Formula for driver current requirement ?

mahleriana · 18th October 2011, 05:23 PM

Hello,
I am looking for the formula that calculates the driver stage minimum quiescent current requirement regarding the following stage (triode or pentode)
Thanks in advance !
Eric (France)

oshifis · 18th October 2011, 06:31 PM

The quiscent current is determined by the charging current of the input (Miller) capacitance of the output stage at the highest frequency at the highest amplitude sinusodal signal possible at the grid of the output stage.

mahleriana · 18th October 2011, 07:01 PM

which translates into ?

DF96 · 18th October 2011, 07:12 PM

You asked for a formula. Here it is:

I >> A Cag 2 pi f Vpk

where
I is quiescent current
A is voltage gain of next stage
Cag is anode-grid capacitance (valve plus holder plus strays)
f is highest signal frequency
Vpk is peak signal voltage at grid of next stage

Strictly, the input and Miller capacitance should be Cgk+(A+1)Cag but ACag is close enough in most cases. You only need a rough estimate, as you then need to exceed it by a decent margin.

oshifis · 18th October 2011, 07:13 PM

I = C * (dU/dt)

boywonder · 18th October 2011, 07:34 PM

Quote:

Originally Posted by DF96

You asked for a formula. Here it is:

I >> A Cag 2 pi f Vpk

where
I is quiescent current
A is voltage gain of next stage
Cag is anode-grid capacitance (valve plus holder plus strays)
f is highest signal frequency
Vpk is peak signal voltage at grid of next stage

Strictly, the input and Miller capacitance should be Cgk+(A+1)Cag but ACag is close enough in most cases. You only need a rough estimate, as you then need to exceed it by a decent margin.

What's >> above? 100:1? More? Decent margin=?

DF96 · 18th October 2011, 07:37 PM

>> means much greater than. I would say a factor of 2 or 3 in this case.

> means greater than, so a factor of 1.01 would satisfy that.

euro21 · 18th October 2011, 07:52 PM

For example 300B requires (almost) 2mA pp grid current at 20kHz (220V pp driving, A2 mode).

mahleriana · 18th October 2011, 08:28 PM

If I try for a 45 tube :
Cag = 7uuF
Cgf = Cgk = 4uuF
Amplification factor (A) = 3.5
ppV = class A1 centre value cathode bias * 2 = 100v
f = 20.000 Hz

C=Cgk+(A+1)Cag = 35.5 uuF

I = 3.5 * 35.5x10^-12 * 2 * 3.14159 * 100 * 20000 = 0.0015613 = 1.56 mA

Right ?

boywonder · 18th October 2011, 08:52 PM

Quote:

Originally Posted by DF96

>> means much greater than. I would say a factor of 2 or 3 in this case.

> means greater than, so a factor of 1.01 would satisfy that.

Got it, thanks....I know what the symbols mean ......was wondering how much margin is typical/required.

18th October 2011, 05:23 PM	#1
mahleriana diyAudio Member Join Date: Nov 2007	Formula for driver current requirement ? Hello, I am looking for the formula that calculates the driver stage minimum quiescent current requirement regarding the following stage (triode or pentode) Thanks in advance ! Eric (France)

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18th October 2011, 06:31 PM	#2
oshifis diyAudio Member Join Date: Mar 2004 Location: Budapest, Hungary	The quiscent current is determined by the charging current of the input (Miller) capacitance of the output stage at the highest frequency at the highest amplitude sinusodal signal possible at the grid of the output stage.

18th October 2011, 07:01 PM	#3
mahleriana diyAudio Member Join Date: Nov 2007	which translates into ?

18th October 2011, 07:12 PM	#4
DF96 diyAudio Member Join Date: May 2007	You asked for a formula. Here it is: I >> A Cag 2 pi f Vpk where I is quiescent current A is voltage gain of next stage Cag is anode-grid capacitance (valve plus holder plus strays) f is highest signal frequency Vpk is peak signal voltage at grid of next stage Strictly, the input and Miller capacitance should be Cgk+(A+1)Cag but ACag is close enough in most cases. You only need a rough estimate, as you then need to exceed it by a decent margin.

18th October 2011, 07:13 PM	#5
oshifis diyAudio Member Join Date: Mar 2004 Location: Budapest, Hungary	I = C * (dU/dt)

18th October 2011, 07:37 PM	#7
DF96 diyAudio Member Join Date: May 2007	>> means much greater than. I would say a factor of 2 or 3 in this case. > means greater than, so a factor of 1.01 would satisfy that.

18th October 2011, 07:52 PM	#8
euro21 diyAudio Member Join Date: Sep 2004 Location: Budapest	For example 300B requires (almost) 2mA pp grid current at 20kHz (220V pp driving, A2 mode).

18th October 2011, 08:28 PM	#9
mahleriana diyAudio Member Join Date: Nov 2007	If I try for a 45 tube : Cag = 7uuF Cgf = Cgk = 4uuF Amplification factor (A) = 3.5 ppV = class A1 centre value cathode bias * 2 = 100v f = 20.000 Hz C=Cgk+(A+1)Cag = 35.5 uuF I = 3.5 * 35.5x10^-12 * 2 * 3.14159 * 100 * 20000 = 0.0015613 = 1.56 mA Right ?

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