Resistive Load for Amp Testing

Sonnie

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Here's another one with 3 tubes... 6000w/220v...

41890
 

linearphase

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Looks like that design has there different elements with one side of each element shorted together. You could remove the shorting bar and just connect to each element separately then use the rules I gave you above. Using 6000 W/ 220 Volts and assuming those figures are for the total wattage of the elements should be V^2/W= 48400/6000=8.07 ohms. So each element appears to be 24.2 ohms. If you short the unshorted terminals and then connect between the two shorted pairs you should have 8.07 ohms.
I would always test the impedance with an ohmmeter first. Also verify the residual lead impedance of the meter by first shorting the two test leads together. Clean the leads first then wiggle them to get a good reading. If your meter has a zero button use it. Else just subtract the residual reading from the measured value.
 
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linearphase

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This one would be easier to connect to.
note I have no experience with using water heater element as a dummy load. If they have a nichrome wire inside and no windings in it is should be OK. I there are turns they will have a least some inductance so the impedance would rise with frequency. You would have to use an impedance meter to measure this.
 

Head_Unit

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- Yeah @linearphase is right, those devices should really be checked to see if they are truly resistive over whatever frequencies are being tested.
- I agreee with @RichB it would be awesome to do reactive testing. Yeah the size of the elements can become problematic. I miss the Power Cube! Browbeat the company where I used to work to get one and it gave the engineers a lot of insight. My dream was to set up something to do capacitive/resistive/inductive load at maybe 70 Hz (few harmonics in common with 60 Hz hum) but events conspired against that.
 

RichB

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- Yeah @linearphase is right, those devices should really be checked to see if they are truly resistive over whatever frequencies are being tested.
- I agreee with @RichB it would be awesome to do reactive testing. Yeah the size of the elements can become problematic. I miss the Power Cube! Browbeat the company where I used to work to get one and it gave the engineers a lot of insight. My dream was to set up something to do capacitive/resistive/inductive load at maybe 70 Hz (few harmonics in common with 60 Hz hum) but events conspired against that.

I liked the power cube and discussed it with Gene (Audioholics) and his only criticism was the lack of distortion measurements, which I think is fair.

Here is an article from Audioholics on amplifier measurements:

Audioholics Amplifier Measurement Standard | Audioholics

- Rich
 

Sonnie

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Doesn't make sense for me to spend 15K on the PowerCube... no reason to look at or even consider it at this point... ain't gonna happen.
 

Sonnie

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I ordered one of the 6000w/220v units from Amazon.

I think this will be my best bet... 8.07 ohms to 4.03 ohms to 2.02 ohms should be close enough.

How much difference will .07 ohms make in measurement watts? Looks like maybe .0008 to .0009 ... so on a 100 watt amp, it would be 99.92 watts instead of 100? Not sure if I'm figuring that correctly or not, but it would be even less on 4 ohms and 2 ohms.

Of course when we start talking 6000 watts... obviously there wouldn't be any worries with the larger powered amps. Testing my Behringer amps that are collecting dust would be super easy to do.
 

Sonnie

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So the first one I received tested 8.7 ohms... not 8.07. I will order a few more and see if they all turn up being 8.7 or if they vary. I may have to simply keep ordering until I find enough that test consistently.
 

Sonnie

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I went to Home Depot tonight... found a combo of elements that should work. I can get to 8 ohms with 2x 3800w/240v on some of them. I just need to order more until I get the right ones... there are several that are dead on 16 ohms... a few at 16.1, 16.2... and in parallel I can get to 8 and 8.1 ohms. paralleling those with a 2000w/120v at 7.8 ohms gives me 4 ohms. So getting to 4 ohms is not an issue. I'll just have to get the right combos to get where I need to be for each channel, but it is definitely doable.
 

Sonnie

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Here's a list of common water heater elements I came up with when I was researching the dummy load.

