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Reference Voltage Schematic

M

Marte Schwarz

Jan 1, 1970
0
Hi,
I found a schematic for a low voltage reference in LM311 datasheets from
National, TI and ST. It looks like:

VCC
|
\
/ R1
\ 3.9k
/
|
,-----+
| |
| |
| |/c Q1
+---| 2N3708 (Si), 2N797 (Ge) or 2N1304 (Ge)
| |>e
| |
| |
| +-------Uref
| |
| |
| |/c Q2
'---| 2N2222
|>e
|
|
|
gnd

followed by a LM311 as a noninverting operational amplifier (!?!)

I want to understand this and need a little help. As I play a few with spice
it seems to give a voltage about a few ten mV but not really like a
reference source. What is the benefit of this schematic and how does it
work?

Marte
 
H

Helmut Sennewald

Jan 1, 1970
0
Marte Schwarz said:
Hi,
I found a schematic for a low voltage reference in LM311 datasheets from
National, TI and ST. It looks like:

VCC
|
\
/ R1
\ 3.9k
/
|
,-----+
| |
| |
| |/c Q1
+---| 2N3708 (Si), 2N797 (Ge) or 2N1304 (Ge)
| |>e
| |
| |
| +-------Uref
| |
| |
| |/c Q2
'---| 2N2222
|>e
|
|
|
gnd

followed by a LM311 as a noninverting operational amplifier (!?!)

I want to understand this and need a little help. As I play a few with
spice it seems to give a voltage about a few ten mV but not really like a
reference source. What is the benefit of this schematic and how does it
work?

Marte

Hello Marte,

This circuit may be as bad as it looks nowadays.

Prerequisite:
A Si-transistor has a Vbe of about 0.7V with a tempco of
about -1.8mV/degree.
A Ge-transistor has a Vbe of about 0.36V* with a tempco of -1.xmV/degree**.

The concept of this circuit:

Vbe of Q2(2N2222) is at about 0.7V with a tempco of -1.8mV/degree
regardless of Q1.
The base of Q1 is 0.36V* below this 0.7V at 0.34V (0.7-0.36).
The net effect is that the voltage Ve_Q1 at the emitter of Q1 is 0.34V with
a tempco of 0mV/degree if the Vbe_tempco of both transistors is the same.

Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

The opamp circuit around requires a potentiometer to adjust the output
voltage because of the wide variation of Vbe.

My advice: Don't use this circuit for a commercial application because
Ge-transistors are dead since 30 years.
I also have never seen Ge-transistors in SMT.

Best regards,
Helmut

* Just a guess. The real value may have a wide variation in the datasheets.
** -1.xmV means -1.5mV or something like that. Ideally it should be equal
the tempco of the Si-transistor.
 
M

Marte Schwarz

Jan 1, 1970
0
Hi Helmut,
This circuit may be as bad as it looks nowadays.
;-)

A Si-transistor has a Vbe of about 0.7V with a tempco of
about -1.8mV/degree.
A Ge-transistor has a Vbe of about 0.36V* with a tempco
f -1.xmV/degree**.

Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

Right, all clear. With Ge transistors it is clear. With Si transistors the
Voltage will be very low and sensible to Vcc.
The opamp circuit around requires a potentiometer to adjust the output
voltage because of the wide variation of Vbe.

May be they wanted to tell, that LM311 may also be useable as a opamp and
lead to its ability of common mode input below gnd.
My advice: Don't use this circuit for a commercial application because
Ge-transistors are dead since 30 years.
I also have never seen Ge-transistors in SMT.

What's about Schotthy Diode instead of Germanium transistor? Or LED instead
of 2N2222?

Marte
 
F

Fred Bloggs

Jan 1, 1970
0
Helmut said:
Hello Marte,

This circuit may be as bad as it looks nowadays.

Prerequisite:
A Si-transistor has a Vbe of about 0.7V with a tempco of
about -1.8mV/degree.
A Ge-transistor has a Vbe of about 0.36V* with a tempco of -1.xmV/degree**.

