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Old 19th June 2012, 02:03 PM   #1461
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Old 19th June 2012, 02:11 PM   #1462
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Idss is a device parameter.
Id is the device operating current.
They are not the same.

If one builds a B1 or a DCB1, one chooses to run the two devices at an Id that happens to be the same value as the Idss, if one can keep the Tj at the specified 25șC.

In this regulator we select a device with a certain Idss. That is not the same as the operating current Id.
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Old 19th June 2012, 02:16 PM   #1463
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Sorry my ignorance, good to know AndrewT really appreciate.
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Old 19th June 2012, 02:31 PM   #1464
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1.309V
Dummy load 10R3
Vout 5.23
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Old 19th June 2012, 05:14 PM   #1465
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Quote:
Originally Posted by Salas View Post
The BC550 contributes 0.6V, the 1N4007 you used another 0.6V, the rest is all the resistance you got in series times your associated K117 current. If the things don't add up normally for K117GR IDSS range then there is something wrong with it.
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Is Q103 the jFET with 0.6mV across D to S?
If that is so, then the Idss of 4.3mA measured at 10Vds will be much different from Id when 0.6Vds is applied.
Expect Id ~ 50%Idss, i.e. around 2mA to 2.5mA.

The output voltage will be Vbe of the BJT plus the Voltage across the Vref string.
The Voltage across that Vref string will be due to the Id passing that jFET of about 2mA to 2.5mA.
If the Vref string were a diode + a Zener (5.6V) + fixed resistor (1k0) then the voltage drop due to 2mA of current flow is ~ 600mV for diode + 5.6V for Zener (if Pz >10% of Pmax) + 0.002 * 1000 ~= 8.2V
One can short out (link across) or add as many components to the Vref string as needed, to get the desired Vout.

Just measure the in circuit Id of the jFET by measuring the voltage drop across the fixed resistor. That resistor can be 10r. 2mA and 10r would read ~20.0mVdc using a 199.9Vdc scale of a dmm, good enough accuracy for our purpose.

It really is that simple, some simple arithmetic combined with an understanding of what the circuit is doing.
Voltage drop across Vref resistor (11R78) 3,20V + 0,6V diode + 0.002 * 1000 = 5.8V but I get Vout 5.23?
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Old 19th June 2012, 05:21 PM   #1466
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What is the measured voltage across the fixed resistor?
What is the value of the fixed resistor? Or if it's a parallel pair, what are the values?
From there you get the actual current passing and can use that to substitute for the 0.002 value that I guessed at for a 4.3mA Idss running @ 600mVds.
What is the actual diode Vdrop (Vf)?

I don't understand the 11r78 giving 3.2V for Vdrop. That indicates ~270mA of current flow through the Vref string. That can't be right !
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Old 19th June 2012, 06:16 PM   #1467
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Quote:
Originally Posted by AndrewT View Post
What is the measured voltage across the fixed resistor? Now after 1 hour of warm up 3.14VWhat is the value of the fixed resistor? 11R78 Or if it's a parallel pair, what are the values? 22R+33R in series paralleled with 15RFrom there you get the actual current passing and can use that to substitute for the 0.002 value that I guessed at for a 4.3mA Idss running @ 600mVds.
What is the actual diode Vdrop (Vf)? 0.76V
I don't understand the 11r78 giving 3.2V for Vdrop. That indicates ~270mA of current flow through the Vref string. That can't be right !

Last edited by merlin el mago; 19th June 2012 at 06:20 PM.
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Old 19th June 2012, 06:22 PM   #1468
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AndrewT see attached pic.
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Old 20th June 2012, 09:12 AM   #1469
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Quote:
Originally Posted by AndrewT View Post
............I don't understand the 11r78 giving 3.2V for Vdrop. That indicates ~270mA of current flow through the Vref string. That can't be right !
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Originally Posted by merlin el mago View Post
AndrewT see attached pic.
it still can't be right.

How can 270mA flow through the Vref string?
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Old 20th June 2012, 10:19 AM   #1470
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For sure there is something wrong is what I'm looking for all the time. Isn't normal a resistor of near 12R at Vref to get the desired 5.25Vout when calculator says 800R aprox.
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