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