The concept of this circuit:

Vbe of Q2(2N2222) is at about 0.7V with a tempco of -1.8mV/degree
regardless of Q1.
The base of Q1 is 0.36V* below this 0.7V at 0.34V (0.7-0.36).
The net effect is that the voltage Ve_Q1 at the emitter of Q1 is 0.34V with
a tempco of 0mV/degree if the Vbe_tempco of both transistors is the same.

Ve_Q1=Vbe_Q2-Vbe_Q1=0.7-0.36=0.34V

Ve_tempco=Vbe_tempcoQ2-Vbe_tempco_Q1=-1.8mV-(-1.xmV)=-0.ymV.

The opamp circuit around requires a potentiometer to adjust the output
voltage because of the wide variation of Vbe.

My advice: Don't use this circuit for a commercial application because
Ge-transistors are dead since 30 years.
I also have never seen Ge-transistors in SMT.

Best regards,
Helmut

* Just a guess. The real value may have a wide variation in the datasheets.
** -1.xmV means -1.5mV or something like that. Ideally it should be equal
the tempco of the Si-transistor.

That's getting close but not quite there. Using your notation for Q1 and
Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance.
Q2 must saturate and Q1 is diode connected with its Vce=Vbe,Q2 for
practical purposes. The Vce,sat is inherently the difference between two
forward biased PN junctions so that on that basis alone you would expect
the tempco to be an order of magnitude below that of single diode. Also,
when Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with
Q1 providing negative feedback by shunting the supply current from
driving the base of Q2 into its CE circuit. The net effect is to produce
a Vce,sat with regulated base drive. It is not necessary for Q1 to be
Germanium, an Si will work too but the Vout is of lower magnitude.
 
H

Helmut Sennewald

Jan 1, 1970
0
----- Original Message -----
From: "Fred Bloggs" <[email protected]>
Newsgroups: sci.electronics.design
Sent: Saturday, February 17, 2007 2:41 PM
Subject: Re: Reference Voltage Schematic

That's getting close but not quite there. Using your notation for Q1 and
Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance. Q2
must saturate and Q1 is diode connected with its Vce=Vbe,Q2 for practical
purposes. The Vce,sat is inherently the difference between two forward
biased PN junctions so that on that basis alone you would expect the
tempco to be an order of magnitude below that of single diode. Also, when
Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with Q1
providing negative feedback by shunting the supply current from driving
the base of Q2 into its CE circuit. The net effect is to produce a Vce,sat
with regulated base drive. It is not necessary for Q1 to be Germanium, an
Si will work too but the Vout is of lower magnitude.

Hello Fred,

Sorry, your assumption is wrong. The basic idea is to compensate the tempco
of Vbe1 wth the tempco of Vbe2 by using Q2 in its active region with Vce >
Vce_sat.
This can only be achieved with a Germanium transistor for Q1 with a much
lower
Vbe than the Vbe of Q2.

Best regards,
Helmut
 
G

Gibbo

Jan 1, 1970
0
Helmut said:
----- Original Message -----
From: "Fred Bloggs" <[email protected]>
Newsgroups: sci.electronics.design
Sent: Saturday, February 17, 2007 2:41 PM
Subject: Re: Reference Voltage Schematic



Hello Fred,

Sorry, your assumption is wrong. The basic idea is to compensate the tempco
of Vbe1 wth the tempco of Vbe2 by using Q2 in its active region with Vce >
Vce_sat.
This can only be achieved with a Germanium transistor for Q1 with a much
lower
Vbe than the Vbe of Q2.

Best regards,
Helmut

Why does the National data sheet show Q1 as being either Si or Ge then?
 
H

Helmut Sennewald

Jan 1, 1970
0
Gibbo said:
Why does the National data sheet show Q1 as being either Si or Ge then?


Hello Gibbo,

Don't believe all what's printed.

Best regards,
Helmut
 
G

Gibbo

Jan 1, 1970
0
Helmut said:
Hello Gibbo,

Don't believe all what's printed.

I believe very little of what's printed until I see it with my own eyes.
But I do believe the circuit does indeed operate as FB says it does.
 
