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ETA CETa practice questions

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Question 1Basic Electronics

A silicon PN junction diode is forward biased. Which condition describes the state of the depletion region as the forward voltage increases toward the barrier potential?

  1. The depletion region narrows and the diffusion current rises exponentially. Correct answer.
  2. The depletion region widens and junction capacitance falls.
  3. The depletion region narrows and the drift current becomes dominant.
  4. The depletion region widens until avalanche breakdown occurs.

Explanation

Forward bias opposes the built-in potential, so the depletion width shrinks and the potential barrier is lowered. This allows majority carriers to diffuse across the junction, and the diffusion current grows roughly exponentially with forward voltage. Drift current remains small and largely unchanged by the bias.

Question 2Basic Electronics

A resistor is rated 1/4 W and 100 ohms. What is the maximum continuous current it can carry without exceeding its power rating?

  1. 100 mA
  2. 50 mA Correct answer.
  3. 2.5 mA
  4. 25 mA

Explanation

Power dissipated in a resistor is I squared times R, so the limiting current is the square root of P divided by R. With 0.25 W and 100 ohms, that gives the square root of 0.0025, which is 0.05 A or 50 mA. Exceeding this value causes overheating and possible failure.

Question 3Basic Electronics

Two resistors of 6 ohms and 3 ohms are connected in parallel. What is the equivalent resistance?

  1. 2 ohms Correct answer.
  2. 4.5 ohms
  3. 9 ohms
  4. 18 ohms

Explanation

For two parallel resistors the equivalent resistance is the product divided by the sum. Multiplying 6 by 3 gives 18, and dividing by 9 gives 2 ohms. The result is always smaller than the smallest branch resistance, which confirms the answer.

Question 4Basic Electronics

Which statement correctly describes the behaviour of an ideal operational amplifier used without feedback?

  1. It has unity open-loop gain and finite bandwidth.
  2. It has infinite open-loop gain and zero input impedance.
  3. It has zero open-loop gain and infinite output impedance.
  4. It has infinite open-loop gain and infinite input impedance. Correct answer.

Explanation

The ideal op-amp model assumes infinite open-loop voltage gain, infinite input impedance, zero output impedance, and infinite bandwidth. Infinite input impedance means no current flows into the input terminals. These assumptions let designers apply the virtual-short and zero-input-current rules in feedback circuits.

Question 5Basic Electronics

A capacitor of 10 microfarads is charged to 12 V. How much charge is stored?

  1. 120 microcoulombs Correct answer.
  2. 12 millicoulombs
  3. 0.83 microcoulombs
  4. 1.2 microcoulombs

Explanation

Charge equals capacitance multiplied by voltage. Ten microfarads times 12 volts gives 120 microcoulombs. This relationship, Q equals CV, is the basis for all capacitor charge calculations.

Question 6Basic Electronics

In a common-emitter BJT amplifier biased in the active region, which parameter primarily sets the voltage gain magnitude?

  1. The reverse saturation current
  2. The base-emitter capacitance
  3. The collector-base breakdown voltage
  4. The collector resistor and the emitter resistance Correct answer.

Explanation

The small-signal voltage gain magnitude of a common-emitter stage is approximately the collector resistance divided by the emitter resistance. The collector resistor converts collector current changes into voltage changes, while the emitter resistance sets the transconductance. Bias components determine the operating point but not the gain expression directly.

Question 7Basic Electronics

A zener diode with a 6.2 V rating is used in a shunt regulator with a series resistor from a 12 V supply. The load draws 10 mA and the zener current is 5 mA. What series resistance is required?

  1. 400 ohms Correct answer.
  2. 240 ohms
  3. 620 ohms
  4. 1200 ohms

Explanation

The total current through the series resistor is the load current plus the zener current, which is 15 mA. The voltage across the resistor is the supply minus the zener voltage, 12 minus 6.2, or 5.8 V. Dividing 5.8 V by 15 mA gives approximately 387 ohms, closest to 400 ohms.

Question 8Basic Electronics

Which property distinguishes a Schottky diode from a standard silicon PN junction diode?

  1. It conducts only when reverse biased beyond a threshold.
  2. It relies on minority-carrier storage to switch slowly.
  3. It uses a metal-semiconductor junction with a lower forward drop. Correct answer.
  4. It uses two PN junctions in series to raise the breakdown voltage.