Water heater elements ordered by resistance

(Note: tempco of Nichrome resistance is positive, at roughly
.04%/degree C. So the wire temperature would have to increase
significantly to change more than a couple percent.)

3500W 240V nom. 16.46 ohms
2 in parallel should be very close to 8 ohms
VERY common and cheap ($6-10 each): high watt density screw-in -
Camco #02282/02283
Utilitech #362342
Everbilt #15010
Eastman #60122n
Rheem/Ruud #SP10552KH, UV12896 ($6.16 each direct from Ruud!)
Less common, more expensive: Medium/low watt density foldback style -
Camco #02523, #02912/02913
Rheem/Ruud #AP12903, SP10869KL, UV12903

3800W 240V nom. 15.16 ohms ($10 at HD, qty discounts)

4500W 240V nom. 12.8 ohms (very common, $10-11 @ Lowe's, $12 @ HD)

5500W 240V nom. 10.47 ohms ($12 @ HD, $15.48 @ Lowe's) -
4 Lowe's (Utilitech brand) units measured ~9.9 - 10.2 @ room temp

1440W 120V nom. 10 ohms ($12 at HD)

1500W 120V nom. 9.6 ohms ($7.28 @ Lowe's, $9-10 at HD, qty discounts)

6000W 240V nom. 9.6 ohms ($33 @ HD, Camco #02612/02613 $12.11 @ Amazon)

6500W 240V nom. 8.86 ohms (Dernord brand $30 @ Amazon)

1650W 120V nom. 8.73 ohms (e.g. Grainger #2E754, Camco #02482) $17 on eBay

5000W 208V nom. 8.65 ohms (Camco #05652, Ruud #AP10867NL-5)

1750W 120V nom. 8.23 ohms (Camco #02182/02183) ($13 @ Amazon)

6000W 220V nom. 8.07 ohms (Aliexpress, $13-18)

5500W 208V nom. 7.87 ohms (e.g. Camco #02862) $26 + ship @ eBay

6000W 208V nom. 7.21 ohms (Camco #05692)

2000W 120V nom. 7.2 ohms ($10 at HD, qty discounts)

9KW 240V element nominally 6.4 ohms ($41 at HD, for tankless)

3500W 120V nom. 4.11 ohms ($34 at HD, for tankless)
Do you have an ohm meter (or multimeter) that measures .xx (two decimal points) for resistance?

Trying to find one that doesn't break the bank.
 

Sonnie

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Double checked ... I can only get to 8.1 ohms. May try to combine a larger wattage with the 3800 and see what I can get.

I was able to get the 4 ohm load by connecting a 1750 watt with a 2000 watt. Connecting those two 4 ohms yielded 2.1 ohm. I may still be able to get to 2 ohm, as I have a couple of other elements that are .1 ohm less that I can probably replace and make it work out.
 

Sonnie

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Here is what we have thus far... and it looks like what we will go with. While not dead on 8 and 4 ohms... even the other type resistors have a tolerance of +/- 1.5%... which means they could vary from 7.88 to 8.12 ohms. I'm not sure there are any speakers with a ruler flat 4.00 or 8.00 ohm impedance. Consistency of measurements are what I believe are more important.

42354


42368
 
Last edited:

hibikijin

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I am not sure regarding the wiring you mean, let me try. Following the Ohms law you could use the patch cables to combine the 8 ohm individual resistors, either in parallel, in serie or both, to the desired resistance (I use resistance, since it will not be an inductive load to use impedance, as in the case of a loudspeaker) 2, 4 ,8 and maximum wattage. When in parallel the maximum power/wattage supported will be the power rating of the individual resistor, in series it will be doubled, trippled etc depending on the number of resistors in parallel.

Two 8 ohm resistors in parallel = a 4 ohm net; four 8 ohm in parallel = 2 ohm
Two 8 ohm resistors in serie = 16 ohm net;
so combining 2 of these 16 ohm net in parallel, i.e., four 8 ohm resistors, provides an 8 ohm net with double (if I am not wrong) max wattage of each individual 8 ohm resistor. The dissipated power (watts) is given by P (power dissipated, in watts) = V ˆ 2 (voltage squared) ÷ R (resistance of the combined resistors), and the voltage will depend on the volume.. e.g. for a 300 W output on a 8 ohm resistive load, I measured around 49V on my amp.