H

Helmut Sennewald

Jan 1, 1970
0
----- Original Message -----
From: "Marte Schwarz" <[email protected]>
Newsgroups: sci.electronics.design
Sent: Saturday, February 17, 2007 12:27 PM
Subject: Re: Reference Voltage Schematic

Hi Helmut,


Right, all clear. With Ge transistors it is clear. With Si transistors the
Voltage will be very low and sensible to Vcc.


May be they wanted to tell, that LM311 may also be useable as a opamp and
lead to its ability of common mode input below gnd.


What's about Schotthy Diode instead of Germanium transistor? Or LED
instead of 2N2222?

Marte

Hello Marte,

Both methods may be at least as good as the combination of a Ge-transistor
and a Si-transistor.
The ladder may be better because it gives a higher voltage. You have to
"play" with LEDs
of different color and from different manufacturers of course.
Just try it in an oven.

Best regards,
Helmut
 
M

Marte Schwarz

Jan 1, 1970
0
Hi Fred,
Q2, you can see that Vout is Vce,sat of Q2 and nicely at low impedance. Q2
must saturate

right here. But guess Vce,sat with a few mV in case of two Si-transistors?
The Vce,sat is inherently the difference between two forward biased PN
junctions so that on that basis alone you would expect the tempco to be an
order of magnitude below that of single diode.

But I get the tempco with about .x mV/K at a voltage of a few ten mV instead
of 500 to 600 mV from a simple si-diode. I would say that wouldn't be a
real benefit then.
when Q2 saturates, the BE junctions of Q1 and Q2 are in parallel, with Q1
providing negative feedback by shunting the supply current from driving
the base of Q2 into its CE circuit.

At least here the simulations with LTSPICE disagrees with you. Making a DC
sweep with V1 creates a nearly linear function of Ib(Q2)
The net effect is to produce a Vce,sat with regulated base drive. It is
not necessary for Q1 to be Germanium, an Si will work too but the Vout is
of lower magnitude.

In my simulation it looks like that changing the more powerfol transistor to
the upper side (Q1) and the "smaller" one (I took 2N3904 for simulations
now) as Q2 produce more stable results. I think this stabilases Helmuts
theorie, isn't ist?

Marte
 
J

James Arthur

Jan 1, 1970
0
Why does the National data sheet show Q1 as being either Si or Ge then?

It doesn't, at least not in my copy. See http://www.national.com/ds/LM/LM111.pdf

The circuit only makes sense if Q1 is germanium (e.g. 2n797), in
which case Q2 robs its own base drive via diode-connected Q1. The
output, then, is Vbe(Q2) - Vbe(Q1).

The datasheet's "Precision Squarer" (pg. 14), and "Low Voltage
Adjustable Reference Supply" (pg. 15) support this interpretation,
each clearly needing a reference voltage on the order of 300mV.

The "Precision Photodiode Comparator" (pg. 16) appears to be in
error when it states "at comparison, the photodiode has less than 5mV
across it..."

As for a modern-day version, yes, you might make do with an LED and
a silicon transistor:

..
.. Vcc >-+-------,
.. | |
.. .-. |
.. | | |
.. R1 | | |
.. '-' |
.. | |
.. | |/c
.. o-----|
.. | |>e
.. --- +--------> U(out)
.. LED \ / .-.
.. --- | |
.. | | | R2
.. | '-'
.. | |
.. === ===

Keep in mind that d(Vbe)/dT is not constant; you can adjust it by
changing i(c), possibly achieving better cancellation / temperature
compensation. Also, Vce(sat) has a small positive tempco, which can
be handy too.

Cheers,
James Arthur
 
G

Gibbo

Jan 1, 1970
0
M

Marte Schwarz

Jan 1, 1970
0
It doesn't, at least not in my copy. See
look at Page 15 on the top.

Sorry, i mised the sentence before :) But look at ST's LM311 datasheet
there is a silicon transistor used.

