Quick ESR answer needed

Oh, Fred, You're just so brilliant! Swoon some more ;-)

...Jim Thompson

"Phil Allison" = " a totally ASD Fucked Stinking ARSEHOLE and Dog Fucker "

You've left the phrase "capacitor manufacturers" unqualified. Do you mean "all", "most", "some", or "a few" manufacturers? Can you offer any examples (web pages or pdf files) to support your assertion?

I think they understand full well. They buy an instrument that can measure the vector impedance of a capacitor, and they report the real part of that impedance at a given frequency, or they provide a plot of the real part vs. frequency, calling it ESR. If their instrument says that the real part of the impedance of one of their capacitors at a given frequency is some value, why shouldn't they believe it? The model (the ESR model) they are using is just an ideal capacitor in series with a single resistor, and the value of that resistor in the model is just the real part of the impedance, as measured. This value varies with frequency as I explain below.

The cap I measured with an 800 ohm ESR was .27 uF, not .1 uF. I dug out a .1 uF cap and its measured ESR at 20 Hz is 2200 ohms.

Assuming the LSR of a capacitor is what you have shown as R1 in your more complicated model (reproduced below), I would agree that no .1 uF ceramic cap will have a model value for R1 of 800 ohms (or even 2200 ohms). I've never made any such claim. But if you derive the parameters for a .1 uF cap using your model, and then do the algebra to reduce it to a two terminal impedance at 20 Hz, with a real and an imaginary part, the real part will be about 2200 ohms (if your cap has the same dielectric as mine). But whatever dielectric, the real part of the 20 Hz impedance (the ESR) will be much larger than R1.

Let me re-iterate. These high ESR's (which are not the same as the LSR) only occur at very low frequencies. I understand that your business involves very fast pulses and low impedances, but if you were to design a 20 Hz notch filter with the .1 uF capacitor I just measured, and you wanted to analyze the performance, you would have to use a 2200 ohm resistor in series with an ideal .1 uF capacitor (at 20 Hz, for example, and if the capacitor model, the ESR model, consisted of an ideal capacitor in series with only a single resistor) to account for the Q lowering losses in the capacitor. The R1 value (the LSR, probably 10 to 40 millohms) in *your* model would not be the correct value to account for the losses at 20 Hz, if the model used was just an ideal capacitor in series with a single resistor, which is the ESR model used by the manufacturers.

There is absolutely nothing wrong with the more accurate model you have given in another post:

---------+ | L1 | | R1 | +--------+----- etc, if you're | | compulsive C1 C2 | | | | | R2 | | | |

---------+--------+------

But its R1 value is not the same at low frequencies as the ESR given by the capacitor manufacturers, although it is essentially the same at high frequencies. However, you can calculate the ESR at any frequency by reducing all those parallel and series branches to a single two terminal impedance. The ESR is the real part of that result.

Suppose you derive values for L1, R1, C1, C2, R2, etc., for the .1 uF capacitor and then let the frequency be 20 Hz. Then calculate all those parallel and series branches, finally arriving at a single impedance between the external terminals; a simple textbook exercise. The impedance will have a real and an imaginary part, and that impedance would also be the impedance of an ideal capacitance of some value in series with a resistance of some value, the "equivalent" series resistance. At 20 Hz, this "equivalent" series resistance will be much larger than R1.

If you use your model with its 5 parameters directly in the analysis of the 20 Hz notch filter at 20 Hz, you will get the same result as when first reducing to a single ideal capacitor in series with an "equivalent" series resistor, if the reduction is done at 20 Hz. (You better get the same result, or there's a mistake in the algebra.) If you want the performance at another frequency, then you could use your model directly, or do the reduction to a capacitor in series with a single resistor at the new frequency, in which case the "equivalent" series resistor will have a different value, but it will give the correct result at this new frequency. So the "equivalent" series resistor will have a different value at different frequencies.

It's the very presence of C2 and R2 in your model that gives rise to an "equivalent" series resistance that varies with frequency. If C2 and R2 were absent, then the ESR would equal R1 (and LSR) at all frequencies.

And at high frequencies (fast pulses) where you like to operate, if you do this same reduction, the "equivalent" series resistance will be very close to the value of R1.

This means that the ESR at high frequencies as reported by the capacitor manufacturers is a useful number for you; it's essentially the same as the LSR for that capacitor.

Your model has the advantage that the parameters don't vary with frequency, and could be used in a spice analysis directly. The ESR model has a resistance that varies with frequency, but it's easier for the manufacturers to provide this information since all they have to do is measure the real part of the impedance, and in a lot of cases all the user needs is an approximation for the losses at a single frequency.

Yep. How many 2n7002s are you using?

I'd like one. Is there an educational discount?

Two. And a transmission-line transformer, for output isolation.

Here's a 50-volt pulse into 50 ohms:

If you really believed the 2n7002 datasheets, you wouldn't even try it.

OK, will do.

Sorry, no.

John

If those manuals include schematics I would love one too. Cheers, Harry

Ah. Humor. No, they don't.

John

Nice looking waveform.

You used two in parallel to pull up, and two more in parallel to pull down? Or one each up and down?

Followed by a 50-ohm source-termination resistor? That's 25 volts at the end of a terminated 50-ohm line -- 500mA into the output, or 1A into a short? What's the power rating of your 50-ohm resistor?

Two fets, and both pull down, and drive a 1:1 transmission-line transformer into the load; that makes the gate drive simple. Our output is low impedance, so delivers 50 volts, 1 amp, into the customer's 50 ohm load. The scope pic is really 50 volts into a 50-ohm load, namely a 40 dB attenuator at the scope sampling head.

Our front-panel outputs are true 50-ohm sources, programmable from -4 to +12 low/high open-circuit levels, but that's a fairly awful (ie, component-rich) GaAs thing.

I sure wish somebody would make a nice "pin driver" chip: programmable Vl:Vh, lots of swing, clean and fast, reasonably affordable. GBL had one, back when they were still alive, but it was like $120 a pop or something silly. I'm thinking that something could be done from a diode steering network and a couple of really fast TI opamps... gotta try that some day.

John

So, if there's a long 50-ohm coax, with an open output, you get a 100-volt pulse lasting twice the delay through the transmission line?

What's the maximum pulse length with this setup?

Yup, and the aftermath would be really ugly. But, as far as I know, it won't destroy the fets.

There's graphs in the manual, sec 5.11. The cores saturate around 200 ns or so at 50 volts. If you fire all 5 channels at 50 volts, that's

John