kw7y
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« on: April 22, 2015, 08:05:19 PM » |
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I just fired up my new 2 x 813s Triode Connected push pull Modulator, driving a 10K resistor instead of the PA. It works well except the efficiency is low. The HV is about 2500V. Idle current is set at about 60mA (150W). Efficiency at 400W RMS output is 42% as calculated by plate voltage times cathode current.
Does this seem typical for this configuration?
Thanks
- Paul KW7Y
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WBear2GCR
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« Reply #1 on: April 22, 2015, 10:44:07 PM » |
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Triode connected always gives less power out.
Also 150w of quiescent is a bit high, rather class A-ish. I'd bias more to class B, maybe 20-30w max quiescent current.
How are your driving the 813s? Cap coupled? Or?
_-_-
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« Reply #2 on: April 22, 2015, 11:55:34 PM » |
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Do you mean triode as in screen connected to plate, or the 'special class B' connection with screens and grids driven and zero bias?
Power seems good. 380W CCS 490W ICAS
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kw7y
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« Reply #3 on: April 22, 2015, 11:58:05 PM » |
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75W idle per tube.
They are driven with a center tapped transformer to ground. 8ohms to 500:500 ohms, with a 20W Audio IC. 14V P/P in gives me 2KV RMS out in to the 10K load, with two random 813s from the junk box.
- Paul
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kw7y
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« Reply #4 on: April 23, 2015, 12:00:04 AM » |
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Screen grid and control grid connected together.
I have 4 diodes in the cathode providing a little bias to keep idle current below 60mA.
- Paul
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WBear2GCR
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« Reply #5 on: April 23, 2015, 08:18:45 AM » |
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Assuming maximum B+ of 2500v then that is the peak output voltage available, and RMS = 0.707 x 2500, 1750v. So, P = E ^2 / R = ~300wrms. Seems about right. Maybe I'm not right.
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K1JJ
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« Reply #6 on: April 23, 2015, 12:18:36 PM » |
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I just fired up my new 2 x 813s Triode Connected push pull Modulator, driving a 10K resistor instead of the PA. It works well except the efficiency is low. The HV is about 2500V. Idle current is set at about 60mA (150W). Efficiency at 400W RMS output is 42% as calculated by plate voltage times cathode current.
Does this seem typical for this configuration?
Thanks
- Paul KW7Y
Hi Paul, Did you consider cathode current includes screen and grid current in the calculations? If not, then subtract these and recalculate for plate current only for a truer plate-only efficiency indication. When triode-connected, they may be drawing a fair amount of screen and grid current, something like a grounded grid class B RF amplifier does. When I used to run this config, the modulator efficiency was pretty decent. I found a pair of 813s X pair showed no plate color when run full power. If need be, you can add some air to increase dissipation by at least another 40% or so. Lantern chimneys work very FB. T
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Use an "AM Courtesy Filter" to limit transmit audio bandwidth +-4.5 KHz, +-6.0 KHz or +-8.0 KHz when needed. Easily done in DSP.
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kw7y
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« Reply #7 on: April 23, 2015, 12:30:50 PM » |
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The control and screen grids are driven by a 20W amp, so at the most are taking that much power.
I'm not to worried about the poor efficiency as the average power for speech is probably less than 25% of the sine wave I'm testing with.
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w4bfs
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« Reply #8 on: April 23, 2015, 01:23:14 PM » |
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I just fired up my new 2 x 813s Triode Connected push pull Modulator, driving a 10K resistor instead of the PA. It works well except the efficiency is low. The HV is about 2500V. Idle current is set at about 60mA (150W). Efficiency at 400W RMS output is 42% as calculated by plate voltage times cathode current.
Does this seem typical for this configuration?
