# Tag Info

0

If a long wire is used to connect the source to pin3 on LM386, the wire will essentially act as an antenna, picking up atmospheric static and the interference due to the 50/60 Hz electricity supply ( a.k.a 50/60 Hz hum ). A solution to this would be to add a pull-down resistor as stated before. Another possible solution would be to use a shielded cable from ...

3

The problem with your circuit is the 2N3906 transistors are wired backwards. In this configuration they have very low gain, but the real problem is that reverse bias between the Base and Emitter exceeds the breakdown voltage, which is not modeled in the simulator. According to its datasheet the 2N3906 has a minimum Emitter−Base breakdown voltage of 5V, ...

0

I'll address the gate resistor trade-offs. As evident from a MOSFET model, there are two parasitic capacitances at the gate - Cgd, Cgs. Also, along with this you have parasitic inductance of the MOSFET leads and PCB tracks. So effectively, you have an LC circuit at the gate. This produces ringing at the gate terminal. To damp this ringing (high frequency ...

0

the initial equation should be: $$\frac{Vs-Vin}{Rs}+\frac{Vo-Vin}{Rf}=\frac{Vin}{R}$$ assuming Rf is large, this simplifies to: $$\frac{Vs-Vin}{Rs}=\frac{Vin}{R}$$ $$\frac{Vs}{Rs}-\frac{Vin}{Rs}=\frac{Vin}{R}$$ $$R Vs - R Vin = Rs Vin$$ $$R Vs = R Vin + Rs Vin = (Rs+R) Vin$$ $$\frac{Vin}{Vs}=\frac{R}{R+Rs}$$

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 '...

0

A single-transistor CE amplifier with 10pF Cob and gain of 10x will have 110pF input capacitor, ignoring the emitter-base capacitance. With an Rsource of 200 ohms, the 110pF Cin yields 22 nanosecond time constant, or 45MegaRadians 3db. In Hertz, that is 45/6.28 or ~7MHz. Thus about 1 dB droop in the frequency response at 2MHz, due to CMillerInput and the ...

1

I think this is a good example of an XY problem. Instead of focusing on minimizing the input capacitance in the specific configuration you have chosen, try to modify the circuit in a way that removes the need to minimize the capacitance. When you were learning about op-amps, you probably encountered the fact that the open loop gain of the amplifier does not ...

3

after digging deeper in the internet I figured that the collector-base parasitic capacitance, which is about 9 pF for the BC547C, is what kills me - when inserting this capacitance (C3 in the image below) I get the EXACT same curve as measured with the AnalogDiscovery2 in the real circuit. Collector-Base capacitance should be included in the ...

1

Ideally you'll want to look at the loop compensation and see if the load is making it unstable. This is best done with a spice package, or you might be able to get TI's tool to give you the loop compensation. Here it is for the TPS55340 with a regular load: Source: https://webench.ti.com/power-designer/switching-regulator/customize/3

1

You never have to just guess and check. Or at least, you can make educated guesses, and then carefully simulate. You can estimate the input impedance of the TPA3221 by observing two things: first, it's a switching amplifier, so power in = (power out)/efficiency; and second, it's got those great big input caps. So, let $\eta = 0.9$ be the efficiency. ...

-1

The even harmonics, like 2f , 4f , are synchronous (to) the zero crossing of each half-cycle of alternating-polarity and thus (with) perfect 50% duty cycle (it) has no even harmonics, as they cancel out. Whereas 49% or 51% duty per cycle adds even harmonics that are 1% of the input or approx -20dB in power density. ( not exactly as I left out some details). ...

1

A preamp like you show has an output current that is much too low to drive a speaker. A speaker is usually driven by a Power Amplifier that is usually an IC. But if you use a power amplifier IC then its output will always be connected to the speaker so then the speaker will no longer work as a microphone.

0

That's a bit tricky - ask a dozen RF engineers, and you'll probably get a dozen different answers. I think the best "cookbook" resource I've personally encountered is RF Circuit Design by Chris Bowick. I have the second edition, but if you can find a used first edition for cheaper, feel free to grab it. The second edition added two chapters that weren't ...

2

Your PC's audio output is really neither bipolar nor DC-biased – it's pretty certainly AC coupled, i.e. there's a capacitor in series with it! (In effect, a headphone port would be centered around ground potential, due to the headphone being connected to ground on its other end.) That means you can just AC couple it yourself (I recommend doing that to avoid ...

1

Maybe put a high pass filter in to filter out the first frequency component. Then it would have to lock-in to the harmonic. Don't know if this would actually work in your application. Most lock in amplfiiers have a frequency to set to lock into, and you may be able to set it to the harmonic, and it would lock in on that.

1

Your variable resister is for setting up the idle current in the main powermosfet bank .More resistance means more gate source volts meaning more current .Mosfets are very spready so a fixed resister would not be good for a population of amplifiers,hence the variable resister .Once you have set the idle current you could use a fixed resister of the same ...

