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Help understanding slow start circuit and tracking circuit in simple dual rail power supply

carebare47

Oct 21, 2010
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Hello,

I am currently studying electronics and I am writing a report on the project I have undertaken this year, which is a dual rail tracking power supply. I have got a working prototype through trial and error and playing around on simulation software. I understand how most of the circuit works but am having trouble with two parts, the slow start circuit on the positive rail and the tracking circuit.
I know that the slow start circuit works by R9 charging C4, which gradually turns on Q1. The thing I don't understand is how Q1 gradually turning on makes the circuit slowly start up. As Q1 becomes more conductive does it not just bypass R3 and R11? However I think about it it seems like Q1 should slowly turn the circuit off.
Also, the tracking circuit. I get that the OPA445 somehow drives the LM337, but I don't know what signal it gives the LM337 and I don't quite understand how it knows how much of whatever the signal is to give to the LM337.

Any help in understanding this would be greatly appreciated. As I've said I've got a prototype working on breadboard, I just don't fully know how it works.

Many thanks,

Tom
Dual rail psu How it works.png Dual rail psu How it works.png
 

carebare47

Oct 21, 2010
66
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Oct 21, 2010
Messages
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Also could somebody explain what R4 is used for? Something to do with creating an equal resistance with the op-amp, I'm not sure. Also if a higher resolution image is needed I will upload one. Any help will be greatly appreciated :)
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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OK. The first thing you need to understand is that the LM317 produces an output voltage that's equal to the ADJ pin voltage plus the reference voltage (about 1.2V).

So the voltage at the ADJ pin determines the output voltage, and there is a fixed offset between them.

(In your schematic, the ADJ pin is labelled with an upside down 'T' which is probably supposed to represent 0V. It should be labelled ADJ.)

The current flowing into the ADJ pin is quite low and can usually be ignored.

The standard adjustable regulator circuit just has a voltage divider across the output, with the ADJ pin connected to the centre of the voltage divider.

Because the LM317 ensures that there's 1.2V across the top resistor, the current through the resistors is known, and this produces a known voltage across the bottom resistor; the output voltage is 1.2V greater than that voltage. That's how the simple regulator works.

The voltage divider resistors in the basic regulator circuit correspond to R5 and R3+R11 in your circuit. The LM317 ensures that there will always be 1.2V across R5, and this causes a constant current flow in R5, which can be calculated using Ohm's Law:

I = V / R
= 1.2V / 240 ohms
= 0.005 amps
= 5 mA.

This 5 mA current flows through R5, R3 and R11 - assuming the slow start circuit isn't there. In this case, the ADJ pin voltage is equal to the voltage drop across R3+R11 caused by 5 mA current flow. For example if R3+R11 = 2k, the ADJ pin voltage will be:

V = I R
= 0.005 amps × 2000 ohms
= 10 volts

And the output voltage is always 1.2V higher than the ADJ pin voltage, so it will be 11.2V.

Q1 is being used as an emitter follower. It's upside down, because it's a PNP, but that's how it's used. Depending on the voltages on Q1, some of the 5 mA current from R5 can flow into Q1's emitter and out its collector, and this will drag down the ADJ pin voltage.

In other words, Q1 acts as an emitter follower and pulls its emitter down towards the 0V rail, with a constant voltage of around 0.7V between the emitter and the base. Since it's a PNP, the emitter is more positive than the base.

So the voltage on Q1's base determines the voltage on its emitter, with a 0.7V offset. And the voltage on Q1's emitter determines the output voltage, with a 1.2V offset. So Q1's base voltage determines the output voltage, with a 2V (approx) offset. The output voltage will be about 2V higher than Q1's base voltage.

The output voltage will also be limited by R3+R11 because the total current through R5 is set to 5 mA (from 1.2V across a 240 ohm resistor). Q1 can pull the output voltage low by sucking some of that current and pulling ADJ lower, but even with no current flowing into Q1, the ADJ voltage is limited by the voltage produced across R3+R11 by a current of 5 mA flowing through it. So Q1 can only reduce the output voltage below the voltage determined by R3+R11.


OK, to the soft start circuit. Initially on power-up, C4 is discharged, so Q1's base is at 0V, so Q1 sucks current and pulls ADJ down to about 0.7V. So the output voltage will be about 2V.

With 0.7V base-emitter voltage, there will be 0.7V across R9, which causes a current flow in R9 of I = V / R = 0.7V / 47k = 15 µA. This current flows into the slow start capacitor C4 and it starts to charge up.

As this happens, Q1's emitter voltage (and the output voltage) increase; because the charge current is set by Q1's base-emitter voltage (roughly constant at 0.7V) and R9, the charge current is roughly constant so this increase should be roughly linear, not the traditional R-C charge/discharge curve shape.

As the ADJ voltage increases, the current through R3+R11 increases, leaving less current for Q1. At a certain voltage (determined by R3+R11), the R3+R11 current will reach 5 mA and there will be no current left for Q1. Q1 now stops having any effect; the output voltage is determined by R3+R11. A tiny bit of current flows through R9 and into C4, and the C4 voltage slowly approaches the ADJ voltage, then everything is steady-state.

If you suddenly increase the output voltage setting by increasing R3+R11, the soft start circuit will kick in again. Initially the output voltage will jump up by 0.7V (Q1's base-emitter voltage) then another slow-start ramp-up will occur as C4 charges to the new voltage.

If the output voltage is reduced, or the supply is turned OFF, D7 ensures that C4 is quickly discharged to fairly close to 0V, ready for the next slow start.


I've run out of time to explain the tracking circuit, but it's pretty simple. It's based on an op-amp inverting amplifier. The op-amp's non-inverting ("+") input is tied to 0V so it tries (through the negative feedback loop) to keep its inverting ("-") input at 0V as well. The inverting amplifier stage's input is the right hand side of R4 (which is the positive output) and its output is connected to the ADJ pin on the negative regulator. Its gain is determined by the ratio of R2+R13 to R4.
 

carebare47

Oct 21, 2010
66
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Oct 21, 2010
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That was a wonderfully detailed reply, thank you very much for the time you put into it and for the amount you managed to teach me in one go! Your response has been incredibly useful, thank you again.
 
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