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13

The BE junction can be reverse biased, and your PNP will be off. The limit is the BE-junction breakdown voltage, as for any PN junction. As long as you don't exceed that breakdown voltage, you'll be safe. The maximum reverse voltage the BE junction can withstand is usually listed in the datasheet in the Absolute Maximum Ratings section. Here follows an ...


10

Murata has a pretty interesting explanation on its FAQ section about what is going on. I will quote the most relevant parts here for future reference. The characteristic of change in capacitance according to the applied voltage is called "DC (direct current) bias characteristic." The mechanism of DC bias characteristic In the high dielectric constant ...


10

The 'ground' on R3 becomes the 'zero reference' for the amplification. This should be connected to Vcc/2. Otherwise you will get strange results (what you are seeing). There is some good information here about how to generate a virtual ground for this type of single-supply circuit: http://tangentsoft.net/elec/vgrounds.html


10

Most of the hobbyist circuits you find by googling are crap originating from dubious tinkering and then copied from each other. The bias level of this circuit is not very predictable from BJT to BJT (meaning it may clip at high level one way or the other depending on the capsule sensitivity and actual sound pressure level) and the distortion is relatively ...


9

The TRRS connectors are wired like this, in order: Tip: Left output Ring: Right output Ring: Ground/return Sleeve: Microphone input That way the headphones will work plugged into a normal TRS jack, too. The microphone input is meant to connect to an electret mic with a circuit like this: The bias voltage, resistor, and DC-blocking capacitor are all ...


9

Three reasons: First, operating in the forward orientation only allows operation at a single voltage, nominally about 0.7 volts for a silicon diode. Diode construction can be tailored to produce a wide range of breakdown voltages, with a consequent choice of different regulator outputs. Second, your V-I curve overstates the sharpness of a forward-biased ...


8

The square wave starts out at a non-zero value. The sine wave starts out at zero. The time constant of 0.1uF and 5K is 500usec. If you look carefully at the blue line in your simulation you can see it trending downward. You need to either run the simulation for several milliseconds or figure out how to set the initial conditions so that the capacitor ...


8

Is there any reason this wouldn't work the way I think it would? You cannot expect that circuit to work because the guitar signal ground connection has to connect into the op-amp circuit ground node and without that connection you are just going to get noise. The impact of this is that you are not therefore isolating the two receiver circuits as you ...


7

The input signal is present on the node at the right of R3, so R3 must be duplicated for each channel (otherwise you would be joining all the inputs together!). R1, R2 and C2 can be common to all channels. There are two possible issues with using a common voltage divider - bias current, and crosstalk. OP295 op-amps have a maximum bias current of 20nA ...


7

As Olin said, circuit shown in #1 and #3 are completely open loop. So the bias stability of the circuit is less and it can even lead to thermal runaway. The bias stability can be improved by including a negative feedback mechanism in these circuits. Circuits #2 and #4 does that. Negative feedback in circuit #2: Assume that the collector current increases. ...


7

In general 3V reverse bias is okay for most jellybean transistors- typically the rating is 5V or so, and the actual breakdown more like 9V. This is not necessarily true for some RF transistors, which may have a much lower breakdown voltage. There is, however a practical problem which may occur with your circuit. If the 15V supply comes up before, or stays ...


7

The purpose of the diodes is to set a bias voltage between the transistors' bases, which sets a small idle current through the push-pull. This makes it work in class-AB and lowers crossover distortion. However the diodes should be thermally coupled to the transistors, to prevent thermal runaway. Also, emitter resistors should be used for this reason. Anyway....


6

Circuit analysis is generally broken into two parts: DC and AC analysis. For the DC analysis, all signal sources are zeroed, i.e., only the DC sources are considered. For DC analysis, a capacitor is an open circuit and an inductor is a short circuit. The solution for the circuit, under these conditions, is the Q-point; the "quiet" point. It is the value ...


6

The simulator first finds the DC solution and uses the results as the initial conditions for the transient simulation. Zeroing the signal source and assuming the signal source has no DC offset voltage, the voltage across the capacitor in DC steady state is 45V with the right-most terminal the more positive. Looking closer at your pulse generator statement,...


6

Yes you can directly connect these pins to GND or V+. The reason you did not find this explicitly stated is that it is standard that a high-impedance digital input can be connected directly to a logical 0 or 1. When a resistor was needed this would be stated explicitly. Using a resistor instead of a hard connection might be convenient when you need to ...


