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Two part question! I'm building an audio dummy load featuring an 8Ω, 1%, non-inductive 300W beast of a resistor. This is connected to the speaker terminals of an amplifier under test. In addition, I'd like to scope across the resistor (using a female BNC connector to attach a scope) and have a switch which toggles a 10:1 voltage divider, to compensate for higher voltages that my scope may not be able to handle (much like the 10X setting on an oscilloscope probe). My circuit looks something like this:

Dummy load

  1. I realise using small value resistors for R1 and R2 would mean less current going through the load, so high values are preferable so as to not interfere with the measurements. What, if any, are the pros and cons of using smaller or higher value resistors here? Why not go for much higher values - say, 360kΩ and 40kΩ - for even greater measurement precision?

  2. How careful do I need to be about the power ratings of R1 and R2? Am I correct in assuming that with the values I currently have, that the ratio of current going through the parallel branches is so tiny that even puny 1/4W resistors would barely be taxed?

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    \$\begingroup\$ Note that if R1+R2 summed to 800 0hms, they would only reduce the load by 1%. (and they would then consume 1% of the 300W power. At 8kohms, that's 0.1% and 0.3W; any higher "precision" is illusory. So I'd use 18k and 2k, and make the 18k 0.5W to be conservative. \$\endgroup\$ – Brian Drummond Sep 22 '17 at 19:47
  • \$\begingroup\$ I hope that is not a bridge amplifier... \$\endgroup\$ – Trevor_G Sep 22 '17 at 20:25
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    \$\begingroup\$ I hope you know that a speaker is definitely not non-inductive. :) \$\endgroup\$ – pipe Sep 23 '17 at 12:22
  • \$\begingroup\$ Yup as @pipe mentions a resistor is not a great substitute for a speaker as a load. \$\endgroup\$ – Trevor_G Sep 23 '17 at 13:22
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The capacitance of your cable will form a RC lowpass filter with the resistors. With such high values, you should check if you still have the bandwidth you need.

Look at the pF/m cable capacitance in the cable docs. If you don't have it, pick a datasheet for a cable of similar diameter and impedance.

At audio frequencies, impedance matching won't be a problem, so most likely there is no need to take this into account.

I forgot: if the measurement device at the other end of the cable has input noise current (ie, not a scope, but say, a BJT opamp) then high value resistors will increase noise. But high value resistors also make input protection a lot simpler, since they will limit the current...

Note if you want a 1:10 attenuation, why not use your scope's 1:10 probe?

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An oscilloscope's probe has a small capacitance.

Having a super high impedance on what you are measuring will mean that the signal will get degraded because it can't source or sink enough current to charge and discharge the cap.

Also keep in mind that the resistors in the divider have their own precision rating of 1%. So using higher value resistors won't help that much.

The total resistance over the divider is 25000 times higher, so the power rating can be 25000 lower when exposed to the same voltages: 300W/25000 < 1/4W

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Good answers posted here already but I would add..

You would probably be better off using .8R and 7.2R resistors for your dummy load, if you can find/make them, and tap off there to your switch.

Or better use something active to buffer/divide the signal to minimize the effect of the scope connection on the output of the amplifier you are testing.

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  • \$\begingroup\$ Great idea on the .8Ω and 7.2Ω resistors - an elegant solution there to "tap" one tenth of the resistance. However, unfortunately I've already invested in 4 of these bad boys - phwooar. \$\endgroup\$ – abza Sep 22 '17 at 20:08
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What, if any, are the pros and cons of using smaller or higher value resistors here? Why not go for much higher values - say, 360kΩ and 40kΩ - for even greater measurement precision?

The disadvantage of using high value resistors in a voltage divider is it makes the output impedance higher and hence makes the output voltage more sensitive to loading.

Lets run some approximate numbers.

At audio frequencies we can regard a coaxial cable as a capacitor.

You voltage divider has an output impedance of just under 20K.

A scope input in hi-z mode usually has an input impedance of 1 Megohm in paralell with low 10s of picofarads.RG58 coax has a capacitance of about 100pF per meter. Lets say you have a couple of meters of coax and say 250pF total for cable, scope, PCB traces etc.

So first lets look at the near-DC case. A 20K impedance driving a 1 Megohm load gives a gain error of about 2%.

