Voltage regulators - die pictures

[magic]

Anyone an idea what these structures do:

The green/red stripes?
Perhaps some resistors or capacitors

[Noopy]

Anyone an idea what these structures do:

The green/red stripes?
Perhaps some resistors or capacitors

Yes, these green/red stripes.

I would have stated the stripes are resistors but with the squares in the LT1084 and the LT1085 perhaps that gives some transistors... 

[Noopy]

Hi all!

Today I have a LM300 for you:




You can barely read LM300.




A nice old design.




The bright green circuit generates the voltage reference. The reference is based on the zener D1 which causes a constant current through Q4, Q6/Q5, R1, R2 and Q7. At the base of Q8 the voltage is 1,7V.
The transistors Q6/Q5 look like a current mirror but the current in the left emitter of Q5 is also flowing into the reference path.
The current through the left path of the differential amplifier and the current mirrored from the reference path are flowing through the collector of Q5 and are controlling the transistor Q2. Q2 is supplying the regulator. Low output voltages and high voltage drops increase the supply currents. The zener D2 limits the effect of the voltage drop. Probably that stabilizes the bias point.
The red part is the regulator if you short Pin 3 and Pin 2. Without the short you can also use the Transistor Q12 as a driver for an external power transistor.
The dark green part is the overcurrent protection.
The unnamed diode seems to protect the regulator from high voltages at the output.




You can find every part of the schematic on the die.
Some of them are quite interesting to identify:




The unnamed diode is a parasitic part in the collector area of Q10 created by the 1,4k resistor.




Why is the 520-resistor not connected directly to the Pin 3 metal contact? 
The 20k-resistor is quite small but has an n-type overlay. The junction gives us the zener D2.




The zener D1 is located near the ground connection in a p-type area.




Between the left collector of the pnp Q2 and the base of the npn Q1 there is a small resistor (=high value) which we can´t find in the schematic. Probably the resistor supports start-up.


https://www.richis-lab.de/voltageregulator04.htm


 

[magic]

The transistors Q6/Q5 look like a current mirror but the current in the left emitter of Q5 is also flowing into the reference path.
The current through the left path of the differential amplifier and the current mirrored from the reference path are flowing through the collector of Q5 and are controlling the transistor Q2. Q2 is supplying the regulator. Low output voltages and high voltage drops increase the supply currents. The zener D2 limits the effect of the voltage drop. Probably that stabilizes the bias point.

I don't think that's accurate. Q2 current is defined by Q4 and Q3: whenever Q2 collector current exceeds collector current of Q4, Q3 is pulled up and its emitter reduces voltage across Q1, therefore reducing Q2 base current.
Q6 and the left part of Q5 split the current of the reference divider in proportion to their areas between Q4 and Q3/Q1. This ensures that as long as the zener reference is active, the Q4/Q3/Q1 circuit is powered up and actively regulating Q2 current and that Q4 current and therefore Q2 current and therefore D1 current are held constant, or maybe drift predictably with die temperature.
The right side of Q5 is simply a common base transistor over the noninverting input transistor Q8. Q8 can draw whatever current it wants from Q5 and that current is then supplied to Q5 by Q3 emitter. Q8 operating current is defined by Ohm's law across the 2.2kΩ tail resistor, minus whatever is the current of the inverting input, determined by Q2. That's weird but I can't find any alternative explanation.

Between the left collector of the pnp Q2 and the base of the npn Q1 there is a small resistor (=high value) which we can´t find in the schematic. Probably the resistor supports start-up.

Startup requires pulling current from Q2 base. I don't think this resistor can do that, because where would that current go from the collector node? I think startup is the job of R9 and R7.

[Noopy]

Nice to hear from you, magic! 

Q2 current is defined by Q4 and Q3: whenever Q2 collector current exceeds collector current of Q4, Q3 is pulled up and its emitter reduces voltage across Q1, therefore reducing Q2 base current.

You are right, whenever Q2 collector current exceeds  collector current of Q4, Q3 is pulled up and reduces the current supplied by Q2.
I´m not absolutely sure but nevertheless I would say the dominant regulation path is the current flowing through Q5 and R9 and Q3 does only some equalizing or stabilizing or something like that.

The right side of Q5 is simply a common base transistor over the noninverting input transistor Q8. Q8 can draw whatever current it wants from Q5 and that current is then supplied to Q5 by Q3 emitter. Q8 operating current is defined by Ohm's law across the 2.2kΩ tail resistor, minus whatever is the current of the inverting input, determined by Q2. That's weird but I can't find any alternative explanation.

 
Q8 can sink whatever it has to sink. The voltage at the right emitter of Q5 will adjust itself so that the current can flow through it.
Q5 sums the current of Q8 and the current of the reference path.
I assume the current supply Q2 is depending on the error signal to give better regulation performance.

