Hot answers tagged

6

It makes no difference whether it is a power amplifier or a DC supply feeding a resistive load via a variable resistor. The variable resistor is, in effect, your amplifier. And, as engineers, we learn that when the variable resistor has the same resistance as the resistive load, maximum power is wasted in the variable resistor. See the maximum power transfer ...


5

Such devices are available but in almost all cases you must be licensed by the appropriate authority, such as the FCC in the USA, to operate such equipment. The requirements for unlicensed operation almost always has an upper limit on the radiated power so that interference with licensed service does not happen. The 433 MHz band is shared by a number of ...


5

That would be because of power dissipation of a transistor is P=U*I. If it is fully off, I=0, and if it is fully on, U=0, so ideally the transistor dissipates no power. But it dissipates most power when it is halfway on, as there is both voltage over it, and current through it.


4

If there is no load, then there will be no series attenuation. Beads come in all sizes and ratings. for Z(f) and DCR . In order to be effective the circuit closed loop impedance must be at least 1 or 2 decades below the bead at lower frequencies. "Any wire" cannot achieve that. It is intended to be a low loss at DC or low frequencies and lossy at ...


4

DSSS vs OFDM: these are different standards (a/g/n vs b, which is obsolete). This situation where you'd need to detect the differences never arises: a beacon from an access point is either of these standards, and that defines what the whole network does. Since the channels for these different modes don't even match, there can't be interoperability. A card ...


3

The other answers are also correct but I missed the relation to how efficiency is defined, so here's my additional answer: For most amplifiers, the total powerconsumption consists of two parts, a constant part (for biasing etc.) and a variable part which depends on how much power the amplifier is delivering: (1) \$P_{total} = P_{constant} + P_{out}\$ If we ...


3

The value you choose for a blocking capacitor mainly depends on how low you need your low frequency cut-off to be. Macom chose 100 pF for their demo circuit in order to obtain a low-frequency cut-off well below 1 GHz. If you only need a 2-GHz low-frequency cut-off, and (say) you're willing to accept 10 ohms reactance from your blocking capacitor at that ...


3

Class A will transfer max current and dissipate least power when the voltage drop is lowest. This is why Class A is just about 50% efficient on average and Class D 98% or so. For RF they might use Crest Factor Reduction (CFR) methods to improve efficiency by clipping, filter harmonics or other methods for ODFM.


3

The circuit you show isn't a useful filter. The current from the voltage source is nicely bandpass filtered, but in the real world that's not terribly useful. A useful model for an RF filter would be something like this, with a source that has a real source impedance (represented by R1), and a load that also has some real impedance, not to mention that the ...


2

The first table is only usable at an RF power of -10dBm, and only at an LO frequency of 5.25 GHz. You would expect the power of a 0RF spur to stay constant, regardless of the input RF power, so it's not correct to try to use the first table to take 3.4dB with respect to a different RF input power. Given that the LO to IF isolation is given as typical 32dB, ...


2

If this may interest you, after further elimination of all other possible causes, I got stuck into the output connector on some bare boards to reproduce a similar difference, but this time with simply a 50 Ohms 0402 resistor as the first shunt element to act as a load at the end of the micro-strip line. This also gave vastly different results, above 10% ...


2

Since this is a distributed structure, in principle we can't even define a potential difference between the two ends of the transmission line. In practice, it's a useful approximation to take the ground nodes at the two ends of the line as equal, and that's what the simulator does.


2

If the signal is changing faster than the bandwidth of the bandpass filter, then the following circuitry won't see the effect as the signal changing -- they'll see the effect in the frequency domain. For instance, if the signal is changing between 5MHz and 4MHz, randomly every 500ns, but staying phase coherent (meaning that there's no time-domain jump ...


2

Like Brian Drummond, I'm not aware of car key fobs using IR. A significant change in key fobs is the move from push-button fobs, where the user presses a button sending a burst of RF from the key fob to the car locking or unlocking it, to keyless entry fobs, where the user just stands near the car which unlocks without needing to press a button. Walking away ...


2

RF does not require line of sight. Works better in daylight too. It is also harder to crack as it requires more complex hardware than a $5 learning IR remote. IR devices use IR LEDs, not lasers. TV remote controls originally used ultrasonic acoustic transmission in 1950s, and switched the ultrasonic transducers to IR LEDs and receivers in the 1970s.


