Noob here - Modifying an attenuator

[Cactus]

Hi, I'm almost a complete novice at this electronics game but increasingly my interests seem to be drawing me into this mysterious world. My first project I thought I was jumping in at the deep end with was building a fuel injection ecu from a kit. It was a success but really all it proved was I could use a DMM and stick hundreds of components on to a board. I went on to build a chip amp and to repair my mates tv thanks to the university of youtube. All successful but it is really just a case of monkey see monkey do. I don't really understand it. One of my next projects will be to repair my AV amp. I have lost the front left speaker output signal for some reason. I have an old crt oscilloscope and a usb scope that i play around with and it is the scopes that i am trying to get to grips with at the moment. And so the learning curve continues

Ok thats for the background, now the question:

I have a couple of 20:1 hantek ht201 attenuators for use with my scope when looking at signals from vehicles, but I would like them to be 10:1. Tearing one down I can see a small board with some caps and a couple of resistors. If i can post a pic I will. Can I make this into a 10:1 attenuator simply by removing the resistors and replacing them with resistors half their respective values.

[Kleinstein]

To change to approximately 1:10 one would have to change the resistor to twice the value and the capacitor to half the value. The resistor part is reasonably easy, but the capacitor part could be tricky finding the right value.

[w2aew]

Just a silly question, why not simply use a 10x probe?

[Cactus]

Thanks so much for the reply Kleinstein and so soon too!  I guess to find the value I will have to take the caps off the board and measure them? Is that right or are there some pitfalls to this idea?

I'm going to say straight away I probably have a flawed understanding of how this circuit works. What I think happens is the resistors control the voltage out ie 20v in = 1v out, and the caps I presumed were there to smooth things out. Which is why I am finding doubling the resistance counter intuitive.  If I double the resistance would it not be the case that 20v in becomes 0.5v out. I guess i'm confused. 

Forgot to say I also have the book electronics for dummies which I feel I may have to read over again 

w2aew - not silly at all, and to answer your question: I sometimes connect to car engine sensors using 3m long bnc to 4mm banana jack leads connected to back probes or at  push insulation piercing probes. I could use a normal 10x scope probe with a bnc to banana converter but it all just seems to get a bit clunky. An attenuator at the scope would solve this.

Some may be wondering why don't I just buy one? Well I wouldn't learn anything, it seems wasteful if it can be modified for less expense, and I find this sort of stuff fun to do. The learning bit is the most important though.

[Cactus]

Just checking your channel out w2aew in particular #9 - great stuff thanks!

[PA0PBZ]

I'm having a hard time to understand how this works. I see what I guess is a 500 Ohm resistor in the signal path, followed by a '5602' resistor to ground. Assuming DC and ignoring the caps I can't see why this would do 20:1. What is the value of this '5602' resistor to make it work?

[Cactus]

The resistor values are 1M ohm and 56k ohm.  Do you mean what value should the 5602 be replaced with?

[PA0PBZ]

The resistor values are 1M ohm and 56k ohm.  Do you mean what value should the 5602 be replaced with?

No, I wrongly assumed that the '105' resistor was upside down and should read '501' so 500 Ohm and could not understand how it would do 20:1 with a 56K resistor. I guess I am more used to 50 Ohm attenuators and because of that never thought of a 1M resistor. Thanks for the wakeup call

[Zero999]

The resistor values are 1M ohm and 56k ohm.  Do you mean what value should the 5602 be replaced with?

Yes, it's 1M and 56k.

The 56k will be in parallel with the oscilloscope's internal resistance of 1M, making the lower resistor on the potential divider 53k03 so the output signal will be attenuated by 1/1053.03*53.03 = 0.0503.

The capacitor is there to form a potential divider with the cable's capacitance. One important thing to bear in mind is that extending the cable will reduce the bandwidth.