{"text":"The output adjusts so the voltage at the non-inverting terminal ($+$) equals the voltage at 0 V (ground).","correct":false,"explanation":"The non-inverting terminal is set by $V_{in}$, not forced to ground. Negative feedback adjusts the output until the inverting-terminal voltage approximately equals $V_{in}$."},{"text":"The output adjusts so the voltage at the inverting terminal ($-$) opposes the voltage at the non-inverting terminal ($+$).","correct":false,"explanation":"Negative feedback is sometimes mistaken for opposite input voltages. It actually drives their difference towards zero, making the inverting and non-inverting terminal voltages approximately equal."}]},{"id":"754b9beb-d091-4a05-8f30-9b09c0e45143","cardType":"cloze","points":null,"front":"In a non-inverting amplifier, if the calculated output voltage exceeds the supply rail voltage, the op-amp is said to be in [ ].","back":"In a non-inverting amplifier, if the calculated output voltage exceeds the supply rail voltage, the op-amp is said to be in saturation.","mcqOptions":[{"text":"saturation","correct":true,"explanation":"An op-amp output cannot exceed its supply rails. When the gain equation predicts a larger magnitude, the output reaches its limiting rail and the op-amp enters saturation."},{"text":"cut-off","correct":false,"explanation":"Cut-off usually describes a transistor carrying negligible current, not an op-amp reaching its output limit. Exceeding the available supply-rail voltage causes op-amp saturation."},{"text":"breakdown","correct":false,"explanation":"Breakdown implies damaging electrical stress or uncontrolled conduction, rather than normal output limiting. An op-amp whose demanded output exceeds a supply rail is in saturation."},{"text":"oscillation","correct":false,"explanation":"Oscillation means the output varies repeatedly without the intended input behaviour, often because of unstable feedback. Rail-limited output caused by excessive demanded voltage is saturation."}]},{"id":"b9357ccd-83a2-4f9d-95cc-76720dba7ef3","cardType":"basic","points":1,"front":"If $R_f = 10\\text{ k}\\Omega$ and $R_1 = 10\\text{ k}\\Omega$, what is the voltage gain?","back":"2 (calculated as $1 + \\frac{10}{10}$)","mcqOptions":[{"text":"2 (calculated as $1 + \\frac{10}{10}$)","correct":true,"explanation":"The resistor ratio is $10/10=1$. Substituting this into the non-inverting gain equation gives $1+1=2$, so the output amplitude is twice the input amplitude."},{"text":"1 (calculated as $\\frac{10}{10}$)","correct":false,"explanation":"This calculation omits the unity term from the non-inverting gain equation. Although $R_f/R_1=1$, the total gain is $1+1=2$."},{"text":"0 (calculated as $1 - \\frac{10}{10}$)","correct":false,"explanation":"This subtracts the resistor ratio, confusing negative feedback with a negative term in the gain formula. The correct calculation is $1+10/10=2$."},{"text":"11 (calculated as $1 + 10$)","correct":false,"explanation":"This ignores division by $R_1$ and treats $R_f$ alone as a dimensionless gain. The required ratio is $10/10=1$, giving a gain of 2."}]},{"id":"256d9945-b9d7-4d20-aaa9-d88486e8b3cb","cardType":"cloze","points":null,"front":"A non-inverting amplifier circuit with a gain of 1 ($R_f = 0$) is known as a [ ] or buffer.","back":"A non-inverting amplifier circuit with a gain of 1 ($R_f = 0$) is known as a voltage follower or buffer.","mcqOptions":[{"text":"voltage follower","correct":true,"explanation":"A voltage follower is a non-inverting amplifier with unity gain, so its output voltage follows the input voltage while providing isolation between circuit stages."},{"text":"voltage comparator","correct":false,"explanation":"A comparator switches between output levels according to which input voltage is greater. The unity-gain non-inverting circuit described is a voltage follower, which reproduces the input voltage."},{"text":"summing amplifier","correct":false,"explanation":"A summing amplifier combines multiple input signals using a resistor network. With unity gain and a single non-inverting input, this circuit is instead called a voltage follower."},{"text":"current follower","correct":false,"explanation":"The circuit is designed to reproduce the input voltage, not to create a defined copy of an input current. It is therefore called a voltage follower."