Termination impedance of an oscilloscope

[calculus_cuthbert]

oscilloscope impedance

Hi,

I had a question about the termination impedance of the oscilloscope. If the termination impedance of the oscilloscope is given as 50 ohms, it is equivalent to putting a shunt resistance of 50 ohms to ground at the output. Is this correct?

Thank you.

 

[flatulent]

oscilloscope termination

It means that the impedance looking into the coaxial connector on the front panel is 50 ohms. This is not the impedance at the probe tip unless you have a 1:1 probe. This 50 ohms will load down the circuit you are trying to measure.

 

[dkumar]

so how can i emulate this oscilloscope termination impedance in circuit simulations??

What i tried was to terminate the output of my circuit to 50Ohm impedance to ground but i believe that's not what should be done. is there any way i can take the effect of this 50Ohm termination in to account in my simulation or it doesn't really matter much.


thanks and looking forward to an early reply

thanks
-dk

 

[karesz]

... This is not the impedance at the probe tip unless you have a 1:1 probe...

Sorry,
an 50 Ohm input can you not apply for an usual probe, then their need an input of 1MOhm!
Your 50 Ohm makes practically a short fort your probes output, exatly does an (unneeded) attenuation 1millon (Ohm) to 50 Ohm=20`000, what you dont need or wish:-(...
K.

Added after 3 minutes:

In simulation, if you have DC coupling, this is an equivalent of a "normal" 50 Ohm resistor.
Quetion is:
must you have a 50 Ohm system?
Is this a high-speed circuit?
If, then - of course you must have a termination (you must know/decide) what is it (an AC or DC termination) & so simulate it...
K.

 

[FvM]

I understand, that you want to simulate the effect of connceting a 50 ohm oscilloscope input to your circuit, e.g. to an amplifier
output. The basic idea is, that it exactly behave like connecting a 50 ohm resistor directly to the output. In the RF or high-speed
instrumentation field, the output would have 50 ohm internal impedance, too. So connecting the oscilloscope creates impedance
matching and reduces the output voltage to 50%. If the output amplifier can't drive 50 respectively 100 ohm (it means e.g. 200
respectively 100 mA at 10 V), an unmatched series termination resistor is often a solution. With a 450 ohm resistor, you get e.g. a
10:1 signal attenuation and a mostly acceptable load.

 

[JoannesPaulus]

What i tried was to terminate the output of my circuit to 50Ohm impedance to ground but i believe that's not what should be done. is there any way i can take the effect of this 50Ohm termination in to account in my simulation or it doesn't really matter much.

Oh, it does matter! What you did is correct. Of course, your output driver must be able to drive a 50Ω load.