8

The classes were originally arranged to describe the conduction angle for a single quadrant of a power amplifier. In class-A, active conduction occurs throughout all \$360^\circ\$ of the period. In class-B, active conduction occurs for \$180^\circ\$, or one-half of the period. In class-C, active conduction occurs for \$\lt 180^\circ\$ (but more than \$0^\...


7

The diodes are part of a RF switch that switches between RX/TX modes of the transceiver (See also my answer to another question). This can be accomplished by changing the DC potential at the RX LINE side of the diodes. You can see a RF filter of the TX LINE side consisting of R150 C174, C176 and L40. I'm sure there is a similar filter on the RX line side (...


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 ...


5

That the amplifier needs a DC supply, not AC. The + G - is a strong give away. An AC power input would likely have a pair of ~ symbols. Also an AC input would need at least a bridge rectifier and a big capacitor. I spot neither near the power supply connector.


5

Yes, LM386 is designed to supply far more current than most op-amps. However, LM386 should not be viewed as an op-amp. It is designed to drive low-resistance loads like a speaker, while op-amps have far more applications. LM386 does have differential inputs (inverting and non-inverting). And it does have a single output, like an op amp. But its internal ...


4

The collector of \$Q_{19}\$ cannot be more than a diode drop above the bottom rail, because its base is tied to the collector. \$D_5\$ then prevents the collector of \$Q_{12}\$ from being more than two diode drops above the bottom rail. Since the \$Q_6\$ (I was wrong to write \$Q_8\$ before, in comments) base is tied there too, this prevents the base of \$...


4

They are pin diode switches, when TX is enabled, current flows via R150,L40 forward biasing the diodes and effectively making them transparent to the RF power. Remove the bias (Or better reverse bias the diodes) and they present as small capacitors to the RF effectively isolating the PA from the aerial on RX. I expect there is a similar arrangement ...


4

It all depends on how much distortion you can tolerate, and any other side effects that may occur when saturating the amplifier (harmonics, etc.). You might be fine running several dB over the P1dB point. Or maybe you need to stay several dB below. You might just need to experiment. If your FMCW radar can tolerate some distortion, then you might be able to ...


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 ...


3

This should explain your questions: - EDITED SECTION I was clearly wrong about C5 - upon further inspection it has nothing to do with being part of a tank circuit so apologies for that. And L1 is therefore not a tank inductor but just a collector load of about 240 ohms at midband broadcast FM frequencies. Here's another attempt to fix my earlier stupidity:...


3

They both must have opposite direction and must differ from each other and the 0 must be on load. What makes you think that the output currents of the NPN and PNP transistor have a different direction? You should be careful with the direction of currents and also separate the direction of a DC current (bias current) and the direction of a change in current (...


3

Have you plotted the stability circles? Do a load pull and plot these, they contain useful information. Low frequency instability is often down to one of a few things: Source impedance goes high Z at low frequency, unterminating the input: Either provide a low frequency termination or an input pad of a few dB or so to reduce the mismatch at the input. A ...


3

That board uses an INA-02186 device, which is not a bipolar transistor but a complete MMIC gain block encapsulated in a typical transistor package. This "performance figure" comes from a random ebay seller (to whom, by the way, I'm not related in any way): which, as you can see, is nothing but a snapshop from the INA-02186 datasheet: The INA-02186 is pre-...


3

Class A = both transistors are ON all the time. Class AB = both transistors are ON at idle, then up to a certain output current. When output current is higher than a certain limit, one of the transistors turns off. Class B = either one transistor or the other is ON, but not both. The transistor that is ON is determined by output current polarity. Class C ...


3

When the base is driven at the positive peak, the emitter-follower's emitter current is at a maximum because it must supply both the current sink current as well as the current through \$R_1\$. But this is also at the same point where the \$V_\text{CE}\$ is at a minimum. So there is a power minima here. When the base is driven at the negative peak, the ...


3

Found a datasheet for the NTE1337. It is a replacement for the STK086. The datasheet for the NTE1337 includes this drawing: It shows the package size, hole sizes and locations, and the pin spacing (as well as pin dimensions.) The measurements seem to match those that @KingDuken found, but are easier to read. You will need all that it has to offer since ...