}]},{"id":"8a5bc6e9-4c3f-4374-babb-018914f69dc4","cardType":"basic","points":2,"front":"An amplifier has $R_f = 20\\text{ k}\\Omega$, $R_1 = 5\\text{ k}\\Omega$, and $V_{in} = 0.5\\text{ V}$. Calculate $V_{out}$.","back":"2.5 V (Gain $= 1 + 4 = 5$; $V_{out} = 5 \\times 0.5$)","mcqOptions":[{"text":"2.5 V (Gain $= 1 + 4 = 5$; $V_{out} = 5 \\times 0.5$)","correct":true,"explanation":"For a non-inverting amplifier, gain is $1 + \\frac{R_f}{R_1}=1+\\frac{20}{5}=5$. Multiplying the input by 5 gives an output of $2.5\\text{ V}$."},{"text":"2.0 V (Gain $= \\frac{20}{5} = 4$; $V_{out} = 4 \\times 0.5$)","correct":false,"explanation":"This uses the inverting-amplifier gain magnitude and omits the added 1. Non-inverting gain is $1+\\frac{20}{5}=5$, so the correct output is $2.5\\text{ V}$."},{"text":"10 V (Gain $= 20 \\div 0.5 = 40$; $V_{out} = 40 \\div 2$)","correct":false,"explanation":"This incorrectly combines the input voltage with resistance values when finding gain. The resistance ratio gives a gain of 5, producing the correct output of $2.5\\text{ V}$."},{"text":"0.1 V (Gain $= 1 + \\frac{5}{20} = 1.25$; $V_{out} = 0.5 \\div 5$)","correct":false,"explanation":"This reverses the resistance ratio and then divides by an unrelated gain. Using $1+\\frac{R_f}{R_1}=5$ gives the correct output of $2.5\\text{ V}$."}]},{"id":"b7d671a2-751a-4a92-adbd-1731ba86eebf","cardType":"basic","points":1,"front":"Why is it usually unnecessary to convert $R_f$ and $R_1$ from $\\text{k}\\Omega$ to $\\Omega$ when calculating gain?","back":"Gain depends on the ratio $\\frac{R_f}{R_1}$, so units cancel if they are the same.","mcqOptions":[{"text":"Gain depends on the ratio $\\frac{R_f}{R_1}$, so units cancel if they are the same.","correct":true,"explanation":"Dividing two resistances expressed in the same units cancels those units. Therefore, values in $\\text{k}\\Omega$ can be used directly to calculate the dimensionless gain."},{"text":"Gain depends on the difference $R_f-R_1$, so units cancel if they are the same.","correct":false,"explanation":"Subtracting resistances does not cancel their units, and non-inverting gain is not based on their difference. It uses the ratio $\\frac{R_f}{R_1}$, whose matching units cancel."},{"text":"Gain depends on the product $R_fR_1$, so units cancel if they are the same.","correct":false,"explanation":"Multiplying the resistances would produce squared resistance units rather than cancelling them. Gain depends on the ratio $\\frac{R_f}{R_1}$, so equal units cancel through division."},{"text":"Gain depends on the ratio $\\frac{R_f}{R_1}$, so units cancel even if they are different.","correct":false,"explanation":"The ratio is correct, but differently scaled units cannot be ignored without conversion. Units cancel numerically only when both resistance values use the same unit, such as $\\text{k}\\Omega$."}]},{"id":"9f8897a4-aba1-4a5d-822f-7649630a6352","cardType":"basic","points":2,"front":"An engineer requires a gain of 6 and has $R_1 = 2\\text{ k}\\Omega$. Calculate the required value for $R_f$.","back":"10 kΩ (since $6 = 1 + \\frac{R_f}{2}$, then $5 = \\frac{R_f}{2}$)","mcqOptions":[{"text":"10 kΩ (since $6 = 1 + \\frac{R_f}{2}$, then $5 = \\frac{R_f}{2}$)","correct":true,"explanation":"Rearranging the non-inverting gain equation gives $R_f=(6-1)(2\\text{ k}\\Omega)=10\\text{ k}\\Omega$. The subtraction accounts for the added 1 in the gain formula."},{"text":"12 kΩ (since $6 = \\frac{R_f}{2}$, then $12 = R_f$)","correct":false,"explanation":"This omits the added 1 from the non-inverting gain formula. Using $6=1+\\frac{R_f}{2}$ gives $R_f=10\\text{ k}\\Omega$, not $12\\text{ k}\\Omega$."},{"text":"5 kΩ (since $6 - 1 = 5$, then $R_f = 5$)","correct":false,"explanation":"This finds the required resistance ratio but mistakes it for the resistance itself. Multiplying the ratio 5 by $R_1=2\\text{ k}\\Omega$ gives $R_f=10\\text{ k}\\Omega$."},{"text":"2.5 kΩ (since $6 - 1 = 5$, then $R_f = \\frac{5}{2}$)","correct":false,"explanation":"This divides by $R_1$ instead of multiplying by it. Since $\\frac{R_f}{2}=5$, the correct rearrangement gives $R_f=10\\text{ k}\\Omega$."}]},{"id":"c1af0980-06bd-457e-86c9-c9f35c20110e","cardType":"cloze","points":null,"front":"If the supply rails are $\\pm 12\\text{ V}$ and the gain calculation results in $+15\\text{ V}$, the actual output voltage will be [ ].","back":"If the supply rails are $\\pm 12\\text{ V}$ and the gain calculation results in $+15\\text{ V}$, the actual output voltage will be +12 V.","mcqOptions":[{"text":"+12 V","correct":true,"explanation":"An op-amp cannot produce an output beyond its supply rails. The calculated positive output therefore saturates at the positive rail, giving an actual output of approximately $+12\\text{ V}$."