How to calculate resistors of bypass transistor?

Question

In the following circuit, I would like to understand the importance of the current limiting resistor at the base of Q1 (5K Ohm), and the function of the 500 ohm resistor at the base of the NPN transistor. Do they affect the maximum current that we specify for the regulator? I don't want it to catry more than 0.4A. I'll use 7812 regulator.

Are they optional? Because some other circuits on the internet don't use them like the following one:

I would appreciate it if you tell me a rule of thumb to calculate their values. The equations don't have to be accurate. Just easy approximate calculations are good for me.

Edit : Resources of images:

1. ST Datasheet
2. https://www.eleccircuit.com/boosting-regulator-current-for-ic-78xx-by-mj2955/#comment-162345

Answer 1

Do they affect the maximum current that we specify for the regulator?

Here is answer to your request ... Note that I don't have TIP73 in my database.
It was replaced by TIP3055.

DC Analysis with interactive simulator microcap v12

First case : Iload = 0

Second case : Iload = 10 A

You can see that the currents through the resistors are very light ...

The current through LM117 is 50 times lower at full load.
It is from 25 mA to 190 mA. With 20 V input, it is 204 mA.

Here is the simulation showing the input current to the LM117.
It shows the dependance of this current to the value of R11 ...
Curve with R11 = 5 kOhm is in "red". Curve with R11 = 10 kOhm is in "green".

Here is the power diagram for the output transistor Q3.

Note that the output voltage Vo is changed (the voltages on the simulations are for 10 A).
The maximum current should be ~ 7 A for a output voltage of 12.6 V.

NB: if TIP3055 replaced by a Darlington 2N6284 (160 W) ... 2N2905 is ok.
Be careful for powers across all components.
Short-circuit is not well designed.

Answer 2

No one has addressed some of the elephants in the room about that circuit (it's not viable for your use-case, so don't use it):

• An (obsolete) TIP73, at \$V_{\small{CE}}=4\:\text{V}\$ and \$I_{\small{C}}=5\:\text{A}\$ can have \$h_{\small{FE}}=20\$ (that's the datasheet minimum.)

  

  That's assuming you can keep its case temperate at \$25^\circ\:\text{C}\$.

  You've not specified a maximum load current. And that is a very important specification.

• You have specified the input as \$21\:\text{V}\to 22\:\text{V}\$ and the output voltage as \$12\:\text{V}\$. This means that the difference can be taken as \$10\:\text{V}\$. This will then be the TIP73's (obsolete) operating \$V_{\small{CE}}\$.

  Looking at the safe operating area chart:

  

  The maximum is \$I_{\small{C}}=8\:\text{A}\$. So, to be safer and to stay in keeping with explicit specifications on the datasheet, I'll take it as given that a new design specification of \$I_{{\small{C}}_\text{max}}=5\:\text{A}\$.

  Also, this means dissipation can exceed \$50\:\text{W}\$. To be safer, set a design specification for power dissipation of this device must be sufficient to safely meet \$60\:\text{W}\$.

  The absolute maximums, assuming \$25^\circ\:\text{C}\$ case temperature can be attained, say \$80\:\text{W}\$. So already this may be a serious issue as this is walking up very close to the absolute maximums for the device -- which should be avoided at all cost! Conservatively, a design should be half that much, or less. And it's already beyond that point. So this is no longer a conservative design.

• From the above, \$I_{{\small{B}}_\text{max}}=\frac{5\:\text{A}}{20}=250\:\text{mA}\$. And your schematic specifies a 2N2905 (another obsolete part) to supply this magnitude as continuous duty. It's \$V_{\small{CE}}\$ will be almost the same as for the TIP73, less one \$V_{\small{BE}}\$. So this means almost up to \$2.5\:\text{W}\$! And that device won't handle it. Not even if you can get the TO-39 can and find a tiny heat sink to stick on it. (I have some of those still laying around. The won't handle it.) Their absolute maximum specification says \$600\:\text{mW}\$, which is a darn-sight less!!

  And even if it could handle it, which it won't, the base current could be as much as \$5\:\text{mA}\$ given that much collector current. And \$5\:\text{mA}\cdot 5\:\text{k}\Omega=25\:\text{V}\$, which would be the worst case voltage drop across the \$5\:\text{k}\Omega\$ base resistor.

In short, the design is not only obsolete but it is not even right. There is no way anyone in their right mind could have designed that schematic with even so much as a load current of \$5\:\text{A}\$. The absolute most, and this would avoid considering operating temperatures above \$25^\circ\:\text{C}\$ -- unlikely -- might be \$1\:\text{A}\$ load current. But the 7812 is already specified to handle that current, alone.

(And there's no current-foldback protection added, either.)

So this entire idea was insane for your use-case. The added topology provides no inherent protections and it can't be used to extend the current compliance capabilities of the 7812, anyway. So there's no point considering it, further.

Find something else.

And write better specs.

Answer 3

Your second diagram isn't even a working circuit.

The colorful diagram from ElecCircuit is an intermediate circuit intended as a block diagram to understand the changes that are needed to modify and improve the previous circuit into a better circuit. It is not a working circuit itself. The web page explains what the modifications do, how they work, what they achieve, and finally the next diagram shows a full working circuit, that is identical to the datasheet circuit, except for minor differences in the resistor values, with the accompanied text how to calculate these.

Also there are various different ways of implementing outboard pass transistors. As many regulators have differences how they work internally, it might be best to refer to 7805 datasheets how to implement the outboard pass transistor, instead of relying on LM317 datasheet.