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# Tag Info

25

There are downfalls with choosing very large resistors and very small resistors. These usually deal with the non-ideal behavior of components (namely Op-Amps), or other design requirements such as power and heat. Small resistors means that you need a much higher current to provide the appropriate voltage drops for the Op-amp to work. Most op amps are able ...

25

Gain/Bandwidth product, you want maybe 50KHz bandwidth at 60dB (1,000 times), so you need somewhere around 50MHz, gain/bandwidth product (And more would lower HF distortion)... Make it 80dB and now you need 500MHz GBP, which is getting difficult if you want low noise down near DC (And is getting really bad news to stabilise at low gain). Also consider that ...

17

Don't bother. 2.5 W is unlikely to hurt a 2 W speaker. It will just cause distortion. Somewhere in your audio chain there must surely be a volume control. Turn down the volume to the point where the sound from the speaker isn't distorted. The 2.5 W rating of a power amp is what it can put out, given the right signal and load. Put in lower volume, and ...

16

An op-amp is not required. Given the minimal requirements given, a voltage divider will do. Gain = 0.5. simulate this circuit – Schematic created using CircuitLab If you need to buffer the output, you can just place a unity gain op-amp at the output of the voltage divider.

15

It's simple: manufacturers make what customers will buy! It's the same reason why Ferrari won't put a trailer attachment at the rear of their cars... Price is a very important part of this, of course, and price is tied to silicon area, process, yield, and of course packaging. For example, an opamp with +/-12V supplies and 1A output current will dissipate ...

14

One reason is that the transistor gain is degraded at high frequencies. To pick a specific example, the ON semiconductor BC546 has a gain-bandwidth product (GBP) of 100MHz at 1mA collector current (see figure 6 in the linked datasheet). This means that at a frequency of 27MHz, the current gain (beta) is about 100MHz/27MHz = 3.7, not 100. At 27MHz, stray ...

13

There is the matter of GBW(gain-bandwidth product), so a single stage is improbable with good performance. It's not enough to just squeak through in bandwidth you also want enough gain to reduce the distortion and get accurate reproduction with flat response (though arguably distortion at more than about 10kHz is unimportant for human hearing). Of course you ...

12

Your schematic should be as following: simulate this circuit – Schematic created using CircuitLab Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally. Now let's take a look at the ...

11

I just posted this as an answer to another question, but I can reuse it here :-) (without the part about the non-inverting amplifier). edit Oops, I hadn't noticed the question is from the same person as the other one. Let's look at the most simple feedback situation: The opamp will amplify the difference between $V_+$ and $V_-$: V_{OUT} =... 11 There are two problems with having the -Vs of a 741 connected to ground. First, as others have said, the output cannot go below ground. In fact, with a 741 it can't even get very close to ground, only to within maybe 1V or 1.5V of ground. This is called the output swing and is usually rated with some kind of a load resistor. Take care that the load ... 11 It's just a convention. Decibels always refer to power, and power is proportional to voltage squared and current squared. The math works like this: \begin{align} A_{v,dB} &= 10\cdot\log \frac{V_o^2}{V_i^2} \\ &= 10\cdot\log \left(\frac{V_o}{V_i}\right)^2 \\ &= 10\cdot 2 \cdot\log \frac{V_o}{V_i} \end{align}  EDIT: Power is ... 11 Okay, after a lot more research, I think I've gotten to the bottom of this. Actually I'm certain it's only approaching the bottom, as I've found this topic area quite deep, but I think I've gotten close enough to shed some light. A basic misconception A turning point in my understanding was when I realized that the equation I led off with in the OP:  G =... 10 Most likely a common-mode voltage violation. See Figure 15 of the datasheet. Unfortunately it does not give you the limits for a single power supply, but shows you that common-mode and output voltage limits go hand in hand. Actions that may solve your problem: Use dual power supply. Lift inputs above ground. Use a lower gain setting. EDIT: As pointed ... 10 As you say, the decibel is a unit of power ratio. G\ [\mathrm{dB}] = 10 \log_{10}\left(\dfrac{P_1}{P_2}\right)$. When the input and output impedances are equal and then we can express the gain in terms of voltage as$ G\ [\mathrm{dB}] = 20 \log_{10}\left(\dfrac{V_1}{V_2}\right) $I wouldn't call this the "voltage gain in decibels." I'd rather say ... 10 If you ignore the input capacitor and look at the input impedance into your op-amp circuit from the left of the 200k resistor, the input impedance is the 200k resistor. This is because the op-amp is configured as a virtual earth amplifier. In other words there is 200k loading your capacitor and the 3db high pass cut off point is when Xc = 200k ohms (in ... 10 Short: No!!!! Longer: Without a very specific description of a real world requirement this sort of question is useful at best as a thought experiment. Single stage: So - yes - it can be configured, but it would not be useful. Without looking it up - 10^6 gain = 120 dB which would be in excess of the open loop gain. A bad start. With an input ... 9 There are many kinds of potentiometers. As a rule of thumb, for audio, pots generally range from a resistance of about 10K to around 1 megohm. I.e. where a pot is going to be needed, you design that part of the circuit for some higher impedance like that. Controlling two signals with one control is done with a "dual ganged" potentiometer. This is ... 8 I'm going to say something that you might initially think of as totally wrong: If the voltage gain of a circuit is 6dB, the power gain is also 6dB To produce 6dB voltage gain requires a voltage gain of 2 and 20log(2) = 6.02dB To produce 6dB power gain requires a power gain of 4 and 10log(4) = 6.02dB This means you can talk about gain and not worry ... 8 First we must start with some basic assumptions, we assume that the inputs of the circuit are fed from ideal voltage sources and that the op-amp is an ideal op-amp. The latter means that the circuit is linear and we can apply the principle of superposition to calculate the two gains.. For the inverting gain we treat things as if the + input of the circuit (... 8 Many op amp circuits are designed so that they would yield a known finite gain if constructed using ideal components including an infinite-gain op amp. In practice, such circuits will always be constructed with non-ideal components, and their behavior will not quite match what would have resulted from ideal components. Consider a very basic amplifier: ... 8 Why (microphones) preamps designs tend to limit opamp gain to max 60 dB? A good overall picture of the whole range of what microphones and other audio devices produce: - Picture taken from here. As can be seen a studio microphone (depending on type) can produce a range of outputs from -60 dBm (relative to 600 ohm hence 0 dBm = 0.775 volts) to -20 dBm. ... 7 Would a schematic help? Yes This is your circuit as I read it: First thing, the divider by itself gives about 330 µV, because the AA battery should be 1.5 V, not 1V. And, including the input resistance of the amplifier, it shouldn't be that affected since it's 2.2k in series with the input resistance of the op-amp and the feedback resistor. So, when ... 7 Gain is just the ratio of two quantities. When talking about circuits, it's almost always the ratio of some measurement of the output of a circuit or component to its input. As Olin said, the most common case is the voltage gain. That is, the ratio between the voltage amplitude of the output signal to the input signal. Another common specification is power ... 7 In your specific op-amp circuit, the voltage on junction of Rf and Rin is the same as the voltage on the non-inverting input. This has to be so - it's called a virtual earth. Given that fact, this means that your signal (Vin) sees an input impedance of exactly Rin. It also means that your output (without connecting to anything else) has to drive an output ... 7 With two voltages (input, output) I would expect that it is likely a "voltage follower". More than that, are you sure about the part number? When speaking about classical voltage opamps the real closed-loop gain for the unity gain amplifier (follower) is$Acl = \dfrac{Ao}{1+Ao}$with Ao: Open-loop gain. Fort a typical value$Ao=10^5$(100 dB) we ... 6 To ensure repeatability in production. Beta varies a great deal between examples of the same device, and units would have very different characteristics if they were not designed so that they were independent of parameters like transistor beta. 6 Response is low at low frequencies because$C_1$blocks low frequency signals from reaching the input of the amplfier. Response is low at high frequencies because of parasitic capacitances within the MOSFET. A typical small-signal model of a MOSFET looks like this: (Image source: QSL.net) Typically, the most important parasitic for limiting the high-... 6 As @FakeMoustache hinted in a comment to your question, the explanation lies in the behavior of a reverse-biased PN junction, because that's what Q1's collector-base junction is in your circuit. From a macroscopic point of view any reverse-biased PN junction acts like a parallel-plate capacitor whose capacitance (called transition capacitance$C_T\\$) ...

6

Don't bother using MOSFETs. Instead, use FET-based devices such as the CD4066. Furthermore, if you only need 2 gain settings, you only need 1 switch. As an example simulate this circuit – Schematic created using CircuitLab will give a gain of 10 or 100. (Note that your example, if it worked, would give gains of 11 or 101). Also note that, whenever ...

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