×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

compensated divider

compensated divider

compensated divider

(OP)
Hello Forum,
I'm toiling just with a Frequency compensated voltage divider followed by a G=1 buffer. The divider is 100k//150p and 900k//16p,1Vpp up to 10 MHz. Now I adjust it so that the sine-output of the opamp is 0.9*Vin. Channel 1 of a 10:1 probe directly connected to the BNC at the divider input, ch 2 at the output of the opamp. The frequency response is ok. If I now connect a 1:1 probe with approx. 460 Ohm to the BNC and compare the waveform at this 1:1 probe tip (ch1) with the amp output (ch2) the frequency response is worse. Shouldn't a correctly adjusted divider have the real value of 1Meg ? Is the input capacitance masked by the 10:1 probe? If I now connect a simple coax cable to the BNC input the output is even rising a little.Can anyone tell me what's the right procedure to adjust such a divider?
Regards

RE: compensated divider

What's this? "...1:1 probe with approx. 460 Ohm..."

If your test equipment isn't much higher Z than your circuit, then it will have an effect on the circuit.

RE: compensated divider

In a divider the purpose of the compensating capacitor - note just the one capacitor - is to correct for the RC low pass frequency roll-off that would otherwise occur from your 900K and the input capacitance of your buffer amplifier. You do not need to place a shunt capacitor across the 100k resistor - this will make things worse.

If the input capacitance of the buffer amp is,for example 100pf, you would only need a 10pF in parallel with the 900K to make the frequency response flat.

The compensating capacitor required varies directly as the ratio of the divider multiplied by the buffer (or oscilloscope) input capacitance. For example for a 10:1 divider:
input capacitance = 100pF, compensator cap = 10pF
input capacitance = 10pF, compensator cap = 1pF.

Throw away the 150pF across the 100K and reduce or trim the compensator cap to around 10pF.
 

RE: compensated divider

I hit the send button too quickly after editing - add this bit to my previous post!

You may have to experiment with the compensation capacitor value unless you have an idea of the input capacitance or can measure it. For starters I suggest you throw away the 150pF across the 100K and reduce or trim the compensator cap to around 10pF.
 

RE: compensated divider

(OP)
@VE1BLL
It's a Hameg 1:1 / 10:1 probe and it has this resistance in the 1:1 range. Check your probe if you don't believe it.
@BrianG
The divider is a 1:0.9 which means the 100k is up and the 900k at ground. The input capacitance of the amp/stray I estimated at 16p because I needed 150p across the 100k to get a flat response.
I think the problem is this 16p that spoils the HF-behaviour.
Probably the layout needs more care to eliminate the problem.
Regards and thanks for all replies.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources