ESR homebrew revisit ? good design ?

Thanks to all for the ESR meter advice.

I had settled on building this *free* ESR meter schemat ...

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however something troubled me about the schematic

Almost everyone alluded to ESR meters using some pulse/wave/function passing through CAP to determine the ESR.

and this plan does not appear to have a pulse/wave/function being fed through the test CAP or am i missing something ?

so is this ESR meter really an ok->good ESR plan ?

If it is not obvious i am looking for cheap (free) ESR schematic/plans to build

thanks for any helpful replies, robb

Reply to
robb
Loading thread data ...

IC1-B is a "square wave" oscillator. Since there are no values on the drawing, no way to determine the intended frequency and duty cycle of the oscillator.

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in the 
address)

"In theory, there isn\'t any difference between theory and practice.  In 
practice, there is."  - Yogi Berra
Reply to
DaveM

being

on the drawing,

the oscillator.

Thanks DaveM,

Ah ..... of course the omnipotent (well versatile, as i am learning) opamp. I knew i must be missing something as someone said this schemat is not too bad.

i only made it about three pages into the datasheet originally but when i read your reply i went to page 19 there was an exmple of using as an oscillator.

thanks Dave for the help, robb

Reply to
robb

IC1-B is a square wave generator due to the hysteresis design with the feed back to the (+) input. C3 makes up for the timing frequency used.

The circuit is using the raising edge of the square wave as the time domain for acquiring readings. Initial raise will only be seen in the IC1-D circuit due to the small size of C4. R7 will discharge it quickly.

In the bridge it self, if the Test cap has some (R) in it, R9,R9 will not be at abs 0 volts. The offset is quickly reproduced and amplified via the IC1-C. Because this is the starting of the square wave from the generator, C4 will see a sharp raise in the signal and thus C4 being small like it is, will produce a nice short pulse of the amplitude that translates to an ohm reading via the IC1-D. After this initial starting pulse, there will be a continuous reference as a ramp how ever, this will have little effect. A perfect cap with 0 ESR, should never allow the sharp raise of the generate be seen at C4. All you should get is a 0 to 4.5 volt ramp at best. But in a cap that has some ESR, the starting ramp will not be at

0(R9+R10) and this is where the IC1-C will amplify it greatly at the same time the initial raising of the square wave. DS3 anc C6 will simply hold a reference for you, between cycles.

I know that is long winded but I think at times it's needed.

--
"I\'d rather have a bottle in front of me than a frontal lobotomy"

"Daily Thought:

  SOME PEOPLE ARE LIKE SLINKIES. NOT REALLY GOOD FOR ANYTHING BUT
  THEY BRING A SMILE TO YOUR FACE WHEN PUSHED DOWN THE STAIRS.
http://webpages.charter.net/jamie_5"
Reply to
Jamie

Assuming I did the math right, IC1-b is producing ~ 95 kHz into the bridge via TR1 & 2, which is about what is claimed (100kHz) in the write up.

You can go here to see how the op-amp is configured and the frequency computed:

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The circuit works well, and is real easy on the wallet.

Ed

Reply to
ehsjr

being

formatting link

Tanks Ed, I am amateur occasional electronics hobbyist trying to learn more.

i expect this to be quite easy on my wallet as i plan to use all salvaged components,(because it makes it more interesting for me) i have found all the components i need on a modem/hard drive controller /video card boards now for the even more fun part....

thanks for help, robb

Reply to
robb

How will you check the caopacitors then ? :->

Thanks to everyone for the interesting thread. I think i'll have a look a building one just for fun.

Cheers

--
Mitch

terminal_crazy@sand-hill.freeserve.co.uk
www.sand-hill.freeserve.co.uk/terminal_crazy
Reply to
Terminal Crazy

I don't like it, that meter is more wishful thinking than anything else. We had a meter design contest going here, but it was ruined by the non-producers, rejects, delusional narcissists, and trouble-makers, so you're not going to get anything useful from this newsgroup.

Reply to
Fred Bloggs

That's right, Fred, and we're still waiting for your design to be submitted. You can either submit it, or continue to be a non-producer. You contributed nothing, yet you damned everyone else's design ideas until you killed it off.

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

Hilarious. Here I sit, piggybacking on some local signal with my laptop, in the back room of a coffee shop in Brooklyn, surrounded by middle age men speaking Italian and playing cards. I was almost tempted to ask one of them to translate the words above the circuit diagram... almost.

Reply to
gearhead

I know next to nothing about this, so I don't have an answer. I do have a very simple question. At 100 kHz, wouldn't say a .1 uF cap have high enough reactance to swamp the ESR? Seems like you'd need megahertz, maybe even hundreds of megahertz, to be sure the resistive component of a small ceramic cap's impedance would exceed its capacitive component by order(s) of magnitude. But maybe then ESL would interfere... I'm just a babe in the woods guys, don't toast me.

Reply to
gearhead

when I was at semco (capacitor manufacturer), the lab instruments tested

1 pf using only 1 Mhz.
--
"I'd rather have a bottle in front of me than a frontal lobotomy"

http://webpages.charter.net/jamie_5"
Reply to
Jamie

On a sunny day (Tue, 25 Dec 2007 10:19:46 -0800 (PST)) it happened gearhead wrote in :

It is not the frequency, it is the rise time of the square wave. If the wave rises in 1 femto second, the frequency can still be 1 Hz only. For a capacitor, to charge in a short time the formula Q = C x U = i x t shows that the voltage across the cap is U = (i x t) / C For normal values of 'i' and very short 't', U will be close to zero.

