New answers tagged impedance-matching
0
Read about PI_network matching.
Also read on inline stubs and shunting stubs.
Check into using the Smith Chart for matching.
Surely the American Radio Relay League has literature on matching.
2
MY vna is 50 ohm source impedance, what values and which type of
network would you use?
First convert 4 - j22 to a parallel capacitor and resistor (8.258 pF and 125 ohms) Try this calculator.
Assume an LC pad to will match 50 ohm to 125 ohm (may need a slight mod)
Recognize that the LC pad needs too little capacitance compared to 8.258 pF
Resonate out most ...
0
The 10 ns rise time you mentioned seems rather slow for a part that specs the prop times as 6 ns (max), and 1 ns (min). The data sheet does not give a rise/fall time spec, so let's use the prop time as a surrogate for rise/fall time. That means we need to design the interfaces to properly handle a 1 ns rise or fall time.
1 ns in a typical board material ...
0
The impedance (odd mode impedance to be exact) of most twisted pair cable is around this value - its not an exact figure - it doesn't have to be - see my calculation below.
100 ohm is close enough!
Calculate VSWR for given load and source
0
You mean, why not 220 ,100 or 75 or 50 Ohms?
It was conveniently close to what some wire pair gave for impedance , based on ratios of conductor to insulator diameter for certain Ribbon wire, twisted pair etc.
I think 93 ohms was common at one time for low capacitance coax on arcnet or something like that. Ribbon cable was 120 ohms and many others, due to ...
answered Jul 31 at 21:04
Tony Stewart Sunnyskyguy EE75
94.5k2 gold badges34 silver badges139 bronze badges
0
No (not for what you want to achieve)
The datasheet lists HRout (headroom to either supply) which is 1.5V. So max. output voltage will be 28V - 2*1.5V = 25V. But you need to watch out for the output impedance which is 0.17 Ohms. So if your load resistor is e.g. 0.17V, you will get less than 25V/2=12.5V at the output.
No, but it would be correct if this amp ...
3
15 ns rise time and assumming a prop time of 2 ns/foot gives us a distance of 7.5 feet for the rise time, sometimes referred to as the spacial extent of the edge. Rule of thumb that many designers use is that you can ignore impedance discontinuities if they are less than 10% of the spacial extent of the edge. So 10% x 7.5 feet = 0.75 feet = 9 inches. This ...
1
Connectors do matter, as not all connectors provide impedance match. So there will be reflections at the connector. The connector just needs to be good enough so it does not distort the signal too much. Which means that the length of the connector parts that have mismatched impedance is small enough.
I believe screw terminals could work in your case. 2 MHz ...
2
Impedance matching starts to matters when the following three criteria are satisifed, all relative to each other:
the propagation time is slow enough
the wavelength is short enough
the connection length is long enough
In other words, when you can no longer assume the wave propagates down the line instantaneously and that the potential at both ends of the ...
2
This simple circuit can be solved by inspection instantaneously either by using Thévenin as correctly highlighted by Wheatley but also by using the fast analytical circuits techniques or FACTs. Simply consider the circuit for \$s=0\$ and determine the dc transfer function. Then turn the stimulus off - short the input source - and "look" through the ...
2
You can use the Thevenin equivalent circuit to see what your capacitor "sees".
Your capacitor will see an input voltage of Vin*R2/(R1+R2) in series with a resistor equal to (R1||R2). Hence your load will affect your voltage amplitude and also your cut-off frequency. If it is a problem, you may use an op-amp configured as a voltage follower between ...
0
If you want to check if the equation is applicable (and what is "R" ?), you should use the definition for the cut-off frequency wo. For this purpose you need the transfer function.
It is a first-order lowpass - hence, the general expression applies:
H(jw)=Ao/(1+jw/wo).
Ao: Transfer function for w=0
wo=1/T with T=time constant of the RC product ...
1
You can match the output with transmission lines and/or air cored transformers, and there are plenty of circuits you can use or adapt. But your main problem is going to be matching the input to the gate. At 600MHz, you are going to need several watts into maybe just a couple of ohms, with an inductive reactance doing its best to stop you. Good luck with that....
0
The first time I've built a high impedance microphone circuit I thought it was malfunctioning, but then found that it had to be put inside a shielded enclosure.
Top 50 recent answers are included