I do no quite understand the reason for the 120 V 240 V references for audio testing, it is only a reference the maximum wattage the resistor will be capable at these Voltages. Well, a 10000 W amp over 8 ohm will have around 282 V at its output (be carefull when handling these loads, this voltage may hurt or kill), and be aware of the cable specs, the current will be around 35 Amps (just for comparison, the current over this water heater, 3800 W at 240V, is around 15A

But it has been quite a long time I used those formulas.
 
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DonH57

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Late to this, sorry...

Resistor networks:
  • Resistors in series add in value, so Rtotal = (R1 + R2 + R3 + ... + Rn). If all resistors are of the same value, then Rtotal = n * R.
  • Resistors in parallel combine as Rtotal = 1 / (1/R1 + 1/R2 + ... + 1/Rn). If all resistors are the same value, then Rtotal = R/n.
  • For simple combinations you can use those two equations to figure out networks of series/parallel combinations. For example, for 8-ohm resistors, one in series with two in parallel = :8 + 8/2 = 12 ohms.
    • Two 8-ohm Rs in series is 16 ohms.
    • Two 8-ohm resistors in parallel is 4 ohms.
    • Four 8-ohm resistors in parallel is 2 ohms.
Output power:
  • P = power, V = voltage, I = current, R = resistance; then,
  • P = V * I = V^2 / R = I^2 * R
  • Manipulate the equation to get the parameter you wish:
    • 28.28 Vrms into 8 ohms is P = 99.970 W ~ 100 W (average power; there is no such thing as RMS power, that is a marketing mistake)
    • 28.28 Vrms into 8.07 ohms is P = 99.103 W, a 0.875% difference (you decide if you care about a <1% error)
Dummy loads:
  • Transformer oil is typically used for dummy loads. It is non-flammable, or at least less flammable, and less reactive with electrical components. Modern loads use a manufactured "oil" or silicon-type liquid.
  • Motor oil is highly flammable and has a lot of impurities.
  • Water is non-flammable but more chemically reactive so I would not immerse standard load resistors in water. Heating coils or water heater resistors are encased in a a sealed tube to prevent water from entering.
  • A good load includes a pressure release for safety in case you end up boiling the oil or whatever. Having a multi-kW dummy load explode is rather unnerving and dangerous, Trust me on this.
  • Resistors have a temperature coefficient that can be large so you may find resistance changes with temperature, including temperature changes due to applied power causing the load resistor to heat up. The oil helps maintain even heating among load resistors and provides a way to help dissipate the heat. For ham radio testing I would blow a fan on my little 1 kW gallon-can load. The big boys, 10 ~ 100 kW, are usually water-cooled through a circulating plumbing system.
HTH - Don
 

Sonnie

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Thanks for the additional information guys.

I am not sure regarding the wiring you mean, let me try. Following the Ohms law you could use the patch cables to combine the 8 ohm individual resistors, either in parallel, in serie or both, to the desired resistance (I use resistance, since it will not be an inductive load to use impedance, as in the case of a loudspeaker) 2, 4 ,8 and maximum wattage. When in parallel the maximum power/wattage supported will be the power rating of the individual resistor, in series it will be doubled, trippled etc depending on the number of resistors in parallel.

Two 8 ohm resistors in parallel = a 4 ohm net; four 8 ohm in parallel = 2 ohm
Two 8 ohm resistors in serie = 16 ohm net;
so combining 2 of these 16 ohm net in parallel, i.e., four 8 ohm resistors, provides an 8 ohm net with double (if I am not wrong) max wattage of each individual 8 ohm resistor. The dissipated power (watts) is given by P (power dissipated, in watts) = V ˆ 2 (voltage squared) ÷ R (resistance of the combined resistors), and the voltage will depend on the volume.. e.g. for a 300 W output on a 8 ohm resistive load, I measured around 49V on my amp.

