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voltage divider for high frequency signals

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ikalogic

Member
Hello

I have a very simple question. How can we reduce the voltage of a 20MHz signal from 24V PP, to 5V PP.

Using resistors based voltage divider will kill the signal with its parasitic capacitance. Same problem with optocoupler.

Any hints? :)
 

ikalogic

Member
I am reading that oscilloscope probes that have 10:1 division simply use 9MOhm resistor in series..? is it save then to use a voltage divider in the range of the 100's KOhms ? (given that my total input capacitance is in the range of 20 pF)
 

Nigel Goodwin

Super Moderator
Most Helpful Member
You use capacitors across the resistors as well - accurately adjusted to cancel out. This is why you have to adjust x10 scope probes.

For your scope example (x10) the input of the scope is 1Meg and 20pf (roughly). To make a x10 probe the resistor needs to be 9 times the scope resistance (so 9Meg), and the capacitor across the 9Meg needs to be 1/9th oof the scope capacitance (so 1/9th of 20pF). However, this is altered by the various stray capacitances, so you have to make it adjustable and trim accordingly.
 

MikeMl

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At 20MHz, the signal is likely generated in a coil. You can tap the coil, or put another winding on it to create a coupled tuned transformer, or add an un-tuned transformer with either a tapped winding or two windings with the appropriate turns ratio...
 

BrownOut

Banned
If the small stray capacitance of a resistive divider is killing your signal, then the circuit has exceedingly high output impeadance. You might have to add an active stage to buffer the impeadance, then feed that signal into a capacitive compensated divider.
 

Rolf

Member
Hello

I have a very simple question. How can we reduce the voltage of a 20MHz signal from 24V PP, to V PP.

Using resistors based voltage divider will kill the signal with its parasitic capacitance. Same problem with optocoupler.

Any hints? :)
A simple two resistor divider will do it, no frequency compensation is needed as long as the freq. stays constant. Making the smaller resistor a 10 turn pot will make things easier.
 

ikalogic

Member
Thanks a lot for those hints.

The frequency will be changing, and the voltage range too. I do like the capacitive compensated divider solution...
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thanks a lot for those hints.

The frequency will be changing, and the voltage range too. I do like the capacitive compensated divider solution...
It's how scope probes work, and how the internal attenuators work as well.

However, as with the other suggestions, it depends a great deal on EXACTLY what is going on, and you've given no clue about that.
 

ikalogic

Member
It's how scope probes work, and how the internal attenuators work as well.

However, as with the other suggestions, it depends a great deal on EXACTLY what is going on, and you've given no clue about that.
Sorry about that.

I am trying to build a mixed mode logic analyzer / scope.

Only one of four inputs is wired to an ADC. i want to be able to sample signals ranging from 0 to 24 VDC, using a 5V ADC. That's why i need to lower the voltage to 5VDC using a voltage divider.
 

crutschow

Well-Known Member
Most Helpful Member
Yes, but you probably need to add a capacitor across the bottom resistor as well, to 'swamp' the stray capacitance.
Scope probes don't do that. It just increases the capacitance load of the probe. Why would you do that here? The stray capacitance is fairly constant. You just need an adjustable capacitor to optimize the compensation.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Scope probes don't do that. It just increases the capacitance load of the probe. Why would you do that here? The stray capacitance is fairly constant. You just need an adjustable capacitor to optimize the compensation.
Have a look inside a scope, at all the capacitors across the input attenuator.
 

ikalogic

Member
Is this the correct way to add the compensation capacitor on the voltage divider?

I tried the configuration above and observed the result on a scope:

I injected a 20MHz sine wave on the input of that divider, with a 27 pF cap.


Without the cap, everything seems OK, the voltage at the central node is half the input signal.

When i add the cap, it seems like due to the high frequency, the voltage at the input decreases, and the voltage at the node become higher than the voltage at the input. As if the cap become a short circuit in high frequency. Well, at least that confirmed what i studied! Maybe the solution was to add another cap on the bottom resistor..?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
You can't just add a random capacitor, it has to be EXACTLY related to the capacitance at the bottom and the attenuator values - which is why it MUST be adjustable.
 
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