},{"text":"+15 V","correct":false,"explanation":"This assumes the ideal gain calculation can exceed the supply voltage. The output is limited by the positive supply rail, so it saturates at approximately $+12\\text{ V}$."},{"text":"-12 V","correct":false,"explanation":"The calculated output is positive, so saturation occurs at the positive rail rather than the negative rail. The actual output is therefore approximately $+12\\text{ V}$."},{"text":"0 V","correct":false,"explanation":"Saturation limits the output at a supply rail; it does not force the output to zero. A positive calculated output above the rails is limited to approximately $+12\\text{ V}$."}]},{"id":"06187042-a0ef-4746-8f64-4f528b8887b5","cardType":"basic","points":1,"front":"What happens to the peaks of an output signal if the amplifier saturates?","back":"The peaks are clipped (stay flat at the supply voltage).","mcqOptions":[{"text":"The peaks are clipped (stay flat at the supply voltage).","correct":true,"explanation":"Once the demanded output reaches a supply rail, it cannot increase further. The waveform therefore remains flat at that voltage, producing clipped peaks and distortion."},{"text":"The peaks are amplified (continue rising beyond the supply voltage).","correct":false,"explanation":"An op-amp cannot drive its output beyond its supply rails. Instead of continuing to rise, the peaks stay flat at the limiting rail and become clipped."},{"text":"The peaks are inverted (change sign at the supply voltage).","correct":false,"explanation":"Saturation does not reverse the signal’s polarity when it reaches a rail. The output simply stops increasing, so the peaks become clipped at the supply voltage."},{"text":"The peaks are attenuated (shrink uniformly below the supply voltage).","correct":false,"explanation":"Saturation does not reduce the whole waveform uniformly. Only the portions demanding voltages beyond the rails are limited, causing the affected peaks to be clipped."}]},{"id":"0ec8a2b3-7a19-4550-bae2-63b8beae256e","cardType":"cloze","points":null,"front":"In a non-inverting amplifier, $V_{in}$ and $V_{out}$ always have the same [ ] (or sign).","back":"In a non-inverting amplifier, $V_{in}$ and $V_{out}$ always have the same polarity (or sign).","mcqOptions":[{"text":"polarity","correct":true,"explanation":"A non-inverting amplifier produces a positive output for a positive input and a negative output for a negative input, so input and output retain the same polarity."},{"text":"frequency","correct":false,"explanation":"An amplifier may reproduce the input frequency, but frequency is not another word for sign. The defining non-inverting relationship is that input and output have the same polarity."},{"text":"magnitude","correct":false,"explanation":"The output magnitude is generally larger than the input magnitude because of the amplifier’s gain. What remains the same is the polarity, meaning the sign is not reversed."},{"text":"amplitude","correct":false,"explanation":"The output amplitude changes according to the gain and is equal to the input only for unity gain. Input and output always share the same polarity in this configuration."}]},{"id":"4d7fb125-e937-4097-8f07-e5bf9cc76cc7","cardType":"basic","points":1,"front":"What is the primary function of a voltage follower (unity gain buffer)?","back":"To isolate circuit stages without changing the signal voltage.","mcqOptions":[{"text":"To isolate circuit stages without changing the signal voltage.","correct":true,"explanation":"A voltage follower has very high input impedance and low output impedance. It prevents one stage from loading another while reproducing approximately the same signal voltage."},{"text":"To amplify circuit stages while increasing the signal voltage.","correct":false,"explanation":"A voltage follower has a voltage gain of 1, so it does not increase the signal voltage. Its main purpose is to isolate stages and prevent loading."},{"text":"To invert circuit stages without changing the signal amplitude.","correct":false,"explanation":"A voltage follower is non-inverting, so it preserves the signal’s polarity rather than reversing it. It isolates circuit stages while leaving the signal voltage unchanged."},{"text":"To filter circuit stages by removing the signal voltage.","correct":false,"explanation":"A voltage follower is not primarily a filter and should reproduce the input signal voltage. Its function is to isolate stages through impedance matching without changing that voltage."}]},{"id":"266c1a1b-96e0-468e-b1ed-52a1d99f9450","cardType":"basic","points":2,"front":"Compare the input terminal for $V_{in}$ in inverting vs. non-inverting amplifiers.","back":"

