J
Jon Kirwan
I originally posted this in sci.electronics.basics, but perhaps should
have posted it here.
....
I recently looked at the general schematic for the LM334 on National's
datasheet and with a quick sweep of my arms came up with a design
Iset/Ibias of 8, not 16 as they show on page 5. I'm off by a factor
of two.
My logic went like this. 1/2 of the I from V+ flows via Q6 to the R
rail. 1/4 via Q4 and 1/4 via Q5. Q5's 1/4*I flows via Q1 to the R
rail, too. So now up to 3/4*I into the R rail. Q4's 1/4*I passes
through two paths. The Ic(Q2)=Ic(Q1)/2... but Ic(Q1)=1/4*I, so that
is 1/8*I, leaving the other 1/8*I for Q3's Vbe conduction, which also
flows to the R rail. So the R rail gets 7/8*I and the V- picks up
1/8*I. Multiplying through by 8 to get rid of the divisor, I see a
factor of 8 for Iset/Ibias... not 16.
Can someone do a quick description about how to arrive at something
more like 16 from the schematic? I'm missing a clue (or two.)
To be handy, here's the sheet:
http://www.national.com/ds/LM/LM134.pdf
Thanks in advance,
Jon
have posted it here.
....
I recently looked at the general schematic for the LM334 on National's
datasheet and with a quick sweep of my arms came up with a design
Iset/Ibias of 8, not 16 as they show on page 5. I'm off by a factor
of two.
My logic went like this. 1/2 of the I from V+ flows via Q6 to the R
rail. 1/4 via Q4 and 1/4 via Q5. Q5's 1/4*I flows via Q1 to the R
rail, too. So now up to 3/4*I into the R rail. Q4's 1/4*I passes
through two paths. The Ic(Q2)=Ic(Q1)/2... but Ic(Q1)=1/4*I, so that
is 1/8*I, leaving the other 1/8*I for Q3's Vbe conduction, which also
flows to the R rail. So the R rail gets 7/8*I and the V- picks up
1/8*I. Multiplying through by 8 to get rid of the divisor, I see a
factor of 8 for Iset/Ibias... not 16.
Can someone do a quick description about how to arrive at something
more like 16 from the schematic? I'm missing a clue (or two.)
To be handy, here's the sheet:
http://www.national.com/ds/LM/LM134.pdf
Thanks in advance,
Jon