Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now

Hot answers tagged

7

Let's return to the basics: simulate this circuit – Schematic created using CircuitLab Output Stage Above is a fragment of the circuit in the earlier schematic you were examining. \$V_\text{CC}=12\:\text{V}\$ and the output at the emitter of \$Q_1\$ is assumed to be centered at about \$6\:\text{V}\$ and the output swing should be limited to peaks ...


6

Simply redraw the circuit and you will see from where this base current is coming from. simulate this circuit – Schematic created using CircuitLab As you can see the \$Q_2\$ base current is coming from \$\textrm{Vee}\$ voltage source. And flows this path: +Vee--->R1---> Q2 base-emitter junction--->-Vee And remember that in the electronic ...


4

Class-A amplifier design overview After watching those two videos, Single transistor, 1W and Adventures in a one transistor audio amplifier, I had to illustrate a slightly more sane design approach. To start out, here's the basic idea about how to drive a speaker (without having to find and use a flux-limiting, gapped, audio transformer -- which you will ...


4

It can be done. Remove Rs and C2. Let your speaker be RL. RE1 and RE2 should be smaller, because they seriously limit the voltage swing over RL. Prepare to waste continuously about 9W, half of it in the speaker and possible RE and the other half in Q1 if you want the maximum theoretical audio output power. I bet 2N3904 doesn't stand one second, you need ...


4

An actual design will usually be forced to balance various constraints. And it isn't always the case that you can follow only one recipe all the time. Often enough, you start out with some assumptions you feel are better and then find some reason why you can't go that way and you have to go re-assess your assumptions and make different trade-off choices. So ...


4

The main reason that the 1.5V emitter voltage design point is chosen is to reduce the effect of temperature variations and device variations on the bias current. The voltage across the base-emitter junction will change by about 2mV per degree C. Since the voltage at the base is assumed constant the voltage across the emitter resistor will increase by 2mV/...


3

Your understanding of the Rbe is correct and the emitter R increases Rin to Re*hFE and thus reduces the exponential input voltage effects by making the input impedance more constant. With V(Re) = 1.5V this reduces the nonlinear transformation of input voltage to emitter current and thus reduces sine distortion significantly. But is overkill for a 12V ...


3

I believe this is the circuit you are talking about In order to get the maximum possible output voltage excursion at the collector without clipping, the transistor needs to be biased so that the quiescent collector voltage Vc is approximately half way between Vcc and Vre. The design has to start somewhere and Iq of 1mA is chosen in this example as a ...


2

The (AC) signal from VA does directly end up at the base of both Q3 and Q4 through the capacitor CB. The DC voltage at the base of Q3 and the base of Q4 cannot be the same so we cannot short Q3 and Q4's bases. If we did that then the whole DC biasing of the circuit would be influenced and it would not work anymore. So we cannot short the bases for DC. But ...


1

Your circuit is Ok from the point of view of the input signal polarity since it adds an offset to the input in order to avoid its saturation when \$V_\mathrm{in}\$ goes below zero: note however, from the picture you show, it seems that you need a voltage gain of at least $$ A_v=\frac{V_\mathrm{out}}{V_\mathrm{in}}\approx\frac{1.5\mathrm{V}}{50\cdot 10^{-3}\...


1

Use an opamp circuit, non-inverting config, with a gain of 30. I. E. R2/R1 = 29 Connect R1 to gnd through a large cap. Make a resistive divider across your Vsupply to get midpoint of 1.5v Connect your signal through sufficiently large cap to non-inverting pin. Connect resistive divider centre point also to inverting terminal through large resistor (say ...


1

In the complementary pair you've drawn, altering the bias point alters the amount of gain variation you get in the open loop amplifier. In 'class A' operation, the bias current is large, both devices stay conducting, so the open loop gain variation is very small, but not zero. The VBE has to vary to vary the emitter current, and this is non-linear. In '...


1

The power for an amplifier is provided by the supply. The power a supply delivers is its own voltage (that's why we take \$V_\text{CC}\$) times the current drawn from it. For a Class A power amplifier, with no input signal, the DC current drawn equals only the collector bias current, \$I_\text{CQ}\$, because the current drawn by biasing resistors is ignored....


1

Since Hfe varies greatly with temperature and operating current your class A design is going to have quite variable gain. The 2N3904 has high variance of Hfe with operating current, and requires substantial current to get good gain. Since you seem to have established the Zener current to produce the best noise output, it may be a circuit like this would ...


1

OK you mislead us a tad because what starts as a simple transistor amplifier ends up with an unusually high Q ~14.37Hz Bandpass filter. THe impedance of the 5.6uF at resonance is only ~ 1/2 Ohm and with a 100 Ohm collector shunt the impedance ratio or Q is 200. So unless you have exactly the right caps values and input stable frequency ( like Jim Cutler ...


1

Just based on your specs you're going to have current draw issues. You are driving the circuit with a 0.2 V signal and want a maximum gain of 10, meaning a peak output voltage of 2 V. $$ \frac{V}{R} = I,\quad \frac{2}{8} = 250\ \text{mA} $$ This would also assume a 100% efficient system which you won't get anywhere close to in a class A amp. I would ...


1

There are two things to consider when thinking about "how could it work, having the base connected to ground?" - one is the DC question - how could the transistor ever be biased "on" to work, which others have tackled (basically: "ground" here is part way between the two power supply rails so the emitter is negative with respect to the base, so Q2, an NPN ...


1

Current flows from the higher potential to the lower potential and since the base is connected to ground '0V' and the emitter is connected to a negative voltage which is less than 0V(because its negative!!). The emitter base junction will be ON if the emitter voltage is sufficiently below ground.About 0.6V below ground for silicon transistors. Since the ...


1

Converting a class B output stage to class A is pretty straightforward - just add a low-value resistor to one supply or the other. For instance simulate this circuit – Schematic created using CircuitLab causes the internal drive level to stay positive for any output level greater than the Thevenin Equivalent voltage of the load and bypass. It ...


Only top voted, non community-wiki answers of a minimum length are eligible