Inverting: $V_{in}$ goes to the negative ($-$) terminal.
Non-inverting: $V_{in}$ goes to the positive ($+$) terminal.

","mcqOptions":[{"text":"

Inverting: $V_{in}$ goes to the negative ($-$) terminal.
Non-inverting: $V_{in}$ goes to the positive ($+$) terminal.

","correct":true,"explanation":"Applying the signal to the negative terminal reverses its phase in an inverting amplifier. Applying it to the positive terminal preserves its phase in a non-inverting amplifier."},{"text":"

Inverting: $V_{in}$ goes to the positive ($+$) terminal.
Non-inverting: $V_{in}$ goes to the negative ($-$) terminal.

","correct":false,"explanation":"This reverses the two input arrangements. An inverting amplifier receives $V_{in}$ at the negative terminal, whereas a non-inverting amplifier receives it at the positive terminal."},{"text":"

Inverting: $V_{in}$ goes to the negative ($-$) terminal.
Non-inverting: $V_{in}$ goes to the negative ($-$) terminal.

","correct":false,"explanation":"This treats both configurations as inverting amplifiers. Only the inverting configuration applies $V_{in}$ to the negative terminal; the non-inverting configuration uses the positive terminal."},{"text":"

Inverting: $V_{in}$ goes to the positive ($+$) terminal.
Non-inverting: $V_{in}$ goes to the positive ($+$) terminal.