I do no quite understand the reason for the 120 V 240 V references for audio testing, it is only a reference the maximum wattage the resistor will be capable at these Voltages. Well, a 10000 W amp over 8 ohm will have around 282 V at its output (be carefull when handling these loads, this voltage may hurt or kill), and be aware of the cable specs, the current will be around 35 Amps (just for comparison, the current over this water heater, 3800 W at 240V, is around 15A

But it has been quite a long time I used those formulas.
The voltage is shown to calculate the resistive load. For example... 120 volts and 1750 watts equates to 8.23 ohms, but they might test 8.24, 8.36, 8.19, etc. Then 120 volts and 2000 watts equates to 7.2 ohms, although they might test 7.54, 7.68, 7.72, etc.. they are not consistent and will all vary slightly, which is why I had to test so many to get as close as possible to the desired load.

I used the 240 volt / 3800 watt elements because they all tested from 15.9 to 16.1... and it was the closest I could get to an 8-ohm load by way of wiring them in parallel.

I included this info to simply show the specs of what I purchased, even though as stated above, specs were not as important as the actual measurements in this case.

These are two 8-ohm resistors wired in parallel to yield the 3.99 ohms... one is 1750 watts and the other is 2000 watts... so what is my final power wattage? I was thinking that because they are combined, it would be 3750 watts max. NOT that I need that much, as most of the amps I'll be testing will be well below 1750 amps.

42527


Dummy loads:
  • Transformer oil is typically used for dummy loads. It is non-flammable, or at least less flammable, and less reactive with electrical components. Modern loads use a manufactured "oil" or silicon-type liquid.
  • Motor oil is highly flammable and has a lot of impurities.
  • Water is non-flammable but more chemically reactive so I would not immerse standard load resistors in water. Heating coils or water heater resistors are encased in a a sealed tube to prevent water from entering.
  • A good load includes a pressure release for safety in case you end up boiling the oil or whatever. Having a multi-kW dummy load explode is rather unnerving and dangerous, Trust me on this.
  • Resistors have a temperature coefficient that can be large so you may find resistance changes with temperature, including temperature changes due to applied power causing the load resistor to heat up. The oil helps maintain even heating among load resistors and provides a way to help dissipate the heat. For ham radio testing I would blow a fan on my little 1 kW gallon-can load. The big boys, 10 ~ 100 kW, are usually water-cooled through a circulating plumbing system.
HTH - Don
Discussing it among several others, we thought it might be best to use RV coolant, which has a boiling point of 375 degrees... and is supposedly non-flammable.

From a fellow engineer...

Kinda-sorta Thermodynamics 101

First, lets calc the amount of heat needed to heat 46.45liters (12.27 gallons) of 50/50 mix of Poly Glycol/Water from 20 degC to 188 degC (370 dedF). (These are mostly assumed numbers, punch whatever specifics you have into the equations and calculate).

So we’re taking your specified fluid volume, assuming a 50/50 mix of poly/H2O and jacking the temperature of the fluid from ~ room temp (20 degC) to boiling point (188 degC / 370 degF).

cp = 3.473 kJ/(kg degC)

Q = 3.473 kJ/(kg degC) * (1079 kg/m^3 * 0.04645 m^3) * (188 degC - 20degC)
Q = 3.473 kJ/(kg degC) * (50.11955 kg) * (168)
Q = 29242.95 kJ = 8.123 kilowatt - hours

Now convert (2 * 2kW in) * 15 sec:
4kW for 15 sec = 0.0167 kWH
0.0167 kWH << 8.123 kWH

So, no need to worry IF you stick to 4kW in/15sec.
In my case... these should never even get warm.

