I have been trying to understand PMA’s amplifier introduced here: http://www.diyaudio.com/forums/showthread.php?threadid=40543&highlight=
I have a problem with a Vbe multiplier disclosed by Nelson Pass in his patent: US3,995,228. To achieve bias voltage Vb = 9 V, with 2*Vbe = 1.4 V, the equation to be solved is:
(R1+R2+R3)/R2 = Vb/2*Vbe
Setting R1=R3= 600 Ohm, R2 = 543 Ohm
Looking at the schematics: web.telecom.cz/macura/pm_ab_er1.gif , the resistor R7 - corresponding to R2 - is 10 kOhm.
Now, interestingly enough, simulation confirms the value 10kOhm.
Any help would be appreciated.
M
I have a problem with a Vbe multiplier disclosed by Nelson Pass in his patent: US3,995,228. To achieve bias voltage Vb = 9 V, with 2*Vbe = 1.4 V, the equation to be solved is:
(R1+R2+R3)/R2 = Vb/2*Vbe
Setting R1=R3= 600 Ohm, R2 = 543 Ohm
Looking at the schematics: web.telecom.cz/macura/pm_ab_er1.gif , the resistor R7 - corresponding to R2 - is 10 kOhm.
Now, interestingly enough, simulation confirms the value 10kOhm.
Any help would be appreciated.
M
Nelson wrote:
"By your own formula, you would expect the middle resistor R2
to be about 200 ohms, this of course reflecting a simple
Vbe multiplier with two complementary devices and three resistors."
I have made a mistake, the value of 2*Vbe = 2.8 V ( I have a BE junction in series with a diode).
However, it still does not explain the discrepancy.
M
"By your own formula, you would expect the middle resistor R2
to be about 200 ohms, this of course reflecting a simple
Vbe multiplier with two complementary devices and three resistors."
I have made a mistake, the value of 2*Vbe = 2.8 V ( I have a BE junction in series with a diode).
However, it still does not explain the discrepancy.
M
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