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The 9M resistor will have a capacitor in parallel, but it will be much less than the capacitance at the scope (or whatever) input.Thanks.
The way I see it, adding a resistor in series only increases the charing and discharing times of the capacitor, but I dont understand how it decreases the capacitance loading ?
Or as you said, there is less current flowing through the capacitors, how does it decreases the capacitance loading?
The 9M resistor will have a capacitor in parallel, but it will be much less than the capacitance at the scope (or whatever) input.
See Fig. 3 at this site. Read the associated text.
Also see this document, section A.2 and beyond, for the math.
Cp keeps the frequency response of the network flat, which also keeps the pulse response flat.Man, these articles were awesome, specially the last one, it taught me a lot.
As I understood from the last article, only when the probe's bypass capacitor - Cp -(which is in parallel with the 9Mohm resistor) equals 1/9 of the scope's input capacitance - Cs - then Cs and Cp are connected in series, and therefore as you said, the loading capacitnace that is imposed on the DUT is Cs||Cp which is smaller then Cs.
That is why the 9MHz resistor decrease the capacitance loading ?
only because of the Cp connected to it in parallel?
1. It's the combination. As you noted, the resistor alone basically totally isolates the input from the capacitor, at the expense of very poor frequency response (slow pulse risetime). You have to add Cp to compensate for the series 9Meg resistor, so the resulting frequency response is flat.Thank you.
What i'm trying to understand is
1. what causes the imposed loading capacitance to be reduced?
2. and how?
is it the 9Mohm series resistor, or the Cp capacitor?
Ct=9.0pF.Neither. Formula for computing equivalent capacitance of two capacitors in series:
Ct=C1*C2/(C1+C2). If C1 = 10pF and C2 = 90pF; then Ct = 8.18pF
btw-are you a Troll?
1. It's the combination. As you noted, the resistor alone basically totally isolates the input from the capacitor, at the expense of very poor frequency response (slow pulse risetime). You have to add Cp to compensate for the series 9Meg resistor, so the resulting frequency response is flat.
2. Again, as you noted, the resuting capacitance is ideally one tenth the value of Cs.
When you hook a cap up to a voltage source directly, it's going to pull as much current as it can to charge. If the source cannot supply enough current (a very small signal), then the "load" the capacitor puts on the source will cause a significant voltage drop which changes its waveform.
The 9MΩ resistor is acting as a current limiter, so the cap does not pull too much current too fast and change the voltage of the signal. That is what is meant when they say it changes capacitance load.
How the capacitor connected directly to the voltage source causes a significant voltage drop?
IMHO, that was a misleading explanation.Thank you very much!
How the capacitor connected directly to the voltage source causes a significant voltage drop?
Capacitors need to be charged up before their voltage will change. So the capacitor will not allow the DUT voltage to change until it is charged up. It will steal lots of current from the DUT. The high current draw from the DUT distorts the DUT signal because the DUT is working harder and the capacitor's charging distorts the DUT signal because the capacitor delays changes in the DUT voltage.
A large resistor allows a very small amount of current flowing from the DUT to the capacitor to form a large voltage drop between the DUT and capacitor. The DUT only has to supply a small current to produce a voltage drop across the resistor. This capacitor steals less current from the DUT circuit which distorts it less because it is not working as hard, and the resistor allows the capacitor and DUT to have two different voltages with only a very small amount of current flowing between them.
The rising edge looks the same in both circuits because the capacitor charges (rising edge) through the diode, which is a low impedance in the forward direction. When the input falls, the 90pF capacitor has to discharge through the 10k resistor. This discharge time is slower than the fall time of the input, so the diode cuts off (does not conduct).Thank you for this.
Could you explain these two waveforms?
I dont understand why in the 1x waveform, the rise time stays the same, but the fall time gets longer, why is that?