Another guy testing car audio amps up to 10,000 watts has several variations of the tank I am building and uses mineral oil. The has a thermometer in one of the tanks (you can see it in the pics I posted previously) and when I talked with him on the phone, he does more than 15 second max power testing and his tank has never gotten even remotely hot.

None the less, the cap on my tank will be vented. :T
 

hibikijin

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The dissipated power depends on the voltage put out by the amp, and that is dependent on the volume. The P (in Watts) = V**2 (in Volts, voltage squared) / R (in ohms). The watts (max) spec for a resistor may have some lattitude depending on the heat dissipation, in your case a lot.

Once I used a 8 ohm, 2%, 100 W rated resistor to test a 300 W power amp. It was mounted on a heavy metal dissipator, using termal paste, the Voltage on max volume reached 49 V (aprox. 300.125 W). It was hot as hell, melted the cable isolation but it, well, "resisted". Bear in mind that with higher temperatures most resistors will change its nominal resistance.
 

Sonnie

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Yeah... I'll be using 9 and 10 gauge wire for the connections... and shouldn't ever really get these even warm, per the equation quoted above.

I'm thinking that wiring the elements in parallel will combine the wattage.

Connected to one element:

42528


Connected to both elements in parallel (resistance halved):

42529
 

hibikijin

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"I'm thinking that wiring the elements in parallel will combine the wattage."

Yes, I see what you mean.

The total dissipated power will be doubled (for a 2 resistor net) because the resistance will be halved, the total current doubled for the same amp output voltage but each individual resistor will dissipate half of it, that is, the same as for a standalone 8 ohm resistor.
 

Sonnie

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I did verify thru a couple of electrical engineers that whether parallel or series wired, it will use the combined power of all elements connected.

I decided to see what the variance/tolerance is on some of the dummy load resistors out there, so I ordered a few from Parts Express and Amazon... very common brands that are offered thru several sources.

Below are the results which show that rarely are any going to be dead on 8.00 ohms or 4.00 ohm... and I believe I will ultimately have about as close as I'm going to be able to get... as close or closer than any of these, (which it seems is what most are using for amp testing). The small variances I will experience are very negligible on the power ratings.

200 watts / 8 ohms each = 8.08 and 8.02
20210610_121046.jpg 20210610_121110.jpg

300 watts / 8 ohms each = 8.04 and 8.05 paralleled to 4.07
20210610_121617.jpg 20210610_121632.jpg 20210610_121728.jpg

300 watts / 4 ohms each = 4.04 and 3.97 paralleled to 2.04
20210610_122240.jpg 20210610_122301.jpg 20210610_122331.jpg

200 watts / 8 ohms each = 8.04 and 8.04 paralleled to 4.06
20210610_115238.jpg 20210610_115308.jpg 20210610_115344.jpg

200 watts / 4 ohms each = 4.06 and 4.07 paralleled to 2.07
20210610_120219.jpg 20210610_120256.jpg 20210610_120323.jpg
 

DonH57

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@Sonnie : I cannot imagine anyone is going to complain about a small difference.

As for power, where P = power, I = current, V = voltage, and R = resistance, you can calculate the dissipation per resistor with a little bit of Ohm's Law.

V = I * R, V / I = R, V / R = I
P = V*I = V^2 / R = I^2 * R

In series, the same current flows through all resistors, and the voltage splits among resistors. For power, since you know the total resistance and applied voltage, then Vtotal / Rtotal = I and for each resistor (R) then P = I^2 * R.

In parallel, the voltage across each resistor is the same, but the current splits in accordance with each resistance. So a little math gets you the power for each resistor as P = Vapplied^2 / R.

HTH - Don
 
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Sonnie

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@Sonnie : I cannot imagine anyone is going to complain about a small difference.

As for power, where P = power, I = current, V = voltage, and R = resistance, you can calculate the dissipation per resistor with a little bit of Ohm's Law.