Thanks
- Paul KW7Y
Hi Paul, Did you consider cathode current includes screen and grid current in the calculations? If not, then subtract these and recalculate for plate current only for a truer plate-only efficiency indication. When triode-connected, they may be drawing a fair amount of screen and grid current, something like a grounded grid class B RF amplifier does. When I used to run this config, the modulator efficiency was pretty decent. I found a pair of 813s X pair showed no plate color when run full power. If need be, you can add some air to increase dissipation by at least another 40% or so. Lantern chimneys work very FB. T Tom is right and the combined grid currents are on the order of 150 mA .... ignoring these will definitely skew your efficiency calculations ....73 .... John
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Beefus
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kw7y
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« Reply #9 on: April 23, 2015, 02:23:36 PM » |
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You are quite right about grid current (screen I presume) being significant. I just measured the current from the B+ supply. It is measuring 320mA when the Cathode current is measuring 400mA (400W sine wave in to the 10K load). That's 52% instead of (42%) efficiency which is more reasonable.
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W2VW
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« Reply #10 on: April 25, 2015, 02:45:25 PM » |
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Curious how are you deriving the power output figure?
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kw7y
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« Reply #11 on: April 25, 2015, 03:10:43 PM » |
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Using a 1000:1 voltage divider to measure the RMS voltage across a 10K load
- Paul
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W2VW
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« Reply #12 on: April 27, 2015, 02:23:21 PM » |
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Triode connected always gives less power out.
_-_-
Do you mean less efficiency as asked in the post or less power with a given drive?
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W2VW
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« Reply #13 on: April 27, 2015, 02:27:28 PM » |
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Using a 1000:1 voltage divider to measure the RMS voltage across a 10K load
- Paul
Maybe try measuring the peak voltage on a scope and plug that into the proper formula as a cross check. Some measuring devices do unexpected things when looking at RMS or average. Anyway a pair of 813s triode connected will modulate a full gallon at 2500 volts with just over 1:1 mod xfmr ratio. They will be running at similar specifications to 810s as far as dissipation which is OK fine.
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kw7y
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« Reply #14 on: April 27, 2015, 02:38:54 PM » |
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I am using an accurate Fluke HV divider and TRUE RMS meter to measure the AF output power. Measuring the input power using plate voltage times anode current gives gives me 52% at 400W out, which I think is reasonable. My error was to initially measure the cathode current rather than anode current.
In my application which is 2x813's in the rf final at legal limit PEP they have about 600W input, so 300W of audio will be all I need.
- Paul
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W2VW
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« Reply #15 on: April 27, 2015, 04:10:51 PM » |
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Close enough except for the name of the output power.
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w4bfs
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« Reply #16 on: April 27, 2015, 05:54:44 PM » |
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Close enough except for the name of the output power.
Is that munky swang ?
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Beefus
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kw7y
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« Reply #17 on: April 27, 2015, 06:31:17 PM » |
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Close enough except for the name of the output power.
What's wrong with the term AF Power?
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W2VW
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« Reply #18 on: April 27, 2015, 10:35:22 PM » |
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Nothing. RMS power is another story.
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Steve - K4HX
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« Reply #19 on: April 27, 2015, 11:39:30 PM » |
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True RMS meter as in? Manufacturer/model
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kw7y
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« Reply #20 on: April 28, 2015, 12:38:43 AM » |
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The meter measures True RMS Voltage. The power is calculated using the known resistor.
A Fluke 187
- Paul
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Steve - K4HX
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« Reply #21 on: April 29, 2015, 08:55:08 AM » |
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I have a 187. It is supposed to measure accurately to 100 kHz. I was wondering if limited bandwidth might be the cause of the apparent low efficiency. It would appear not. I might run a bandwidth test on mine.
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Steve - K4HX
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« Reply #22 on: April 29, 2015, 11:37:03 PM » |
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Also the meter will give good results reading RMS volts or amps on sinewaves or simple AC waveforms with a relatively low crest factor. The high-end true RMS meters I've used in the past were not DVMs. These used a thermocouple to determine true RMS of nearly any waveform.
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