Electronics > Beginners
50 ohm termination again
Phaedo:
Hi ,
I read this thread :
https://www.eevblog.com/forum/beginners/50-ohm-terminator/
and watched Alan Wolke's video there as well.
I am a tad confused ( as usual )...
Suppose we have a square wave (fast rising edge , i guess ) generated by a digital IC running at 5V.This square wave is connected by a 50Ω coax into the BNC of the scope.
Does it mean that the IC has to be capable of sourcing 100mA (5V/50Ω),if it is getting terminated into a 50-ohm terminator (to get the 5V amplitude) ?
TIA
tggzzz:
Firstly, except for femtoamp and photon counting applications, all electronics is analogue. A "digital" circuit is merely an analogue circuit that interprets analogue voltages (or currents) as digital signals - provided the analogue voltages are within the specified limits.
Don't rely on comments thrown together on bulletin boards; do search out decent literature e.g. application notes. One starting point is http://www.ti.com/lit/pdf/snla043
Initially the transmitter "sees" a load of 50ohm+50ohm in series. If the scope input is 1Mohm//20pF, then after a "long" time the transmitter will see a load of 50ohms+1Mohm in series.
RoGeorge:
--- Quote from: Phaedo on January 17, 2020, 11:28:37 am ---Suppose we have a square wave (fast rising edge , i guess ) generated by a digital IC running at 5V.This square wave is connected by a 50Ω coax into the BNC of the scope.
Does it mean that the IC has to be capable of sourcing 100mA (5V/50Ω),if it is getting terminated into a 50-ohm terminator (to get the 5V amplitude) ?
--- End quote ---
Yes.
langwadt:
--- Quote from: tggzzz on January 17, 2020, 11:53:58 am ---Firstly, except for femtoamp and photon counting applications, all electronics is analogue. A "digital" circuit is merely an analogue circuit that interprets analogue voltages (or currents) as digital signals - provided the analogue voltages are within the specified limits.
Don't rely on comments thrown together on bulletin boards; do search out decent literature e.g. application notes. One starting point is http://www.ti.com/lit/pdf/snla043
Initially the transmitter "sees" a load of 50ohm+50ohm in series. If the scope input is 1Mohm//20pF, then after a "long" time the transmitter will see a load of 50ohms+1Mohm in series.
--- End quote ---
where do you get the extra 50 Ohm from?
tggzzz:
--- Quote from: langwadt on January 17, 2020, 12:04:38 pm ---
--- Quote from: tggzzz on January 17, 2020, 11:53:58 am ---Firstly, except for femtoamp and photon counting applications, all electronics is analogue. A "digital" circuit is merely an analogue circuit that interprets analogue voltages (or currents) as digital signals - provided the analogue voltages are within the specified limits.
Don't rely on comments thrown together on bulletin boards; do search out decent literature e.g. application notes. One starting point is http://www.ti.com/lit/pdf/snla043
Initially the transmitter "sees" a load of 50ohm+50ohm in series. If the scope input is 1Mohm//20pF, then after a "long" time the transmitter will see a load of 50ohms+1Mohm in series.
--- End quote ---
where do you get the extra 50 Ohm from?
--- End quote ---
Good question.
I speedread the OP's question and inferred a 50ohm source termination. Doh.
OTOH, it is an excellent example of "Don't rely on comments thrown together on bulletin boards" :)
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