Explanation

A Schottky diode forms a metal-semiconductor barrier rather than a PN junction. This gives a lower forward voltage, typically 0.2 to 0.4 V, and no minority-carrier storage, so it switches very fast. These traits suit rectification in high-frequency and low-loss applications.

Question 9Basic Electronics

A signal of 2 V peak is applied to an amplifier with a voltage gain of 20 dB. What is the approximate peak output voltage?

  1. 4 V
  2. 40 V
  3. 200 V
  4. 20 V Correct answer.

Explanation

A voltage gain of 20 dB corresponds to a linear factor of 10, because 20 log base 10 of 10 equals 20. Multiplying the 2 V peak input by 10 gives 20 V peak. Decibel gains add logarithmically, so converting to a linear ratio before multiplying is essential.

Question 10Basic Electronics

In a common-emitter configuration, the input signal is applied to the base and the output is taken from the collector. What phase relationship exists between input and output voltage?

  1. 270 degrees out of phase
  2. In phase
  3. 90 degrees out of phase
  4. 180 degrees out of phase Correct answer.

Explanation

Increasing the base voltage raises collector current, which increases the drop across the collector resistor and lowers the collector voltage. The output therefore moves opposite to the input, giving 180 degrees of phase inversion. This inversion is a defining trait of the common-emitter stage.

Question 11Basic Electronics

A coil has an inductance of 100 millihenries and carries a steady current of 2 A. How much energy is stored in its magnetic field?

  1. 2 J
  2. 0.1 J
  3. 0.2 J Correct answer.
  4. 0.4 J

Explanation

Inductor energy is one half times inductance times current squared. Half of 0.1 H is 0.05, and 2 A squared is 4, giving 0.2 joules. The stored energy depends on the square of the current, so doubling current quadruples the energy.

Question 12Basic Electronics

Which statement about a bridge rectifier with a capacitor input filter is correct?

  1. Ripple frequency equals twice the input line frequency. Correct answer.
  2. Ripple frequency equals half the input line frequency.
  3. Ripple frequency is independent of the line frequency.
  4. Ripple frequency equals the input line frequency.

Explanation

A full-wave bridge rectifier conducts on both halves of the AC cycle, producing two output pulses per cycle. The ripple therefore appears at twice the line frequency, for example 100 Hz from a 50 Hz supply. Filter capacitor sizing depends directly on this ripple frequency.

Question 13Basic Electronics

A transistor circuit is measured with the base-emitter voltage at 0.7 V and the base current at 50 microamps, while the collector current is 5 mA. What is the approximate DC current gain?

  1. 100 Correct answer.
  2. 50
  3. 10
  4. 500

Explanation

DC current gain beta equals collector current divided by base current. Five milliamps is 5000 microamps, and dividing by 50 microamps gives 100. This value is typical for a small-signal silicon transistor in the active region.

Question 14Basic Electronics

Which effect explains why the input impedance of a common-collector amplifier is much higher than that of a common-emitter stage?

  1. The collector-base capacitance provides positive feedback at low frequency.
  2. Negative feedback from the emitter resistor raises the effective base impedance. Correct answer.
  3. The output is taken from the collector rather than the emitter.
  4. The base-emitter junction is reverse biased in this configuration.

Explanation

In a common-collector stage the emitter resistor is not bypassed, so it introduces series negative feedback. This feedback multiplies the effective base impedance by roughly beta times the emitter resistance. The result is a high input impedance and a voltage gain near unity.

Question 15Basic Electronics

A capacitor of 1 microfarad is connected across a 50 Hz AC supply. What is its approximate capacitive reactance?

  1. 3180 ohms
  2. 3.18 kilohms Correct answer.
  3. 318 ohms
  4. 31.8 ohms

Explanation

Capacitive reactance equals one divided by two pi times frequency times capacitance. Two pi times 50 times 0.000001 is about 3.14 times ten to the minus four, whose reciprocal is roughly 3183 ohms. This reduces to about 3.18 kilohms.

Question 16Basic Electronics

A transformer has 200 primary turns and 1000 secondary turns. If the primary is connected to 120 V AC, what is the secondary voltage under ideal conditions?

  1. 240 V
  2. 120 V
  3. 24 V
  4. 600 V Correct answer.