2

I‘m trying to understand exactly what an L-section Network does. What you are describing in your question is a loss-less, high-pass, L-pad impedance matching circuit as per this example: - The L-pad is used to match a lower impedance on the left with a higher impedance on the right. Either end can be source of course. In other words; you can match a higher ...


2

I have a little understanding of a regular 433mhz transmitter schematics. It is so simple , I want know can I build an rf transmitter with that circuit with more transmitting power (for example 2W power) ? It's not that you can't but that you aren't allowed to. There are regulations set by telecommunication bodies such as the FCC and ITU. You are permitted ...


2

The coax impedance must be matched and the cable must be able to carry the 1900MHz signal. It might say the impedance on the cable (like 50Ω or 75Ω), if it does then your in luck. If it doesn't then the only way to find out is to test it (which would be very difficult and require specialized equipment). Additionally the repair would need a continuous wire (...


2

Efficiency of class B amplifier. Let’s consider a class B power amplifier which we will assume saturates at the supply rails (in reality the amplifier would saturate a few volts below the power rails). The equations below can be used to calculate the power dissipated in the load, the power drawn from the supply rails, the power dissipated in the transistors ...


1

The simplest solution is to use an IR remote control chip to Rx and IR LED to transmit. That ought work as well as other LED remote control. If you can figure out the protocol for the SPI interfaces involved and make it transparent in the cheap radio from Banggood in your link , then you may have found a good solution.


1

There are tons of project on the net using ESP32 to control these LEDs. This microcontroller has WiFi, which is pretty useful for wireless control.


1

There are various ways to achieve this. In all scenarios I can think of, you would leave the Arduino (or an Arduino, not necessarily the one you have) in control of the RGB lights. You then transmit some data to the Arduino to tell it what colour to set the LED strip to, and to turn it on or off, or start some other procedure. The easiest way to do this is ...


1

the output circuit makes use of a duel opamp (LM 358) but my understanding is too flakey to derive the output voltage from this arrangement. LM358 datasheet page 10:- This tells you that maximum output voltage is Vcc - ~1.2 V = ~3.8V with 5 V supply. The ESP8266's rated maximum I/O input voltage is 3.6 V, however it probably has protection diodes that don'...


1

From the ali pages, the chip seems to be an LT8900 (Some ali vendors claim it has a NRF24L01, which I think is wrong. Check the marking on your modules.). As you asked for info on how to use a SPI interface, I guess implementing a driver for this chip is way over your head. But this guy seems to have made and arduino driver for it, so you could try that. ...


1

In real life, I think it'd be as you said: the electric potential at the lower left node is different than the potential at the lower right node. However, when we mathematically model this physical system, we define the resistance per unit of the transmission line to include the resistance of both conductors (the upper conductor and the lower conductor). ...


1

An accepted formula for calculating the power needed by a receiver operating at a specific data rate is this: - Power (dBm) = -154 dBm + 10\$log_{10}(\text{data rate})\$ Formula from this excellent book by Chistopher Haslett: - The module in question has a data rate throughput as high as 300 kbps so, in that case the receiver input power needed would be: - ...


1

Indeed you are missing cable loss, but that's going to be a few dB at most, not anywhere near 79 dB. What you are missing is noise. The receiver sensitivity is the weakest signal that can be received with zero external noise. But unless you have a cryogenic LNA and a highly directive antenna pointed at something very cold, it's going to be external noise, ...


1

This design will fail miserably with the capacitance of diodes greater than 3.3p and microwave f input being unrealistic as well as the 500k load too much for the very low series capacitance that is daisy-chained. There are very real theoretical limitations to the maximum theoretical number of cascades and that reduces when you add a load. Adding each stage ...


1

It all starts with an understanding of bandwidth. Technically, the bandwidth of a pure AC carrier is zero. The carrier could be 2 GHz, but if it is not modulated, then the bandwidth is still zero. A signal starts to have bandwidth when the carrier is modulated. AM signals do not have zero bandwidth even though the frequency of the carrier is constant, ...


1

Input impedance is 50 Ohm Output impedance is 100 Ohm For an unbalanced 50 Ω to 100 Ω impedance matcher you can use this online calculator: - Rin is the input impedance and RL is the load impedance. To make that circuit balanced, you need to split R1 into two resistors of equal value: -


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