0

Given Data: Vceq = 5V, Ieq = 1mA, Beta = 100; Let Choosing supply as Vcc = 12V , 1A; 1.Calculating collector resistor[Rc]: choosing a collector voltage of around half the supply voltage to enable equal excursions of the signal up and down I.e., Vc = 6V and Ic = 500mA Then Rc = Vc/Ic = (6V/500mA) = 12ohm ------------>(1) Calculating Base Current[Ib] using ...

0

At 1ma, the transconductance is 1/reac = 1/ (26 ohms / Ie_ma) = Ie_ma / 26 ohms, or transconductance = 0.039 amps per volt. For gain = 100x, you need the Rcollector * 0.039 to equal 100. Your total Rcollector will be 100 * 26 ohms. The voltage across Rcollector will be 2.6 volts, leaving 2.4 volts for the transistor's Vce.

1

A thermionic valve (vacuum tube) with two electrodes (anode (plate) and cathode) can be used as a rectifier, but does not amplify a signal. Other tubes with one or more grids between the anode and cathode can amplify signals, as a small change in voltage on the control grid will cause a change in the current passing between cathode and plate.

4

The two circuits are both based on emitter followers. As you may recall, the emitter follower has a gain of about 1 and an offset of one base-emitter voltage, Vbe. The followers are arranged as a complementary pair. In both cases the purpose of the diodes D1 and D2 is to insert a bias that parallels the Vbe of each transistor. (Vbe is about the same as a ...

0

The TIP121 is not a single transistor, instead it is a Darlington with two transistors. Then the bridge rectifier has two diodes in series and the Darlington transistor has two transistors in series. The 100uF capacitor has such a high value that it takes a long time for a continuous loud signal from the amplifier to charge it to about 2.4V when it will ...

0

Power matching is said to be used to match an output impedance to a physical system. ex, a precision led with a 27pF capacitance or a moving coil with a physical atenuation factor to it's free movement (A subwoofer). Also when you have a very large power source and very small loads in parallel which could have very different resistances each causing one to ...

1

It's probably something like this that you are after.... When this type of circuit is used for a bridge connected class AB audio amplifier (swap motor for speaker) it is important to reduce crossover distortion as much as possible by using both techniques of feedback and biasing the transistors slightly on with the diodes. Without the diodes, (bases ...

0

The TPA6111A2 headphones amplifier is not designed for electret microphones as inputs because with enough gain then its input impedance is too low and the external compensation capacitor required will cut high audio frequencies. An LM386 power amplifier works fine with an electret mic since its inputs are a higher impedance than the other inverting amplifier ...

2

you gotta start with better specs in case you hadn’t noticed yet, your BJT’s are always conducting creating a condition called ‘shoot thru’ , shorting out the supply on both sides. There must always be a dead band and thus a deadtime during transition. start with your output impedance model of a motor with back EMF as a motor/generator or simply ...

1

You aren't going to get anything below 0V if your low supply Vee is at ground and not -30V. Swing around 0 can only happen through a transistor stage if total voltage from collector supply to negative supply is bounded greater than your desired swing. This is why your output stage is automatically biasing to a a +-15V swing around +15V. Also, an AB output ...

0

(EDIT: Complete rewrite) The TPA6111A2 is a stereo speaker driver with an inverting input. To use it correctly, the source would need to have negative polarity. The datasheet shows that, labeling the inputs IN1- and IN2-. The LM386 is a single channel with internal biasing that accepts a positive input. Neither chip is really a mic amplifier: they don't ...

4

Any sound picked up by a microphone can be assumed to come from some point distant to the microphone (maybe mm to several metres or more). Take a sound at 1 kHz at 10 metres distant to the microphone. 1 kHz has a wavelength in air of around 0.34 metres and this means that between the source of the sound and the microphone diaphragm, the signal has undergone ...

3

In some cases it will make no difference, as the microphone signal is an AC signal, and in many cases the inversion does very little, if anything, to the audio (there are cases, like sending differently phased signals to different speakers, can do bad things to sound). In other cases, it can make a big difference, often depending on the type of microphone. ...

1

Your first circuit uses an IRF2110 MOSFET driver. The capacitors you refer are there to smooth the voltage that the IRF2110 generates to drive the FET gates. The IRF2110 is used to make a higher voltage in order to drive the gates of the FETs turn the FET fully on. FETs require a certain voltage differnce between gate and source in order to switch fully. ...

0

The caps are needed in this highside drive circuit to hold charge for the Gate of the highside N channel fet .The cap value needed is a function of on time .In these bootstrap circuits the cap can only gain charge when the highside Fet is off .When the highside fet is on charge bleeds away due to the gate source pulldown resister and due to chip losses .

1

I would try a Pi Filter to speaker from each output (L(C//2Ohm)L) to shared speaker so each driver has isolated feedback and LPF to speaker. Then the benefit of this is lower distortion at max power or slightly more power at same max distortion due to lower RdsOn/2 @ 2 Ohm load and reduced reactive power in 2 Ohm speaker at class D frequency. Otherwise the ...

2

Do not connect outputs to ground! I would connect INR to ground through a resistor and a cap, per their "typical applications schematic". If I were feeling particularly paranoid (which I usually am when going off-datasheet) I would connect an 8$\Omega$ resistor between OUTR+ and OUTR-. Then I'd cross my fingers and hope the resistor doesn't see much of ...