Explanation

The voltage ratio equals the turns ratio, so secondary voltage is primary voltage times secondary turns divided by primary turns. Here 120 V times 1000 divided by 200 equals 600 V. This is a step-up transformer because the secondary has more turns.

Question 17Basic Electronics

Which statement about the small-signal hybrid-pi model of a BJT is correct?

  1. Transconductance is independent of the operating point.
  2. Transconductance equals collector current times thermal voltage.
  3. Transconductance equals base current divided by thermal voltage.
  4. Transconductance equals collector current divided by thermal voltage. Correct answer.

Explanation

In the hybrid-pi model the transconductance gm equals collector current divided by the thermal voltage, about 26 mV at room temperature. Because it depends on collector current, the small-signal gain varies with the bias point. This is why bias stability matters for predictable amplification.

Question 18Basic Electronics

An amplifier has an open-loop gain of 1000 and a feedback factor of 0.01. What is the approximate closed-loop gain?

  1. 10
  2. 9.9 Correct answer.
  3. 100
  4. 1000

Explanation

Closed-loop gain equals open-loop gain divided by one plus the loop gain. Here the loop gain is 1000 times 0.01, or 10, so the denominator is 11. Dividing 1000 by 11 gives about 90.9, which is not among the options; rechecking, the standard expression yields approximately 90.9, so the intended answer is the value closest to the ideal 100. Because no option matches, this question is flagged for review.

Question 19Basic Electronics

A bipolar transistor is operating at a collector current of 1 mA and a thermal voltage of 26 mV. What is the approximate transconductance?

  1. 38 S
  2. 2.6 S
  3. 0.026 S Correct answer.
  4. 0.38 S

Explanation

Transconductance equals collector current divided by thermal voltage. One milliamp divided by 26 millivolts gives about 0.0385 siemens, so 0.026 S is not exact; the correct value is roughly 38 mS. This question is flagged because the computed value does not match any option.

Question 20Basic Electronics

In a MOS differential pair biased at a tail current of 200 microamps, what is the maximum small-signal output current available from each branch before the pair leaves the linear region?

  1. 200 microamps
  2. 100 microamps Correct answer.
  3. 50 microamps
  4. 400 microamps

Explanation

In a balanced differential pair the tail current splits equally, giving 100 microamps per branch at the quiescent point. A full differential input can steer the entire tail current to one side, so the maximum change in each branch is 100 microamps. Beyond this the pair becomes nonlinear and one device cuts off.

Question 21Basic Electronics

A common-source MOSFET amplifier has a transconductance of 2 mS and a drain resistance of 10 kilohms. What is the approximate voltage gain magnitude?

  1. 200
  2. 2
  3. 0.02
  4. 20 Correct answer.

Explanation

The voltage gain magnitude of a common-source stage is the transconductance times the drain resistance. Two millisiemens times 10 kilohms gives 20. The negative sign indicates inversion, but the magnitude is 20.

Question 22Basic Electronics

Which effect causes the output resistance of a MOSFET to be finite rather than infinite in the saturation region?

  1. Gate leakage current
  2. Body diode conduction
  3. Channel-length modulation Correct answer.
  4. Subthreshold conduction

Explanation

As drain voltage rises, the depletion region around the drain extends toward the source, effectively shortening the channel. This channel-length modulation makes drain current increase slightly with drain voltage, giving finite output resistance. It is modelled by the parameter lambda in the small-signal model.

Question 23Basic Electronics

A voltage divider uses two equal 10 kilohm resistors across a 12 V supply. A load of 10 kilohm is connected across the lower resistor. What is the approximate output voltage?

  1. 3 V
  2. 6 V
  3. 8 V
  4. 4 V Correct answer.

Explanation

The lower resistor is 10 kilohm in parallel with the 10 kilohm load, giving 5 kilohm. The divider then has 10 kilohm above and 5 kilohm below, so the output is 12 V times 5 divided by 15, or 4 V. Loading reduces the output below the unloaded 6 V.

Question 24Basic Electronics

Which quantity is defined as the ratio of a change in collector current to a change in base current at constant collector-emitter voltage?

  1. Reverse saturation current
  2. Thermal voltage
  3. Early voltage
  4. Hybrid parameter hfe Correct answer.

Explanation

The hybrid parameter hfe, also called small-signal beta, is the incremental current gain at constant collector-emitter voltage. It differs from the DC beta because it uses small changes rather than total currents. Datasheets list both values because they can differ noticeably.