Startup requires pulling current from Q2 base. I don't think this resistor can do that, because where would that current go from the collector node? I think startup is the job of R9 and R7.

R9 and R7 can support start-up, right.
The unnamed resistor can support start-up too: it connects the base of Q1 (= the base of Q2) with a collector of Q2 (low on start-up).

[magic]

What happens to Q3 emitter current when Q2 collector current increases by 1µA?

[floobydust]

LM100 operation explained in AN-1 November 1967 "A Versatile, Monolithic Voltage Regulator". I found it in my 1976 Linear Applications Handbook. I'm not sure if it's on the web in 1973 as magic mentioned in the LM723 post.

https://archive.org/details/manuals-nationalsemiconductor?and%5B%5D=linear+applications&sin=&sort=-date

[Noopy]

I have to think about that (and do some research).
Thanks for your input! 

[magic]

Sure it's there. The note itself is dated 11/1967.

PDF for the lazy

[Noopy]

magic is right! 

Thanks for your explanation! 

[floobydust]

I found the explanation for the PNP lateral interesting, as to this day power PNP transistors on IC's seem to be avoided:

LM100 "... emitter follower Q3, and a level-shifting diode, Q1, have been added  to increase the effective current gain of the PNP transistor, Q2. This device is a lateral PNP which has a low current gain (0.5 to 5) but has the advantage that it can be made without adding any steps  or process controls to the normal NPN integrated circuit process. One collector of the PNP serves as a collector load for the error-sensing transistor, Q9. A second collector supplies current for the breakdown diode, D1. A third collector, which determines the output current of the other two, maintains a current nearly equal to the collector current of Q4 by means of negative feedback to the PNP base through Q3 and Q1..."

[magic]

lateral PNP which has a low current gain (0.5 to 5)

That's some very old tech, IIRC already in the '70s improved lateral PNPs were available with beta on the order of tens.
Another problem with them is low bandwidth (a few MHz) so they are avoided not just in output stages.

There are complementaly processes which produce high quality transistors of both polarities on one wafer, but they cost more. They are typically used in advanced and high speed analog where there is no other way, less so in stupid stuff like voltage regulators.

[floobydust]

You see LDO's with the PNP pass-transistor so I thought 44 years later, they are improved and not a cost hit. But chip amps are still using quasi-complementary output stages, like LM3886 and TDA2040 etc.

LM100 redesign was the LM105/LM305/LM376 which I have never seen out in the wild. Something about the National Semiconductor voltage regulators of the late 1970's - many of them were flops. I suspect litigation or Fairchild was winning.
But the LM340 seemed to turn things around for them, that was a hit.

[magic]

Actually, I have a blown 7.5A LDO here, MIC29750. This one seems to use about 9~10 thousand paralleled lateral PNPs conducting <1mA each and it's huge - 5.5×3.5mm. They also claim some "proprietary superbeta process" and per the ground current spec, beta is indeed about 200 at low currents and down to 70 at 7.5A. Supposedly one can increase beta of lateral PNPs by introducing an additional processing step which selectively adds extra doping to their emitters.

MIC29750 isn't exactly cheap, though

I suspect that those PNP LDOs you see still use more die area than NPN parts with similar current rating and cost more. LM3886 likely used quasi-comp to save die area. When you look at opamps with complementary output stages on noncomplementary process, the PNP usually is larger, sometimes ludicrously so (LT1115/1028).

[David Hess]

You see LDO's with the PNP pass-transistor so I thought 44 years later, they are improved and not a cost hit. But chip amps are still using quasi-complementary output stages, like LM3886 and TDA2040 etc.

They require a more expensive process and beta limitations mean using a Darlington configuration anyway so might as well use an NPN output device.  There are LDOs built on CMOS processes but they are more expensive for the same current because their area is larger.

[Noopy]

We need more power! ...and a LM317. 




LM317K gives you at least 1,5A.




A big heatspreader. 
The TO3 is quite big but has with 4°C/W a higher thermal resistance than the TO220 (3°C/W).




Nice! 




The schematic and the structures are quite similar to the LM317 built by National Semiconductor. The power transistor is a bit more than twice as big.




Some nice pictures of the fuses.
ST contacted  the pads with two needles. Interesting...
You can see the residues of the blown fuses and you can see a small point damaged by the thermal energy.


More pictures here:
https://richis-lab.de/LM317_02.htm

 

[magic]

Zeptobars recetly said he has never seen a genuine ST chip without the ST logo. Whom should I trust?

Considering those test structures and numbers being the same as on ST NE555 and the 555 being almost identical to the SGS version, I think you are right.

