Sometimes it's so fun to quote oneself...
http://www.diyaudio.com/forums/showthread.php?s=&postid=1206159&highlight=#post1206159
http://www.diyaudio.com/forums/showthread.php?s=&postid=1206159&highlight=#post1206159
Not finished with Type E yet? No . . .
I have modified the CCS part (see the schematic) once again. And, changed some parts too. Now I get 17Vdc output and the constanr current of 70mA, which will be shared by the control element (Q2) and B1 buffer.
This will be the final version of Babo shunt Type E.
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I have modified the CCS part (see the schematic) once again. And, changed some parts too. Now I get 17Vdc output and the constanr current of 70mA, which will be shared by the control element (Q2) and B1 buffer.
This will be the final version of Babo shunt Type E.
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because Q3 is almost saturated it has very low gain.
Almost 1mA of the R10 tail current passes R7 through the base.
How can Q3 hold a solid reference voltage at it's base when the voltage drop across R7 is 221mV and varies with every tiny current change in the Zener and/or R6 and/or Q3?
Can you compare performance of your posted schematic with a version that has 630r in parallel to each 4k75?
Almost 1mA of the R10 tail current passes R7 through the base.
How can Q3 hold a solid reference voltage at it's base when the voltage drop across R7 is 221mV and varies with every tiny current change in the Zener and/or R6 and/or Q3?
Can you compare performance of your posted schematic with a version that has 630r in parallel to each 4k75?
AndrewT said:
Andrew,
If I reduce R8 from 4K75 to 630, the system will not work. The reduced collector resistor will make Q3 operate in the active region. Then, the collector current will be increased due to high dc current gain (dc beta) so that almost all dc current will flow through Q3 while there will be almost no dc current flowing through Q4. Then, the system in not able to work.
If I need to change R8 and R9 values, I will do with higher values, e.g. with 7K5 or 10K.
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I don't think your analysis is correct.
You can see in your last post, that 7.89V @ x6 does not equal 7.62V @ x3. Your differential amplifier is crippled because you refuse to balance the emitter currents of the LTP.
Assume the Vbe of the shunt transistor (Q2) to be ~ 700mV. You have measured 853mV here. This forces the same ~ 700mV across the R9.
The tail current through R10 is [7.62V - 0.6V] / 2k7 ~ 2.6mA.
Half of this current should flow through each side of the LTP.
i.e. ~1.3mA through R9.
The value of R9 should be ~ 0.7V / 0.0013mA ~ 540r.
The other ~ 1.3mA will pass R8. I can also be the same value. A differential amplifier will work with R8=0r0, but then it operates with a different Vce from it's partner.
If you want accuracy from your LTP differential amplifier, where the output voltage is the DIFFERENCE between the two input voltages then the pair of transistors should be matched at your 1.3 to 1.5mA emitter current, have the same Vbe at these currents, have the same hFE at these currents, have the same junction temperature, have the same Vce and then one will find that the output error voltage closely follows the correction needed to drive Q2, over a range of input voltages and over a sensible range of ambient and junction temperatures.
Try the comparison I suggested and post the voltages at x3 and x6 like you did in the last annotated schematic.
Once you have the LTP almost balanced and close to correct operating voltages you will need to remeasure all your voltages and determine your operational currents and reselect the optimum resistor values to get the balance that the LTP desires to give it's best performace.
You can see in your last post, that 7.89V @ x6 does not equal 7.62V @ x3. Your differential amplifier is crippled because you refuse to balance the emitter currents of the LTP.
Assume the Vbe of the shunt transistor (Q2) to be ~ 700mV. You have measured 853mV here. This forces the same ~ 700mV across the R9.
The tail current through R10 is [7.62V - 0.6V] / 2k7 ~ 2.6mA.
Half of this current should flow through each side of the LTP.
i.e. ~1.3mA through R9.
The value of R9 should be ~ 0.7V / 0.0013mA ~ 540r.
The other ~ 1.3mA will pass R8. I can also be the same value. A differential amplifier will work with R8=0r0, but then it operates with a different Vce from it's partner.
If you want accuracy from your LTP differential amplifier, where the output voltage is the DIFFERENCE between the two input voltages then the pair of transistors should be matched at your 1.3 to 1.5mA emitter current, have the same Vbe at these currents, have the same hFE at these currents, have the same junction temperature, have the same Vce and then one will find that the output error voltage closely follows the correction needed to drive Q2, over a range of input voltages and over a sensible range of ambient and junction temperatures.
Try the comparison I suggested and post the voltages at x3 and x6 like you did in the last annotated schematic.
Once you have the LTP almost balanced and close to correct operating voltages you will need to remeasure all your voltages and determine your operational currents and reselect the optimum resistor values to get the balance that the LTP desires to give it's best performace.
Andrew, thanks for this explanation. I was having a problem understanding why I do not see equal resistors on the tail of difference amplifiers. Now I understand. The currents are very important in defining the operating point, so they should be equal to balance the difference amplifier. The tail resistors job is just to set the currents to be equal and can themselves be different. I see that Papa has only one resistor on the tail of the jfet differential input to the Aleph J. As he is not worried about keeping these resistances the same then I will try to remember that it is the active components that matters the most.
Thanks again
Jim
Thanks again
Jim
AndrewT said:
I believe that you need more listening than talking in this case.
AndrewT said:
Nonsense . . .
Refer to my last schematic. The tail current through R10 is (16.57V-8.49V)/2K7 = 3mA. Meanwhile, Q3 is operating in hard saturation. The saturated collector current through R8 is 8.48V/7K5 = 1.13mA, and the current through R9 is 0.828V/7K5 = 0.11mA. Then, total current through these two resistors is 1.13+0.11 = 1.24mA, which is less than half of the tail current. This means that most of the remaining tail current is flowing through emitter of Q3, through the base, through R7 and through R6 to the ground.
Clear? The Q3 is operating in the hard saturation, looking at the reference voltage, i.e. the zener voltage of 16.57V-7.62V = 8.95V. I have given this role to Q3.
Well, the negative feedback system, i.e. R12 and R13, forces the base voltage of Q4 to be similar to the base voltage of Q3, leaving small error voltage between them because the base voltage of Q4 is to be slightly greater than the base voltage of Q3 to turn on Q2. Then, the whole system works.
Pls note that we are talking about voltage regulator . . .
AndrewT said:
If I reduce R8 and R9 values as you recommend, then the Q4 can't be turned on. The whole system dosn't work. (Re-read my post#275.) The circuit is simple. Build one and try. You will see . . .
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JimT said:
Ayo . . . who has two tails . . . ?
I know they have two horns and one tail . . .
I have just looked into Papa's Aleph J.
Yes, 2sj109 has one drain resistor while there is no resistor on the other side. Nevertheless, the two JFETs equally share the same current. You know why . . . ? It's because they have the same character and their gates are seeing the same zero dc voltage as designed. Yeah, it's as designed . . .
But, when there is ac signal to the gates, they don't share the drain current equally . . .
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