Even more confused. If I look at an unclipped signal at the load with a scope, and measure RMS V with a good Fluke 77, with math confirmed by my p-p scope reading, and calculate RMS * RMS / 8 ohms, how is this a less than meaningful reading?vibratoking wrote:
I can't imagine how you can make this measurement without a scope, power meter, or some other piece of test equipment. Measuring output power with a DMM is impossible IMO. You'll measure something, that's for sure, but it won't have much meaning.
Measuring output power
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Re: Measuring output power
- JazzGuitarGimp
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Re: Measuring output power
Measure the 6V6 cathode voltage (x) at idle, then double that number (2x). Then apply a 400Hz sinewave at the input and bring the volume up just to the point of clipping at the output. Now measure the grid of each output tube with the scope: the peak to peak signal voltage at each grid should be pretty close to 2x. If it is substantially lower than 2x, then clipping is occurring at the output of the preamp, or at the output of the PI.
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Printed Circuit Design & Layout,
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Re: Measuring output power
Sorry, I am not trying to confuse you. Your method is fine as long as you understand the error that occurs when converting from peak-peak to RMS. In your case, I don't understand why you need the DMM except that you don't want to do the math. Whether you use the DMM or do the math yourself, the possibility of error exists.Even more confused. If I look at an unclipped signal at the load with a scope, and measure RMS V with a good Fluke 77, with math confirmed by my p-p scope reading, and calculate RMS * RMS / 8 ohms, how is this a less than meaningful reading?
I understood that the other poster said he wants to make measurements with a DMM only. In the case of a DMM ONLY, the measurment is very questionable and most likely wasted time.
Also, from Wiki describing the possibility of RMS error:
The peak-to-peak voltage of a sine wave is about 2.8 times the RMS value. The peak-to-peak value is used, for example, when choosing rectifiers for power supplies, or when estimating the maximum voltage insulation must withstand. Some common voltmeters are calibrated for RMS amplitude, but respond to the average value of a rectified waveform. Many digital voltmeters and all moving coil meters are in this category. The RMS calibration is only correct for a sine wave input since the ratio between peak, average and RMS values is dependent on waveform. If the wave shape being measured is greatly different from a sine wave, the relationship between RMS and average value changes. True RMS-responding meters were used in radio frequency measurements, where instruments measured the heating effect in a resistor to measure current. The advent of microprocessor controlled meters capable of calculating RMS by sampling the waveform has made true RMS measurement commonplace.
Re: Measuring output power
I was curious about the output transformer - what primary impedance is it running at, etc ?Randall wrote:I'm using the Triode designed Super TF110 paper bobbin made by Mag Comp. B+ was 373.....
Re: Measuring output power
I do have a scope - 60mHz, dual trace, etc. (see my avatar) I didn't say I want to do it with a DMM only.vibratoking wrote: I can't imagine how you can make this measurement without a scope, power meter, or some other piece of test equipment. Measuring output power with a DMM is impossible IMO. You'll measure something, that's for sure, but it won't have much meaning.
I just want to know I'm using the right numbers for the calculations. And is a True RMS meter accurate at all frequencies?
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Re: Measuring output power
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Re: Measuring output power
Looking at the PI. the both sections look about the same at about 50v p-p before clipping, up til about 10 on volume. Then past that, the first section increases in voltage, while the second section clips hard at the top. I tried a new PI tube, no diff.martin manning wrote:Perhaps one of the preamp stages is clipping prematurely, or your PI is very unbalanced? Use your scope to find out. Also, what is the p-p signal level going into the power tube grids?
The grids of the outputs look very similar with about the same 50v p-p on the grids before one side clips at the top.
My current V's at present line voltage of 120 Vac are:
B+ = 374v
Vp = 357v
Vp-k = 335v
Vsc = 331v
Vk = 21.7v
Last edited by Randall on Tue Nov 12, 2013 9:21 pm, edited 1 time in total.
Re: Measuring output power
Did this. 2x = 43.4v. Grids = ~50vJazzGuitarGimp wrote:Measure the 6V6 cathode voltage (x) at idle, then double that number (2x). Then apply a 400Hz sinewave at the input and bring the volume up just to the point of clipping at the output. Now measure the grid of each output tube with the scope: the peak to peak signal voltage at each grid should be pretty close to 2x. If it is substantially lower than 2x, then clipping is occurring at the output of the preamp, or at the output of the PI.
- martin manning
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Re: Measuring output power
So, with Vk at 21.7 and 270R that's 0.0402 A per tube, which is 114% of max... pretty hot! Vg2 to Vk is 309, and assuming an 8K load, the power output from the load line would be 18.3W. You won't get but maybe 85% of that (SWAG), so I would expect maybe 15W. 50V p-p ought to be enough to drive the grid voltage to zero.Randall wrote:Looking at the PI. the both sections look about the same at about 50v p-p before clipping, up til about 10 on volume. Then past that, the first section increases in voltage, while the second section clips hard at the top. I tried a new PI tube, no diff.martin manning wrote:Perhaps one of the preamp stages is clipping prematurely, or your PI is very unbalanced? Use your scope to find out. Also, what is the p-p signal level going into the power tube grids?
The grids of the outputs look very similar with about the same 50v p-p on the grids before one side clips at the top.
My current V's at present line voltage of 120 Vac are:
B+ = 374v
Vp = 357v
Vp-k = 335v
Vsc = 331v
Vk = 21.7v
Re: Measuring output power
Here is the OT I'm using from triode. http://site.triodestore.com/TF110_Paper ... IAGRAM.jpg
- martin manning
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Re: Measuring output power
With a 6k6 load you're down to 16.1W on the load line, guessing 85% of that would put you at 13.7. Are the power tubes known to be good?
Re: Measuring output power
Apologies in advance for this...
...but are you *sure* you're on the 8 ohm tap? Everything looks OK to your power tubes, and that voltage swing would be roughly right for the 4 ohm tap.
...but are you *sure* you're on the 8 ohm tap? Everything looks OK to your power tubes, and that voltage swing would be roughly right for the 4 ohm tap.
Re: Measuring output power
My cathode biased 6v6 amp has 1k screen resistors, 346v plate, 20v cathode and idles at 37.5ma on each plate. Very close to tweed specs AFAIK. I measure about 9 watts on the output using my scope at the onset of clipping.
The OPT is overspec'd for the application and the rectification is SS. The PI is LT so it can drive more than enough to clip the outputs 1st.
Tubes are new JJs.
The OPT is overspec'd for the application and the rectification is SS. The PI is LT so it can drive more than enough to clip the outputs 1st.
Tubes are new JJs.
If it says "Vintage" on it, -it isn't.
Re: Measuring output power
You know what? I asked myself the same question. I emailed Triode for the spec sheet and according to it, I am using the green wire for 8 ohms. How would I check to see if there somehow is a wiring color mistake? I ask because I am having the same approx. measurements on two amps built side by side with all the same components.Tillydog wrote:Apologies in advance for this...
...but are you *sure* you're on the 8 ohm tap? Everything looks OK to your power tubes, and that voltage swing would be roughly right for the 4 ohm tap.
Re: Measuring output power
Sorry if this has been mentioned before, but are you sure your negative probe is grounded? If you accidentally have just the positive probe to the speaker tip, you will get a little less than half the actual AC voltage. This has happened to me a couple of times, and really caused me to scratch my head.
I'm actually a little surprised people are blaming the Fluke. A RMS or averaging meter should give an accurate output power within a few watts, not like what you're seeing.
I'm actually a little surprised people are blaming the Fluke. A RMS or averaging meter should give an accurate output power within a few watts, not like what you're seeing.