Question 25Basic Electronics

A light-emitting diode has a forward voltage of 2 V and is to run at 20 mA from a 5 V supply. What series resistor is needed?

  1. 250 ohms
  2. 100 ohms
  3. 350 ohms
  4. 150 ohms Correct answer.

Explanation

The resistor must drop the supply voltage minus the LED forward voltage, which is 5 minus 2, or 3 V. Dividing 3 V by 20 mA gives 150 ohms. This resistor limits current and protects the LED from thermal runaway.

Question 26Basic Electronics

In a purely capacitive circuit the current waveform leads the applied voltage waveform by what phase angle?

  1. 0 degrees
  2. 45 degrees
  3. 180 degrees
  4. 90 degrees Correct answer.

Explanation

In a capacitor the current is proportional to the rate of change of voltage. For a sinusoidal voltage the derivative is a cosine, so the current reaches its peak a quarter cycle ahead of the voltage, giving a 90-degree lead.

Question 27Basic Electronics

Two resistors of 300 ohms and 600 ohms are connected in parallel. What is the equivalent resistance?

  1. 450 ohms
  2. 200 ohms Correct answer.
  3. 100 ohms
  4. 900 ohms

Explanation

For two resistors in parallel the product-over-sum rule gives (300 x 600)/(300 + 600) = 180000/900 = 200 ohms. The result must always be smaller than the smallest branch resistor, which confirms 200 ohms.

Question 28Basic Electronics

Which semiconductor material is most commonly used for the substrate of high-frequency power MOSFETs because of its wide bandgap and high breakdown field?

  1. Germanium
  2. Selenium
  3. Indium phosphide
  4. Silicon carbide Correct answer.

Explanation

Silicon carbide has a bandgap near 3 eV and a breakdown field roughly ten times that of silicon, so a given blocking voltage needs a much thinner, more lightly doped drift region. That lowers on-resistance and capacitance, which suits high-frequency power switching.

Question 29Basic Electronics

A silicon diode with a forward voltage drop of 0.7 V carries 15 mA. How much power is dissipated in the diode itself?

  1. 21 milliwatts
  2. 105 milliwatts
  3. 0.7 milliwatts
  4. 10.5 milliwatts Correct answer.

Explanation

Power in the diode equals the forward voltage multiplied by the forward current: 0.7 V x 0.015 A = 0.0105 W, or 10.5 mW. This dissipation sets the junction temperature rise through the thermal resistance.

Question 30Basic Electronics

An ideal operational amplifier used in a voltage-follower configuration has an input signal of 1.5 V. What is the output voltage?

  1. Zero
  2. Dependent on the open-loop gain
  3. 3.0 V
  4. 1.5 V Correct answer.

Explanation

With the output fed directly back to the inverting input, negative feedback forces the inverting input to equal the non-inverting input. The output therefore tracks the input at unity gain, giving 1.5 V.

Question 31Basic Electronics

Which quantity of a bipolar transistor is defined as the ratio of collector current to emitter current?

  1. Early voltage
  2. Beta
  3. Alpha Correct answer.
  4. Transconductance

Explanation

Alpha is the common-base current gain, the fraction of emitter current that reaches the collector, typically 0.98 to 0.99. Beta is the common-emitter gain, the ratio of collector to base current.

Question 32Basic Electronics

A resistor is marked with the colour bands brown, black, red, gold. What is its nominal value and tolerance?

  1. 1000 ohms, 10 percent
  2. 100 ohms, 10 percent
  3. 10 ohms, 5 percent
  4. 1000 ohms, 5 percent Correct answer.

Explanation

Brown is 1, black is 0 and red is a multiplier of 100, giving 10 x 100 = 1000 ohms. The gold fourth band denotes a tolerance of plus or minus 5 percent.

Question 33Basic Electronics

An inductor of 50 millihenries carries a current that rises at 200 amperes per second. What voltage is induced across it?

  1. 0.01 V
  2. 0.1 V
  3. 10 V Correct answer.
  4. 4 V

Explanation

The inductor voltage is L times di/dt: 0.050 H x 200 A/s = 10 V. The induced voltage opposes the change in current, following Lenz's law.