[Noopy]

Perhaps the LM317 has no ST logo because it's a very old and not genuine ST design. It looks very similar to the National Semiconductor LM317 and who knows who really invented this design. Perhaps ST wasn't very proud about the LM317.
Of course I'm just guessing...
Nevertheless I'm pretty sure it's a genuine ST part.

...this one also has no ST logo:
https://richis-lab.de/ECU01.htm

[David Hess]

If they had a shortage, then they might buy chips or completed devices from a competitor and package or mark them as their own.

[gman76]

Wow, that LM300 layout probably done in rubylith. Ah, the days of plot checks.

[Noopy]

If they had a shortage, then they might buy chips or completed devices from a competitor and package or mark them as their own.

I´m pretty sure that is a ST chip because there are the typical ST marker for checking process quality. 

[Noopy]

Let´s take a look into a 7805!




The TDB7805 is a voltage regulator built by Siemens.




The die is quite small leading to a maximum power dissipation of 15W.




The edge length of the die is 1,8mm.




The schematic is quite interesting. The regulator itself is a darlington transistor (red). There is a overcurrent protection (light green), a second breakdown protection (light green, Z2/R3) and a overtemperature protection (grey).
The light blue part is the voltage amplification stage of the error amplifier. It uses the dark blue current source. Apparently they needed a overcurrent protection (orange)...
The bandgab reference (dark green) is the most interesting part.




In the current mirror T13/T14 the negative temperature drift of the forward voltages is canceled out leaving the smaller positive temperature drift of the temperature voltage. The resistors R13/R15 amplify the positive temperature drift so it can compensate the negative temperature drift of the transistors T10/T11. So we get a reference voltage with a very low temperature drift across T10/T11/R13.
The voltage at the output of the TDB7805 is connectet to T10. Since the reference is still constant a deviation of the 5V is transfered to T15 leading to a regulation of the output current.




Taking a closer look we see that there is also a overcurrent protection for the predriver T2 (R3*).




That´s also interesting: There are a lot resistors that are not in use but you can´t find R18 of the feedback voltage devider. It seems the feedback works with 5V and you can switch these resistors in to get higher voltages. 




R13 is quite long to amplifiy the small positive temperature drift of the current mirror. You can swith between two long connection electrodes. In each electrode there is a small via which can be moved to vary the resistance and get a temperature drift as small as possible.


More pictures here:

https://www.richis-lab.de/voltageregulator05.htm

 

[magic]

T17 is not overcurrent protection, it actually amplifies T16 current if there is enough voltage drop at R17. More likely it's an emitter follower and R17 provides bias current to T16.

R13 has to compensate for thermal drift of T10,T11 and also T15 and T16 because the reference voltage is between GND at T10 base.

Since there are four BE junctions in the reference "stack", output voltage is 4·1.25V = 5V. Checks out.

[Noopy]

T17 is not overcurrent protection, it actually amplifies T16 current if there is enough voltage drop at R17. More likely it's an emitter follower and R17 provides bias current to T16.

Thanks for you input, magic! 
That sounds reasonable. T17 is quite big. I didn´t see that.

R13 has to compensate for thermal drift of T10,T11 and also T15 and T16 because the reference voltage is between GND at T10 base.

I don´t think the reference voltage is between GND and T10 base. The widlar reference typically generates the positive tempco part above the current mirror and the negative tempco part is added by a vbe.
Because of that I´m pretty sure the green and the red part add up to the reference and the voltage between T15 base and GND is the error signal controlling the VAS and the linear regulator. (OK, error signal actually not perfectly right but the voltage is no part of the reference)
As T2/T1 the Vbe of T15/T16 is not relevant for the reference...
Because of that there are only two pn-junctions to compensate. With the normal tempcos -2mV/°C and +0,085mV/°C that correlates quite well with the resistors factor R13/R15 which has to amplify the +0,085mV/°C. Four times -2mV/°C would need much more amplification.

Since there are four BE junctions in the reference "stack", output voltage is 4·1.25V = 5V. Checks out.

The 5V could be seen across T10/T11/R13/T14/R15 but that´s not the reference imho.

[magic]

T10 base is connected to the output. It surely needs to be exactly 5V
T15 base voltage decreases with temperature so R13 needs to increase to compensate for it. Otherwise T15 would be turned on too much and pull the output down.

Voltage at R12 and R13 is similar (two diodes above GND) so the ratio of their currents equals the ratio of their resistances. This ratio seems quite high, maybe 10x or 20x. T13 and T14 are identical, so the difference of their Vbe (R15 voltage) is probably at least 50mV at room temperature and its room temperature drift is 50mV/300K = +0.17mV/K.