Marte
 
J

James Arthur

Jan 1, 1970
0
James said:
[snip]

Why does the National data sheet show Q1 as being either Si or Ge then?
It doesn't, at least not in my copy. Seehttp://www.national.com/ds/LM/LM111.pdf

You're right it doesn't (I was working from memory). It't the TI one
that shows both Qs as Si.

http://focus.ti.com/lit/ds/symlink/lm111.pdf

And the circuit does indeed work with both Qs as Si but with (obviously)
a much lower voltage as FB details.

We agree the circuit works per Fred's description when Q1 is
silicon, and we agree the output voltages are quite different for
silicon versus germanium Q1. I measure Uo = 18.18mV with Q2=PN2222a,
Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
V. With Q1=2n5772, Uo = 6.88mV.

However, the application circuits then make no sense, as this
massive a change in reference voltage would obviously have a huge
impact on the "Low Voltage Adjustable Reference Supply" and the
precision waveform squarer I identified earlier. The reference
voltage would then be only a few times the comparator's offset
voltage, for one thing, and massively dependent on Vbe matching, for
another.

Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

Cheers,
James Arthur
 
J

James Arthur

Jan 1, 1970
0
Sorry, i mised the sentence before :) But look at ST's LM311 datasheet
there is a silicon transistor used.

Marte


No, at lest in this datasheet ST correctly shows the voltage
reference application circuit (pg. 9) with a 2n1304, which is
germanium:

http://www.st.com/stonline/products/literature/ds/4848/lm311.pdf

T.I. is alone in showing a silicon transistor, which doesn't work in
their circuits for the reasons I described. What good is a "precision
squarer" circuit with an unpredictable output?

Best wishes,
James Arthur
 
G

Gibbo

Jan 1, 1970
0
James said:
We agree the circuit works per Fred's description when Q1 is
silicon, and we agree the output voltages are quite different for
silicon versus germanium Q1. I measure Uo = 18.18mV with Q2=PN2222a,
Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
V. With Q1=2n5772, Uo = 6.88mV.

However, the application circuits then make no sense, as this
massive a change in reference voltage would obviously have a huge
impact on the "Low Voltage Adjustable Reference Supply" and the
precision waveform squarer I identified earlier. The reference
voltage would then be only a few times the comparator's offset
voltage, for one thing, and massively dependent on Vbe matching, for
another.

Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

You could well be right. I never said it was a good circuit, just ageed
about it *could* (indeed does) work with silicon Qs. Well spotted re
the "dot"!
 
M

Marte Schwarz

Jan 1, 1970
0
Sorry, i mised the sentence before :) But look at ST's LM311 datasheet
No, at lest in this datasheet ST correctly shows the voltage
reference application circuit (pg. 9) with a 2n1304, which is
germanium:
T.I. is alone in showing a silicon transistor, which doesn't work in

Sorry that I did blame the wrong publisher. You're right.

Marte
 
M

Marte Schwarz

Jan 1, 1970
0
Hi James,
We agree the circuit works per Fred's description when Q1 is
silicon,

I don't see, why Helmut should be wrong here.
and we agree the output voltages are quite different for
silicon versus germanium Q1. I measure Uo = 18.18mV

did you really measurements or simulations?
with Q2=PN2222a,
Q1=KTC3198, with drift of roughly 0.3% / C, line regulation = 0.9% /
V. With Q1=2n5772, Uo = 6.88mV.

Try Q1 the more powerful transistor and Q2 the smaller ones. So Uo will be a
little better.
However, the application circuits then make no sense,

I agree, one look on the offset voltage and all the dreams are gone :)
precision waveform squarer I identified earlier.

This may be the next topic: What is a "precision squarer" for?
Since the LM311 was originally a National Semiconductor design
("LMxxx," after all) I conclude T.I. copied National's datasheet, that
some helpful T.I. app engineer noticed the obsolete transistor, and
erred in recommending a substitute Q1. Perhaps he's the same fellow
who left the "dot" off the precision squarer circuit?, which, as
drawn, produces no useful output!

Right, I saw so many application hints in datasheets, with obvious mistakes
;-)

Marte
 
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