V = I * R, V / I = R, V / R = I
P = V*I = V^2 / R = I^2 * R

In series, the same current flows through all resistors, and the voltage splits among resistors. For power, since you know the total resistance and applied voltage, then Vtotal / Rtotal = I and for each resistor (R) then P = I^2 / R.

In parallel, the voltage across each resistor is the same, but the current splits in accordance with each resistance. So a little math gets you the power for each resistor as P = Vapplied^2 / R.

HTH - Don
Unfortunately I'm not much of a mathmagician :dumbcrazy: ...instead I mostly rely on those that can work the formulas.

Not sure what happened over the last 41 years since I graduated high school (imagine that), but I thought I was pretty good at Algebra and Calculus back in the day. Obviously not enough practice since then and too much brain strain to try to figure it out these days.

From my good friend Mark at thecrossovershop.com ...
Resistor Fun Facts:

1: With resistors in parallel, each resistor has the full voltage of the source applied to it.

2: When wired in parallel, each resistor draws the same current it would draw if it alone were connected to the voltage source.

Given: Two 8 Ohm resistors, wired in parallel.

What would the total power dissipation capacity be if one resistor was rated at 1750W and the other 2000W?

== Resistor 1:

V = 120 Vac, R = 8 Ohms, P = 1750W
Therefore, i (current) is:

P = i^2 * R
1750W = i^2 * 8
i = 14.79 Amps

== Resistor 2:

V = 120Vac, R = 8 Ohms, P = 2000W
Therefore i (current) is:

P = i^2 * R
2000 = i^2 * 8
i = 15.81 Amps

To find total W dissipated,

P = (i1 + i2)^2 * 1(1/R1 + 1/R2)
P = (14.79A + 15.81A)^2 + 1/(1/8 + 1/8)
P = (30.6A)^2 * 4

P = 3745.44 W Ans (Rounding errors produced 3745.44 W instead of the expected 3750W).

So a little bit of Joule’s law (P = I*V) and a little bit of Ohms law (V = I*R) gets us the answer we need.

Really just about conservation of Energy & conservation of Charge

And from my good friend Doug at Spectral Measurement...
 

SVChucko

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Do you have an ohm meter (or multimeter) that measures .xx (two decimal points) for resistance?

Trying to find one that doesn't break the bank.

Sorry for the belated reply, I don't check this forum often. I found this LCR meter for $152 + shipping and tax at Newark Electronics. The documentation is almost nonexistent from Newark. But it turns out to be a relabeled Hantek 1833C, and you can download specs, a more complete manual, a firmware update, and software for interfacing to a PC from Hantek's site. Don't be alarmed when it restarts with Chinese labels - that's easy to reset to English. It's a very capable LCR meter, especially for the money, and anyone doing amplifier or speaker tweaking should find plenty of uses for it.
 
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SVChucko

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@Sonnie : I cannot imagine anyone is going to complain about a small difference.

I'm sure someone, somewhere, will complain about a 1-2% difference in load resistance making a power measurement invalid. This is the Internet, after all.

My four dummy "8Ω" loads, made from paired Rheem/Ruud 3500W @ 240V water heater elements, all came in about 8.25-8.3Ω. The inductance was much lower than I had expected to see. For my purposes this was plenty close enough. I presume no one around here is naïve enough to expect an "8Ω" speaker to ever present an 8Ω resistive load to an amplifier.
 

SVChucko

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Sorry for the belated reply, I don't check this forum often. I found this LCR meter for $152 + shipping and tax at Newark Electronics. The documentation is almost nonexistent from Newark. But it turns out to be a relabeled Hantek 1833C, and you can download specs, a more complete manual, a firmware update, and software for interfacing to a PC from Hantek's site. Don't be alarmed when it restarts with Chinese labels - that's easy to reset to English. It's a very capable LCR meter, especially for the money, and anyone doing amplifier or speaker tweaking should find plenty of uses for it.

I'll add that what you don't get for that price is a set of Kelvin test leads. I didn't bother but if you are serious about making low-Z measurements, you will want them.
 
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