KT88 and push-pull ultra-linear

rossdt · 22nd July 2006, 08:51 PM

What do you think about this quiescent point for KT88 in push-pull ultra-linear operation?

Values for TWO tubes - Class A

DC Plate Voltage = 395 V
DC Screen Voltage = 395 V
Cathode Bias Resistor = 200 ohms
Peak AF Grid-to-Grid Voltage = 70 V
Zero-Signal DC Plate Current = 170 mA
Zero-Signal DC Screen Current = 12,5 mA
Effective Load Resistance, Plate-to-Plate = 5600 ohms
Screen Tapped at 40%

I use Cathode Bias without capacitor

Better ideas?
Thanks.

gingertube · 23rd July 2006, 11:39 PM

What I think is that with the values you describe you will get about 30W in Class AB (not Class A) and that you should use separate 400 Ohm resistors on the cathode of each KT88, rather than a common 200 Ohm cathode resistor. All the data sheets and application notes are quite clear on this point.
Cheers,
Ian

rossdt · 24th July 2006, 09:28 AM

Really?
Then, how do I do to work in class A?
Which quiescent point?
Thanks.

jane · 24th July 2006, 11:27 AM

2 x 100mA into a 5600 ohms p-p load will give you ~30W class A.
Check the plate dissipation; I'll suppose 100mA will bring you close to the limit.

PmaxA = [(Iq x 2)^2 x Rpp/4] / 2

pengboon · 25th July 2006, 04:54 AM

Since the topic of cathode resistors is brought up, why do some designs share the resistor between 2, sometimes even 4 tubes?

22nd July 2006, 08:51 PM	#1
rossdt diyAudio Member Join Date: Jul 2006	KT88 and push-pull ultra-linear What do you think about this quiescent point for KT88 in push-pull ultra-linear operation? Values for TWO tubes - Class A DC Plate Voltage = 395 V DC Screen Voltage = 395 V Cathode Bias Resistor = 200 ohms Peak AF Grid-to-Grid Voltage = 70 V Zero-Signal DC Plate Current = 170 mA Zero-Signal DC Screen Current = 12,5 mA Effective Load Resistance, Plate-to-Plate = 5600 ohms Screen Tapped at 40% I use Cathode Bias without capacitor Better ideas? Thanks.

24th July 2006, 11:27 AM	#4
jane diyAudio Member Join Date: Jul 2004 Location: Norway	2 x 100mA into a 5600 ohms p-p load will give you ~30W class A. Check the plate dissipation; I'll suppose 100mA will bring you close to the limit. PmaxA = [(Iq x 2)^2 x Rpp/4] / 2 __________________ Life is hard - Then you die.

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23rd July 2006, 11:39 PM	#2
gingertube diyAudio Member Join Date: Oct 2004 Location: Adelaide South Oz	What I think is that with the values you describe you will get about 30W in Class AB (not Class A) and that you should use separate 400 Ohm resistors on the cathode of each KT88, rather than a common 200 Ohm cathode resistor. All the data sheets and application notes are quite clear on this point. Cheers, Ian

24th July 2006, 09:28 AM	#3
rossdt diyAudio Member Join Date: Jul 2006	Really? Then, how do I do to work in class A? Which quiescent point? Thanks.

25th July 2006, 04:54 AM	#5
pengboon diyAudio Member Join Date: Dec 2005 Location: Singapore	Since the topic of cathode resistors is brought up, why do some designs share the resistor between 2, sometimes even 4 tubes?

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