3

1.6W mean's the peak voltage is about 12V. How is it possible with Vs = 5V? Look at the data sheet application circuit diagram below. I've added a red box and a blue box: - The red box is around L1 (an inductor) and when you have a collector of a transistor being pulled up to Vs, the peak-to-peak voltage that can be attained is nearly twice Vs. So, if Vs ...


3

The Sziklai topology places the output BJTs within a local NFB loop. It's enough that the quiescent current is approximately 20 times less sensitive to the output BJT temperature changes. Partly because of this, you do NOT need to include the power output BJTs on a shared and monitored heat sink. It's the driver BJTs that need to be monitored for ...


3

110 watts into an 8 ohm load implies an RMS current of 3.71 amps. That current, through a single 0.22 ohm resistor implies a power of 3.025 watts. Considering that this power is shared by 4 resistors, the average power per resistor is 0.76 watts. But then one has to also consider the biasing of the output stage and this will cause a certain amount of DC ...


3

Always read the datasheet. The NE5534 has an input bias current of 500-2000nA. (500nA)(22k\$\Omega\$) = 11mV. So you're getting just about exactly the expected voltage across R1. If you put a resistor between your output and the inverting input of the op-amp, then it'll have the same bias current, \$\pm\$ the input offset current (which is 20nA up to a ...


2

In this common source stage, the inductor acts as a current source and capacitor act as a voltage source Nonsense ! The inductor is a low impedance for DC, meaning it allows a DC biasing current to flow while keeping the drain of the MOSFET at a DC level equal to VDD. The capacitor does the opposite, it blocks DC current and only allows AC signal to go to ...


2

I think you need to use ideal baluns for that.


2

When you try to understand how commercial amps work from reputable Japanese designs, and see the purpose of each component, you may learn how to improve quality. Of course there are many details in thermal matching and hFE binning needed as well. Here TR110 acts as the Vbe multiplier which must be thermally attached to the output stages. Tuning is ...


2

Why all books told that Av differential of an differential input stage in power amp is very big? I find "all books" a dangerous statement. It really depends on the input stage if it has a large voltage gain (Av). Also you're talking about Power amplifiers for audio, you should have mentioned that because audio is low frequency. There are also Power ...


2

You have mentioned the expression Ad = Rc/re. Please note that this expression is (1) a very rough approximation for a common emitter amplifier with heavy negative feedback (due to re) and (2) does not apply for a diff. amplifier. Neither for a simple one (two or three transistors only) nor for the one shown by you. This is because the resistance in the ...


2

If my memory serves me well the first stage gain is equal to: $$A_{V1} = gm*r_\pi = \frac{r_\pi Q6}{re2}$$ $$re2 = \frac{26mV}{I_{C2}}=\frac{26mV}{290\mu A} = 90 \Omega$$ $$r_\pi Q6 = (\beta+1)*re6 = 150 * 4.8\Omega = 720\Omega$$ Hence first stage gain is: $$ A_{V1} = 8V/V $$ Q6 stage voltage gain is large but will drop due to \$R_L\$ loading effect....


2

There is a lot of overlap in what manufacturers describe as gain blocks and power amplifiers in the +10dBm to +20dBm output region. It would be an unusual power amplifier with less than +10dBm output, and gain blocks don't often have >+20dBm output power. As well as being lower power, gain blocks tend to emphasise good port matching and good isolation, as ...


2

I'll just write out the first few answers I was working on, before seeing G36's response (which you should definitely read.) The leftmost two form a high pass filter with about \$4\:\textrm{Hz}\$ as the corner frequency. The other two form a low pass filter with more than \$100\:\textrm{kHz}\$ as the corner frequency. Together, I'd have to play with how the ...


2

1 - 18kΩ resistor together with 2.2µF capacitor for a high pass filter. The Fc frequency is equal \$F_C\approx\frac{0.16}{RC} = 4\textrm{Hz}\$ Additional 1kΩ resistor form a low pass filter together with 1.2nF \$F_C\approx\frac{0.16}{RC} = 134\textrm{kHz}\$. And this filter stops any RF signal, so the RF signal is not being amplified by the amp. 2 - ...


2

To keep the quiescent current low, the left hand transistors run at a low current, usually <1mA. To provide usable output current, usually >20mA, to drive the load and feedback resistors, the right hand transistors provide current gain. That's not a 'power amplifier', but you do see that sort of transistor arrangement on the outputs of power amplifiers.


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