A typical example:

1 uF, 1 us time, 1 mA, then after 1 uS the voltage across the cap will be: U = 1 mA x 1 uS / 1 uF = 1 mV U = .001 x .000001 / .000001 = .001 V

For a circuit like this:

Uin ----- C ------- out | R1 | ///

For an input voltage change like this: ________________________________________ | | | | _|

The voltage across the resistor will look like this:

peak = Uin |\ | \ | \ | \ _| \_________________

If the capacitor has internal loss in th2 form of a resistance for example, then there is a voltage divider, and the circuit looks like this:

U in ---R2 - C ------- out | R1 | ///

The peak output is now no longer Uin, but R1 / (R1 + R2) x Uin.

peak = Uin x R1 / (R1 + R2) |\ | \ | \ _| \_________________

Reply to
Jan Panteltje

d
,

Okay, the OP's circuit uses a voltage divider like you described (actually some kind of bridge arrangement) and a peak hold. It makes sense that you need a fast-rising edge, to keep delta V on the cap low enough to maintain a linear relationship to C. But you need so much speed to do this, that I wonder if the cap's inductance will affect the measurement. If so, perhaps one could design a circuit that uses a longer charging time, factoring in the exponential behavior of a capacitor when it charges; do the whole thing much, much slower, pretty much eliminating any inductive effects and reducing errors due to the slowness of the op amps. I guess the circuit specifies fet op amps for their speed, but still, are they fast enough for this. Just glancing at the TL084 on digikey, it looks like they have slew rates in the neighborhood of 15 V/uS...

Reply to
gearhead

It works. It's not wishful thinking. Whether you, or I, or anyone likes it or not is irrelevant. A better idea, or proof that it does not work would be relevant.

Ed

Reply to
ehsjr

On a sunny day (Tue, 25 Dec 2007 12:55:26 -0800 (PST)) it happened gearhead wrote in :

You mean to R?

If the cap is inductive is is also defective :-) Some rolled foil caps, where the contact on the side of the foil gets lose, can make inductors. The circuit detects series 'impedance' in a sense. In a good capacitor series impedance is close to zero.

Reply to
Jan Panteltje

d

hea=3D

ly.=3D

be:=3D

ple=3D

,

All right, let me see if I get this. For the purpose of understanding the circuit, the real capacitor under test is modeled as a resistor in series with an ideal capacitor. By design, the circuit measures series resistance Rs by using it to unbalance a bridge. For this to work, the circuit has to see Rs as having one end connected to the bridge and the other end connected to ground. Thus series capacitance Cs has to be invisible to the circuit. It has to look like a short. The circuit demands the capacitor look like a short by by having the pulse so fast, and the charge taken on by the capacitor so insignificant, that there is effectively no change in the capacitor's voltage. Otherwise, it corrupts the measurement. That's what I meant about the need for the "fast-rising edge." Is the circuit so fast that picofarad caps won't take on some small voltage?

  • By "cap's inductance" I meant _parasitic_ inductance, and the inductance in the cap itself may only be a small part of the total parasitic inductance of leads, binding posts, wires connecting them to the circuit board, and traces. Or even jumpers and alligator clips. This is after all a hobbyist project.
Reply to
gearhead

Hi, Robb -

I used SPICE to simulate the important parts of the circuit composed of the bridge and IC1-C and IC1-D. I used a pulse generator rather than IC1-B. I used voltage-controlled-voltage-sources rather than op-amp models. I looked up an arbitrary capacitor's ESR for this simulation.

I got the following output "indications" vs capacitor (arbitrarily 100uF) ESR:

ESR (ohms) Indication (percent)

0 100 .018 100 (specified value) .18 99 1.8 92 18 52 180 7

A 10uF with 0 ESR also indicated 100%.

So, the circuit appears to work, but it depends on what you are expecting from it as to how well it works. If you decide that anything below, say, 50% is bad, you may not catch the capacitor that causes equipment to fail when it has more than 1.8 to 18 ohms of ESR.

I think the amount of ESR that can be tolerated will depend on the target equipment's sensitivity to ESR. In some cases, you may find that even 1.8 ohms causes malfunction. In other cases you may find that 1K ohms of ESR causes no problems.

To verify the simulation, you could build the meter and test various capacitors known to be new and good. Then add some series resistance to each and record the readings. Then all you would need is experience to know when equipment will fail due to high ESR.

Cheers, John

Reply to
John

On a sunny day (Tue, 25 Dec 2007 22:11:21 -0800 (PST)) it happened gearhead wrote in :

Right, and it uses a peak detector approach, so the slowly decaying waveform part is not used.

OK, after posting the previous reply, it occured to me I could have misunderstood the remark about inductive. Indeed, if you used an almost infinite fast rising edge, any inductance, even the capacitor leads, would be seen, and a cap could be rated bad, while it actually was not. In this circuit however, the pulse comes from the output of an opamp oscillator, and those have reasonably slow rise times, micro seconds rather then femto seconds, and so we do not have to worry about very small inductances.

Indeed there is a useful range of capacitor values where it will work. I think the ESR meter, in this case, is mainly used for electrolytic capacitors, so maybe from 1uF up?

Yes, I think we agree here.

Reply to
Jan Panteltje

Funny, I did the same thing for the Ludens-type meter. It seemed it could benefit from a better opamp in order to work at

100KHz. So I took an TLC272. Also the detector diodes were replaced by 1n5818's. Shunt resistors were lowered from 10 to 4.7ohm to improve headroom a bit in the low-ESR range. Simulation then gives: scale ESR 100 0 97.74 0.1 95.59 0.2 89.56 0.5 80.93 1 67.55 2 44.18 5 26.82 10 13.74 20 4.21 50 1.34 100

Although this is not a binary group, I took the liberty to append a small image of what this would do for a meter scale.

Joop

Reply to
Joop

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