What about higher frequencies? the output impedance of your voltage divider forms a RC filter with the capacitance of the cable/scope. The break frequeny of a RC filter is

$$f_c=\frac{1}{2\pi R C}=\frac{1}{2\pi * 20 * 10^3 * 250 * 10^{-12}}=\frac{1}{\pi * 10^5 * 10^3 * 10^{-12}} = \frac{1}{\pi * 10^{-4}} \approx 3000$$

That is a bit low for audio work. I would be wanting to knock those resistor values down significantly to bring the break frequency up. OTOH we don't want to knock them down too far or we run into power dissipation problems. We probablly want resistors about 20 times smaller than your values pushing the break frequency to somewhere around 60kHz.

Unfortunately 9K resistors aren't a standard value but we can form one by connecting a 4.3K in series with a 4.7K.

Note that 10x scope probes do use a very high value resistor, however they also use compensation capacitors that are tuned to match the characteristics of the cable and scope and in some cases special cables.

How careful do I need to be about the power ratings of R1 and R2? Am I correct in assuming that with the values I currently have, that the ratio of current going through the parallel branches is so tiny that even puny 1/4W resistors would barely be taxed?

You said you bought four 300W resistors. It's not clear to me if you plan to use them to build four seperate amplifier measurement circuits with one 300W resistor each or one big measurement circuit using all four. Lets assume the latter for now.

Your divider chain has a resistance of 200K which is 25000 times the resitance of the 8 ohm dummy load. So when 1200W is running through the dummy load resistors 0.048W is running through the divider chain. No problem.

My suggested chain has a resistance of 10K which is 1250 times the resistance of the 8 ohm dummy load. So when 1200W is running through the dummy load 0.96W is running through the divider chain. Of this 0.4512W is dissipated in the 4.7K resistor and less in the other resistors.

That seems like a workable design to me. 0.6W metal film resistors are readilly available and pretty cheap.

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  • \$\begingroup\$ Great answer! Maybe you can add that the low pass behavior can be compensated with a (trimmer) capacitance in parallel to the high-R of the divider... in you scenario, something adjustable around 25 pF could do. \$\endgroup\$ – Rmano Sep 23 '17 at 13:54
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Good answers but I have this : 1. Connect two dummy resistors in series to make 16 ohms 2. The same way for the remaining two 3. Connect this series circuits in parallel to amplifier making an 8 ohms total dummy load 4. Now you can measure safely connecting the oscilloscope between the two intersections of 16 ohms serial circuits 5. Now you are using a high power resistor bridge circuit mainly used in instrumentation amps.--measuring voltage diference -- and milivolts if the tolerance of resistors is as specified.
Be aware: Oscilloscope must not be grounded!!!

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  • \$\begingroup\$ The last point is rather a showstopper since nearly all scopers are grounded. Also afaict the output of your circuit will be dependent on the tolerance of the resisters, with perfect resistors there will be no output. \$\endgroup\$ – Peter Green Sep 23 '17 at 10:26
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The values of resistors actually decide the amount of voltage dissipated across your load. In your present configuration, the load resistor is in parallel to the overall voltage divider circuit, the equivalent resistance, in this case, will be somewhat closer to 8 ohms thus raising the current value. (You might want to consider this fact while making calculations.)

  1. about using the values of resistors. Using a loaded voltage divider calculator like this you can check your answer. For the present case, let's say you have two resistors of 180 k and 20 k, and a load of 10 ohms is connected. Now the power delivered to load is 3.082993253485863e-8 way too small to have a negligible effect, and the current flowing in amps is 0.00005552470 A. So far as power is considered this is very small and will have a negligible effect on the load.

  2. The power rating of resistors actually depends on the input voltage coming from the source. I am adding a screenshot for the 10 V input signal. enter image description here You can check the result of input from your circuit for any value by the formulas or calculator.

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  • \$\begingroup\$ Please try using meaningful engineering units in place of nonsense like 3.082993253485863e-8 ?? (i.e. \$31\,\mathrm{nW}\$) or 0.00005552470 A (i.e. \$56\,\mu\mathrm{A}\$) \$\endgroup\$ – carloc Sep 23 '17 at 17:37

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