","correct":false,"explanation":"This treats both configurations as non-inverting amplifiers. The inverting configuration applies $V_{in}$ to the negative terminal, while the non-inverting configuration uses the positive terminal."}]},{"id":"47f2d490-3db2-4eb1-8dea-62a01227b43c","cardType":"cloze","points":null,"front":"The voltage gain of an amplifier has no units because it is a [ ] of two voltages.","back":"The voltage gain of an amplifier has no units because it is a ratio of two voltages.","mcqOptions":[{"text":"ratio","correct":true,"explanation":"Voltage gain is defined as output voltage divided by input voltage. Since both quantities are measured in volts, their units cancel, leaving a dimensionless ratio."},{"text":"difference","correct":false,"explanation":"Subtracting two voltages gives a potential difference that is still measured in volts. Gain is instead the ratio of output voltage to input voltage, so its units cancel."},{"text":"product","correct":false,"explanation":"Multiplying two voltages would produce units of volts squared, not a dimensionless quantity. Voltage gain is a ratio of output voltage to input voltage."},{"text":"sum","correct":false,"explanation":"Adding two voltages produces another quantity measured in volts, so it cannot explain the absence of units. Voltage gain is the ratio of two voltages."}]},{"id":"504c9cbc-0ec9-4461-8cc1-5354ab3f8641","cardType":"basic","points":1,"front":"State the effect on gain if $R_1$ is significantly larger than $R_f$.","back":"The gain approaches 1.","mcqOptions":[{"text":"The gain approaches 1.","correct":true,"explanation":"For a non-inverting amplifier, gain is $1+R_f/R_1$. If $R_1$ is much larger than $R_f$, the fraction approaches zero, so the gain approaches 1."},{"text":"The gain approaches 0.","correct":false,"explanation":"This overlooks the constant 1 in the non-inverting gain formula. Although $R_f/R_1$ approaches zero, $1+R_f/R_1$ approaches 1, not 0."},{"text":"The gain approaches $R_f/R_1$.","correct":false,"explanation":"This uses the magnitude of the inverting-amplifier gain and omits the added 1. The non-inverting gain is $1+R_f/R_1$, which approaches 1."},{"text":"The gain approaches $R_1/R_f$.","correct":false,"explanation":"This incorrectly reverses the resistance ratio, suggesting a very large gain. The formula is $1+R_f/R_1$, so a much larger $R_1$ makes the gain approach 1."}]},{"id":"79916254-d706-4f4a-afc7-fd864300657a","cardType":"basic","points":2,"front":"Explain why a non-inverting amplifier is used for microphone signals where the wave shape must match exactly.","back":"It increases magnitude while keeping the output in phase with the input.","mcqOptions":[{"text":"It increases magnitude while keeping the output in phase with the input.","correct":true,"explanation":"A non-inverting amplifier raises the signal amplitude without reversing its polarity. The output therefore follows the same waveform and remains in phase with the microphone input."},{"text":"It increases magnitude while keeping the output in antiphase with the input.","correct":false,"explanation":"An antiphase output is inverted by $180^\\circ$, so it does not match the input waveform exactly. A non-inverting amplifier increases magnitude while keeping the output in phase."},{"text":"It decreases magnitude while keeping the output in phase with the input.","correct":false,"explanation":"This describes attenuation rather than amplification. A non-inverting amplifier is chosen to increase the microphone signal’s magnitude while preserving its phase and waveform."},{"text":"It increases magnitude while keeping the output $90^\\circ$ out of phase with the input.","correct":false,"explanation":"A $90^\\circ$ phase shift would alter the timing and shape relationship between input and output. An ideal non-inverting amplifier increases magnitude while keeping both signals in phase."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"non-inverting amplifier configuration (A-level only)","title":"non-inverting amplifier configuration (A-level only) Flashcards","topbar":{"backHref":"/a-level/physics/aqa/flashcards"},"topicSidebar":"$L39"}]}],"topicId":"a79ca4b4-c1a3-5e82-945b-07a3436b669a","walled":"$L3a"}],"$L3b","$L3c"]