Attenuator.

[Mark Hennessy]

Thanks Mark. Think we may have got our wires crossed?  Anyhoo, didn't think about the IP Z of any test gear. I was more concerned about the attenuator which sits in parallel with the dummy load, changing the value of the load.

Hi Andy,

No crossed wires - I understood that you were primarily concerned about loading down the output of the amplifier (see my penultimate paragraph), and that is why you initially chose such high value resistors, but I was trying to explain that that wasn't the issue - rather, the loading effect at the output of the attenuator would be the problem

In the end I built a 100:1/ 10:1 stereo switchable attenuator using a 68k and 22k both 2% to give 90k and a 22k preset to get 99k and a 1k 10 turn pot. I used what I had, I had no 100 ohm pots. A SPDT SW switches between 10 and 100:1.

These values sound quite reasonable. My suggestion was only an example to give us some numbers to discuss.

"I'd build a simple 100:1 attenuator based on the lowest value resistors I can get away with. The lower you use, the less "loading effect" you'll get." I'm confused about this, surely a higher R or Z "load" is better, hence why scope and DMM have 1M or 10M Z. So if our test gear has a 10k Z and we use a 100 ohm R in parallel then any signal will reduce by 100 times? If we have 10k the signal will be halved, no? I'll admitt I'm struggling to visualise this.

I've knocked up a diagram that should explain this. In essence, there are two interfaces where Loading Effect can take place, and we have to consider them separately (remember that I was focussing on the interface between the output of the attenuator and the input of the test gear).

Basically, it shows that we can model a real amplifier output stage using a "perfect" voltage source (with zero output impedance) and a series resistance. And an input stage can be modelled as a "perfect" amplifier (with infinite input impedance) and a parallel resistance. Simplifications, but very useful.

   

As regards HF losses though a lower Z is better. With 10p of capacitance in a cable say inc stray C, there'd be 3dB loss at 159khz with a 100 ohm Z in parallel, I hope I have that right. But as were measuring audio frequencies, who cares.

Yes indeed. That's the point I was making - keep the impedances low, and stray C is not an issue. But it would have been with your MΩ resistances.

As a conservative rule of thumb, figure on about 100pF per metre for coax cable. It might be less in practice - easy enough to measure it needed.

BTW, I measured the DC res of my soundcard and it's 10k, so the Z will be a bit less than this

Not necessarily. But it might be.

Measuring AC impedance of a simple thing like this is quite easy - would you like me to talk you through it? Hint: you only need a resistor (or variable resistor) and some basic Ohm's law...



In summary, while we've discussed loading effect before, it's worth reiterating that whenever you connect anything to anything, there is some loading effect because the device generating the signal will have an output impedance, and the device receiving the signal will have an input impedance.

This change in signal level varies from negligible to disastrous, and includes values that are necessary and expected (such as 6dB, the loss you get when you have to match impedances).

Usually, when making measurements you want the loading effect to be as small as possible, as you don't want the measurement itself to change the conditions you're trying to observe. I know that much is stating the obvious, of course - we all know that a DMM has a high input impedance (10M or more), and that an AVO could cause problems in some situations.

Continually assessing the possible impact of loading effect is something that is just second-nature to experienced engineers. But I know from my day job that many graduate engineers don't arrive with this ability - it's something that takes a bit of time to cement. So don't feel bad if you get confused! Tomorrow in class I'll be explaining all of this from first principles (and there will be homework and a test!), and I know that while some will get it immediately, others will struggle to understand and apply it.

I hope all this helps,

Mark