6

You can just use an op-amp adder, or just a voltage divider if you can live with higher output impedance and low input impedance. For example (voltage divider): simulate this circuit – Schematic created using CircuitLab R1 is picked arbitrarily, and the remaining two values are chosen so that R2||R3 = 2.02K (so the output voltage changes 1.00V for ...


6

I can't give you equations, but I can tell you that the main purpose of AC Bias applied to the record head of an analog tape recorder is to overcome the inherent hysteresis of the magnetic oxide contained in the binder on the recording tape. I'm no longer competent to give you an extremely detailed explanation - I last worked with analog tape recorders more ...


6

It's to overcome the hysteresis of the magnetic tape. The record head is a small electromagnet pressed against the tape. If a current is applied, it magnetises the oxide particles on the tape. the stronger the current, the more magnetic domains are re-aligned and the stronger the signal recorded. However, a very tiny current will have no effect on the ...


6

This is from an old Sony service bulletin. It mentions SV04S (not F) but part numbers show SV04, SV04S and SV04F as equivalent, so it should work. It's a Vbe multiplier, also known as an adjustable diode or rubber diode. It's made from an NPN silicon transistor (in this case 2SD1585) and two resistors. You could replace both the 1k and 3k3 resistors for ...


6

Film capacitors have a very stable capacitance over DC bias. Look at this: But it comes to happen that film capacitors are not the only ones immune to DC bias: So, the actual question is why do ceramic capacitor have such a bad behaviour with DC bias? And the answer lies in the dielectric. In order to achieve high capacitance values in small capacitor ...


6

We are not here to do your homework for you. There are various ways to approach this. Here is one. As you say, the capacitors are open circuit for the purpose of finding the DC operating point. That leaves two transistors and three resistors. For the exercise, unlike in a real circuit, the gain of the transistors is explicitly known. Once you know the ...


5

You write that the peak base current, with the signal source connected is given by $$i_B = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA $$ But this isn't true (which should be obvious as it's less than the bias current!). What's true is $$i_{R2} = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA \ne i_B$$ The resistor current and base current are not equal. ...


5

Thats a Microelectronic Circuits Sedra/Smith's excercise , I recognized it immediately. The way you do it is this: The Diodes are there to counteract the \$V_{BE}\$ drop of each transistor, meaning that each diode will "cancel" each \$V_{BE}\$ drop, so the voltage at R6 is the same as the voltage of R1, and the Voltage of R4 is the same as the voltage of ...


5

simulate this circuit – Schematic created using CircuitLab I don't think horta's answer will help. From the description, this happens every time the radio starts receiving rather than just when the radio is turned on. DC IS what you want to get rid of. A capacitor has a high resistance to low frequencies, and a coil (the speaker) has a low ...


5

The big difference between #1 and #2 is that #1 is completely open loop. Transistor gain varies widely, so it's nearly impossible to come up with values for RB1 and RB2 so that Vce is near the middle of its range. The emitter resistor in #2 provides some feedback so that the operating point is less a function of the transistor gain. The downside is that ...


5

The ADC input floats to some DC voltage. If you disconnect your voltage divider from the ADC, you will see the voltage on the ADC float to some DC voltage. The ADC input looks something like this: simulate this circuit – Schematic created using CircuitLab R1 and R2 aren't resistors as such, but rather more like leakage paths from the power rails ...


5

The obvious answer is ... that they were designed to do that. Not being facetious here, rectifying diodes are designed for forward operation and high withstand in reverse, regulating diodes are designed for a relatively accurate reverse breakdown voltage to be used in regulation circuits. If we want to continue, you could say that Photodiodes are optimized ...


5

This is fairly simple. Use an isolated supply to produce (let's say) +/- 15 volts, and tie its output common to your output HI. Then use any op amp you like (selected for output current), powered by this floating +/- voltage to provide your guard current. Since you can easily find AC/DC power supplies with 2 kV or more input isolation (try Digikey, for ...


5

The problem was that the input signal went below zero. It can go a little negative without problems but at around a diode forward drop it starts loading down the inputs. Sorry for wasting everyone's time. Should I delete this question?


5

You are taking the U3 drive signal from the wrong place. Figure 1. Marked up schematic. Instead of taking the signal from (2) you should be taking it from (3). The problem is that (2) is a distorted signal as it has to overcome all the distortion caused by D2 / D3 and M1 / M2. You are tapping off part way inside the feedback loop of U2. Your other ...


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