0

Those two 1k input resistors will probably be too low in value. They give an input resistance of 500R (1k//1k) which, unless the signal source resistance is very low, will reduce the amplitude of the input signal quite significantly. I would recommend swapping them for 100k resistors (more typical value) giving an input resistance of 50k. The amount of ...

0

Depending on narrow-band versus broad-band design, your SNR becomes a key design variable because broad-band interfaces that have 50ohm Rsource and 50ohm Rload will cost you 6dB signal level. Yet you have the magic-incantation of "but boss, I matched it".

1

There’s a difference. conjugate matching reduces the VAR LOSS to real power transfer. maximum power when matched results and 50% efficiency from a voltage source. Higher efficiency and gain means less output power but lower input signal when load is say higher impedance than source. At extremes Gain with no load is twice the voltage but no current so no ...

10

Good start, just one more thing to add... R1 cannot be DC-grounded. It must be AC-grounded with a capacitor. You would choose the capacitor value so that its reactance is equal to (or less than) R1 at the lowest frequency that's important to you. For example, if R1 is 1000 ohms, and you are amplifying audio where 20 Hz is the lowest audio frequency, C1 (...

7

Your circuit still has DC gain. You need to add a cap between R1 and GND. The output will have a DC offset. If that needs to be removed, add a series cap between the output and whatever is downstream. Since you have not provided any frequency information, you're on your own to determine the values.

1

In theory this is logically true. In practise no, because we need to know prop delay, dead-time (!) , impedance of commutation switches , load and reactance of everything. This thing is called a “Half-Bridge” also used in Buck-Boost SMPS and have many specific parameters for each load and power range for specs. Then for differential high power from a low ...

0

D20 is a flyback clamp diode. The parts encircled ought to be like a T=500us snubber, while yours is ~ 360ms in schematic and too lossy. 1k//1k//330uF = 165ms snubber is just as bad or worse for loss. Check primary L and compute sqrt(L/C)=2pi f and reduce C significantly and choose RC= 500 us. ( I think)

0

It's a part of core reset circuitry. It should get hot when no load is present and it should be cool when max. load is present. It is possible that you swapped the polarity of the output secondary coil? See dots orientation: EDIT: MY bad. It' not a reset circuitry, rather a RC snubber to damp the MOSFET ringing. Your values are not appropiate.

0

RF port and LO port are inter-changeable. If you imagine that LO-to-IF is balanced, then RF-to-IF is balanced as well. But any output at the IF port does cause unbalance. In normal mixer use, both RF signal level and IF signal levels are small...much smaller than LO signal level. Any signal developed at the IF port causes unbalance that will transfer a ...

0

A & B are common to both sides which use the balanced neutral for isolation and are commutated by LO so a +/-difference frequency and some IMD harmonics result in the output. Any mismatch in balance is seen in the output which includes ferrite distribution tolerances, leakage, stray capacitance, geometric tolerances and some port impedance mismatch. ...

1

The greatest possible amplitude of an undistorted output sine is: min(|L+|, |L−|) = 8 V And the amplitude of the corresponding input sine is therefore: min(|L+|, |L−|) ∕ 100 V/V = 8 V ∕ 100 V/V = 0.08 V = 80 mV The minimum (min) value of the absolute values of the supply rail voltages needs to be taken in order not to exceed the supply range and so to ...

0

A series of two GPIO ports can't supply 10V at significant currents. That's not what GPIO ports are capable of. Also, it's not even clear whether you'd get the sum of voltages – if both GPIOs were fed from a power source with a common ground, you've just shorted one supply to ground with the other. This all is a terrible idea and you might really need to ...

1

The basic problems with your circuit are extreme sensitivity to DC bias point of the upper FET, interaction between DC bias point and gain of the input transistor and lower FET, and low AC gain. I put your circuit into LTspice and made a few changes to improve DC stability and AC gain:- To remove the interaction between M1 and Q1 biasing I AC coupled the ...

2

Class A Complementary FETs will always be a struggle with 1 BJT driver low open loop gain, thus >10% THD even with negative feedback since it is very quadratic. only get 16Vpp from a 24V supply. loss of low-frequency response with a single supply requiring large Cap may have poor characteristics. poor supply ripple rejection (hum etc) too many pots that are ...

0

Max Vout pp = +9 - (-6) = 15 Vpp = 7.5Vp Divide by Vgain =85 to get input B.

0

Ignoring the voltage gain of the Half bridge and just making a feedback loop of the filters; I get a feedback loop with a fixed low gain <1 proportional to the 500kHz carrier with k1 from R ratios and T1,2,3 from RC products. I did not bother with the equations but rather direct into simulation with a slider for signal frequency. e.g. above signal ...

0

The Vbe (DC) controls the Ic (DC) ( from some Vcc) The collector voltage DC = Supply Vcc. The AC coupled input signal modulates this collector current such that the AC coupled load R cannot draw more average signal current than the DC Q current. For linear operation, saturation is not permitted. But it can swing higher than 2Vcc when Vce=Vce(sat) ...

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