Understanding Bandgap Voltage Reference Circuits

[likephysics]

I am not able to understand how a bandgap voltage reference works.
The simplest ckt I found was here - (figure 2) - http://www.national.com/rap/Application/0,1570,24,00.html

You have 2 branches of diodes, in one branch rich (high) current flows and in the other lean (small) current flows.
But how will this help temperature compensation?

 

[es1]

I liked this link
http://amesp02.tamu.edu/~sanchez/689-bandgap-fikret.pdf

"If two quantities with opposite temperature coefficient are added with proper weighting, the resultant quantity theoretically exhibits zero temperature coefficient."

Regarding figure 2: Assume T increases, the balance is the drop in the output of R1 due to decreased Vf of string D[1-8] (decreasing Vout so negative tempco) against the decrease in Vf of the string D[8-13] (increasing Vout so positive tempco).

If all the diodes are the same then the diodes with more current will have their Vf drop less for the same increase in T. The differences in current balance out the different number of diodes in the strings. The strings need a different number of diodes to make the Vout signal.

Hope that helps some.

 

[likephysics]

es1, that helped a bit.

Here's what I understand -

You have 2 branches with one diode in each branch.

The Vbe decreases with increase in temp.
Delta Vbe increases with increase in temp.
These quantities need to be added up to cancel temperature variation.

So you...

1. Find delta Vbe (maybe using a diff. amp)
2. Measure Vbe of one diode (using an opamp)
3. Add the delta Vbe to the measured Vbe (op amp as summer)
4. Amplify this

.

Am I getting it right?

 

[es1]

Is Vbe = Vf on a diode?

I am just wondering if I am looking at the right circuit because in the document I looked at there were only diodes in fig 2, no BJTs.

The delta Vf of the two strings should stay constant because the delta Vf is Vout in fig 2 (the top cathodes of the two strings are shorted together). The point of the circuit is to keep Vout constant over temp.

Are steps 1-4 referring to the circuit in fig 2, the simplest one with two strings of diodes? There is no need to diff amps and summers via opamps. You get the diff and sum for free by KCL at the output of R1 and KVL around the loop that includes both strings of diodes.

 

[es1]

Maybe this will help:

Just for the sake of argument assume nominal Vf of the diode is not dependent on current but that the change in Vf with temp is. This just keeps the math a bit easier.

So then:

Vf=Vf0 + K(If)T

Vf0 = forward voltage at nominal temp and current
K = some gain parameter in V/C and it is dependent on forward current in the diode
T = temp of diode

So then KVL around the loop with the diodes and Vout is:

8(Vf0 + K(If1)T) = 7(Vf0 + K(If2)T) + Vout
Vout = (8*K(If1)-7*K(If2))*T + Vf0 [ We need the Vf0 to fall out so Vout !=0, this is why there are not the same number of diodes in the strings ]

Vout does not vary with T if 8*K(If1)-7*K(If2)) = 0, so pick R1 and R2 to create a If1 and If2 such that 8*K(If1) = 7*K(If2), which should be possible after you characterize the diode.

 

[uart]

Re - (figure 2) - http://www.national.com/rap/Applicat...570,24,00.html

For analyzing any of these circuits the tricky part is actually proving that the Vbe forward voltage has a temperature coefficient of about -2mV/C (at room temperature and typical operating current densities). If we take it as "given" that Vbe is approx 600mV with temp-co -2mV/C then the analysis of the diode string circuit is as follows.

[tex]V_o = 7 V_T \, \ln(\frac{I}{I_s}) - 6 V_T \, \ln(\frac{I}{49 I_s})[/tex]

(note that the current in the right branch is 1/49th of the current in the left branch).

Rearranging,

[tex]V_o = V_T \, \ln(\frac{I}{I_s}) + 6 V_T \, ( \ln(\frac{I}{I_s}) - \ln(\frac{I}{49 I_s} ))[/tex]

Using the subtraction property of log gives,

[tex]V_o = V_T \, \ln(\frac{I}{I_s}) + 6 V_T \, \ln(49)[/tex]

(Where "I" is the current of approx 1mA in the left diode string).

The first term in the above equation is the normal forward voltage of a single diode, so we assume it is approx 600mV with a temp-co of approx -2mV/C.

Now consider the second term, [itex]6 V_T \, \ln(49)[/itex]. At room temperature it evaluates to approximately 600 mV, and since V_T is the only temperature dependency of this term it has a positive temp-co equal to [itex](6 k \ln 49)/q[/itex], which is almost exactly +2 mV/C. (Note that the number of diodes and the relative currents would have been chosen to so as to make this so.)

* BTW. In building any of these bandgap circuits, the Vbe voltage of a BJT actually works better than typical diodes as they obey the ideal logarithmic diode equation more closely.