Reference Voltage Schematic

"Marte Schwarz" schrieb im Newsbeitrag news:45d61f9b$0$23148$ snipped-for-privacy@newsspool1.arcor-online.net...

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.

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

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.

----- Original Message ----- From: "Fred Bloggs" 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?

--
Gibbo

This email address isn\'t real.

Hello Gibbo,

Don't believe all what's printed.

Best regards, Helmut

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.

--
Gibbo

This email address isn\'t real.

----- Original Message ----- From: "Marte Schwarz" Newsgroups: sci.electronics.design Sent: Saturday, February 17, 2007 12:27 PM Subject: Re: Reference Voltage Schematic

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

Hi Fred,

right here. But guess Vce,sat with a few mV in case of two Si-transistors?

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.

At least here the simulations with LTSPICE disagrees with you. Making a DC sweep with V1 creates a nearly linear function of Ib(Q2)

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

It doesn't, at least not in my copy. See

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

[snip]

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

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

--
Gibbo

This email address isn\'t real.

Hi James,

look at Page 15 on the top.

or with a schottky as Q1 as I mentioned before.

Marte

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

Marte

See

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

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 their circuits for the reasons I described. What good is a "precision squarer" circuit with an unpredictable output?

Best wishes, James Arthur

...........^^^^

--James

See

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"!

--
Gibbo

This email address isn't real.

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

Marte

Hi James,

I don't see, why Helmut should be wrong here.

did you really measurements or simulations?

Try Q1 the more powerful transistor and Q2 the smaller ones. So Uo will be a little better.

I agree, one look on the offset voltage and all the dreams are gone :-)

This may be the next topic: What is a "precision squarer" for?

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

Marte