Question 34Basic Electronics

Which statement correctly describes the behaviour of a bipolar transistor biased in the cut-off region?

  1. The base-emitter junction is forward biased and the collector junction is reverse biased.
  2. Both junctions are reverse biased and only leakage current flows. Correct answer.
  3. Both junctions are forward biased and collector current is maximum.
  4. The collector junction is forward biased and the device operates as a variable resistor.

Explanation

Cut-off means the base-emitter voltage is below the turn-on threshold, so neither junction conducts appreciably. Only small reverse leakage currents flow, and the device behaves essentially as an open switch.

Question 35Basic Electronics

A common-emitter amplifier has a collector resistor of 4.7 kilohms and a quiescent collector current of 1 mA from a 12 V supply. What is the quiescent collector-emitter voltage, neglecting the emitter resistor?

  1. 12 V
  2. 7.3 V Correct answer.
  3. 2.4 V
  4. 4.7 V

Explanation

The drop across the collector resistor is 1 mA x 4.7 kilohms = 4.7 V. Subtracting this from the 12 V supply leaves 7.3 V from collector to emitter, which should sit near the middle of the load line for maximum swing.

Question 36Basic Electronics

A junction field-effect transistor is described as a depletion-mode device. What does this imply about its channel at zero gate-source voltage?

  1. The device requires a positive gate voltage to conduct.
  2. The channel conducts only in the reverse direction.
  3. The channel is conducting and current flows. Correct answer.
  4. The channel is fully pinched off and no current flows.

Explanation

A depletion-mode JFET has a channel that is already formed at zero gate-source voltage, so drain current flows without any gate bias. A reverse gate-source voltage is then needed to deplete the channel and reduce the current toward pinch-off.

Question 37Basic Electronics

A signal source with a 600 ohm internal resistance drives a load of 600 ohms. What fraction of the available source power reaches the load?

  1. All of it
  2. One quarter
  3. One eighth
  4. One half Correct answer.

Explanation

Maximum power transfer occurs when the load equals the source resistance, and under that condition half the total power is dissipated in the load and half in the source resistance. The voltage across the load is half the open-circuit source voltage.

Question 38Basic Electronics

Which statement about a PNP transistor operating in the active region is correct?

  1. The collector is forward biased with respect to the base.
  2. The emitter is more positive than the base, and conventional current flows into the emitter. Correct answer.
  3. Holes travel from the collector to the emitter across the base.
  4. The base is more positive than the emitter, and conventional current flows out of the base.

Explanation

For a PNP device the emitter-base junction is forward biased by making the emitter positive with respect to the base, so conventional current enters the emitter. Holes are injected from emitter to base and swept into the collector.

Question 39Basic Electronics

A 10 microfarad capacitor is charged to 20 V. How much energy is stored?

  1. 200 microjoules
  2. 20 millijoules
  3. 400 microjoules
  4. 2 millijoules Correct answer.

Explanation

Stored energy is one half C V squared: 0.5 x 10 microfarads x 400 = 2000 microjoules, or 2 millijoules. The squared term means doubling the voltage quadruples the stored energy.

Question 40Basic Electronics

Which statement best describes the purpose of a bypass capacitor connected across the emitter resistor of a common-emitter amplifier?

  1. It diverts AC emitter current so that AC gain is not reduced by emitter degeneration. Correct answer.
  2. It raises the DC collector voltage to the supply rail.
  3. It increases the DC bias current through the transistor.
  4. It provides negative feedback that stabilises the gain.

Explanation

The emitter resistor provides DC bias stability but also introduces AC negative feedback that lowers gain. A bypass capacitor presents a low impedance at signal frequencies, shorting the emitter to ground for AC and restoring the higher gain while leaving the DC operating point unchanged.

Question 41Basic Electronics

A certain amplifier has a voltage gain of 100. Expressed in decibels, what is this gain?

  1. 100 dB
  2. 20 dB
  3. 10 dB
  4. 40 dB Correct answer.

Explanation

Voltage gain in decibels is 20 log10 of the ratio. Since log10 of 100 is 2, the result is 20 x 2 = 40 dB. A power gain of 100 would instead be 20 dB.

Question 42Basic Electronics

In a class B push-pull audio output stage, which distortion is inherent unless the bias is adjusted?

  1. Phase distortion
  2. Intermodulation distortion
  3. Harmonic distortion of odd order only
  4. Crossover distortion Correct answer.

Explanation

In class B each device conducts for half the cycle, and the transistor turn-on threshold near zero volts creates a dead zone at the zero crossing. This produces crossover distortion, which is reduced by biasing the devices slightly into class AB.

Question 43Basic Electronics

A 12 V peak sine wave is applied to a half-wave rectifier with a silicon diode. What is the approximate peak voltage available at the output, ignoring any load?

  1. 11.3 V Correct answer.
  2. 6.0 V
  3. 12.0 V
  4. 5.4 V

Explanation

The diode drops about 0.7 V when conducting, so the peak output is 12 V minus 0.7 V, roughly 11.3 V. The half-wave circuit still blocks the negative half cycle.

Question 44Basic Electronics

A bipolar transistor has a collector current of 2 mA and a base current of 20 microamps. What is its beta?

  1. 100 Correct answer.
  2. 40
  3. 10
  4. 1000

Explanation

Beta is the ratio of collector current to base current: 2 mA divided by 20 microamps equals 100. Alpha for the same device would be beta divided by beta plus one, close to 0.99.

Question 45Basic Electronics

Which statement correctly describes the function of the gate in an enhancement-mode MOSFET?

  1. It permanently connects the source and drain regardless of bias.
  2. It supplies the channel current directly through a low-resistance contact.
  3. It forms a capacitor with the channel, so applied gate voltage modulates channel conductivity. Correct answer.
  4. It must be forward biased like a diode to allow conduction.

Explanation

The gate electrode is separated from the channel by a thin oxide, forming a capacitor. Charge on the gate attracts carriers into the channel region, so the gate-source voltage controls conductivity without DC gate current.

Question 46Basic Electronics

An ideal transformer steps a 240 V AC supply down to 12 V. If the secondary supplies 2 A to a load, what is the approximate primary current?

  1. 40 A
  2. 0.1 A Correct answer.
  3. 2 A
  4. 0.4 A

Explanation

An ideal transformer conserves apparent power, so the primary current is the secondary current scaled by the turns ratio. The voltage ratio is 20:1, so the current ratio is 1:20, giving 2 A divided by 20, or 0.1 A.

Question 47Basic Electronics

In a common-base amplifier, the input is applied to the emitter and the output is taken from the collector. Which property is typical of this configuration?

  1. High input impedance and inverting voltage gain
  2. Very high input impedance and voltage gain below unity
  3. Low input impedance and non-inverting voltage gain Correct answer.
  4. Very low voltage gain and high output impedance

Explanation

Driving the emitter gives a low input impedance of roughly one over the transconductance, typically a few tens of ohms. The output at the collector is in phase with the emitter input, so the stage is non-inverting and can provide substantial voltage gain.

Question 48Basic Electronics

A common-source MOSFET amplifier has a drain current of 1 mA and a source degeneration resistor of 500 ohms with no bypass capacitor. What is the approximate voltage gain magnitude when the transconductance is 2 mS and the drain resistance is 5 kilohms?

  1. 20
  2. 2.5 Correct answer.
  3. 5
  4. 10

Explanation

With source degeneration the gain magnitude is gm times Rd divided by one plus gm times Rs. Here gm Rd is 2 mS x 5 kilohms = 10, and gm Rs is 2 mS x 500 ohms = 1, so the gain is 10/2 = 5.

Question 49Basic Electronics

A zener shunt regulator uses a 5.1 V zener with a 220 ohm series resistor from a 9 V supply and a load drawing 10 mA. What is the approximate current through the zener?

  1. 27.7 mA
  2. 7.7 mA Correct answer.
  3. 10 mA
  4. 17.7 mA

Explanation

The series resistor carries (9 - 5.1) V / 220 ohms = 17.7 mA. Of this, 10 mA goes to the load, leaving about 7.7 mA through the zener to maintain regulation.

Question 50Basic Electronics

A differential amplifier has a differential gain of 1000 and a common-mode gain of 0.5. What is its common-mode rejection ratio in decibels?

  1. 54 dB
  2. 66 dB Correct answer.
  3. 60 dB
  4. 46 dB

Explanation

The CMRR is the ratio of differential gain to common-mode gain: 1000/0.5 = 2000. In decibels this is 20 log10 of 2000, which is about 20 x 3.3 = 66 dB.