Increasing the current handling capability of linear voltage regulators - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: October 09, 2022, 12:28:30 pm
content: Hello guys, I want to calculate the value of resistor R1 in the figure. But while doing the calculation I face the following problemFirst, see the pictures below,In my calculation, VBE =0.7 = I1* R1 to turn the transistor on. But here, but Here I1 = IREG was taken, so thatR1 * IREG = 0.7 volts.I did not understand this part, In my calculation,At the junction,I1+IB = IREGIREG - IB = I1Now the Beta of the transistor is known and From the output current I can calculate IC and I can find IB from IC.a) So I1 and IREG are not the same. So how did equal this two here?b) Where can I find the maximum value of IREG?c) To calculate the output current I need at least two quantities from ( I1, IB, IREG). How i can determine these two quantities

author: strawberry
author_rank: Super Contributor
timestamp: October 09, 2022, 02:45:44 pm
content: beta and Vbe is temperature dependentwith couple transistors it could be more predictablemaximum value of IREG usually 1A for 7805or 0.65V / 3R = 0.217A + Ib

author: ledtester
author_rank: Super Contributor
timestamp: October 09, 2022, 04:08:17 pm
content: QuoteI'm not sure why you saying I1 and IREG are equal. The formula for R1 was not formatted correctly in the datasheet. it should read:$$R_1 = \frac{\beta V_{BE}}{(\beta+1)I_R - I_O}$$This can be derived using the equations:$$I_O = I_R + I_C \quadI_R = I_1 + I_B \quadI_C = \beta I_B \quadV_{BE} = I_1 R_1$$QuoteNote that the VBE for power PNP transistors will generally be larger.

author: David Hess
author_rank: Super Contributor
timestamp: October 09, 2022, 07:06:31 pm
content: The current gain and Vbe of the transistor vary considerably and the current limit of the regulator is not well defined, so the value of the resistor is not that critical; any value you calculate will be wrong. The circuit shown below performs better with the diode compensating for the transistor's Vbe; now the ratio of two resistors more closely controls the ratio of current between the transistor and regulator.In practice either circuit works fine. I have never had a problem with it.

author: Marian_elf
author_rank: Contributor
timestamp: October 10, 2022, 06:58:57 am
content: Quote from: khatus on October 09, 2022, 12:28:30 pmI'm going a bit backwards just to answer that question because the first thing you should do is to assign a maximum regulator current, and from most TO220 devices you can get about 1,5A with proper heatsinking, besides in the first page of that very same datasheet it is said the current can go up to 1,5A, so that is your regulator current for calculations.The next thing to do is to set the maximum wanted output current, that will allow you to determine the PNP current knowing that IQ1=Iout-Ireg.So if let say you want 3A at the output, then you get IQ1=3-1.5, so the PNP's IC is 1,5A.This helps you in the next thing to do, which is to assign a Hfe value.Current amplification in a BJT is anything but a fixed value, but i suspect going the rabbit hole with temperature and Vce and other variations of Hfe is not the pint here, so you just search in the datasheet the estimated Hfe for 1,5A.The one datasheet i could find for 2N6133 does not even have graphs, but they do provide a minimum and a maximum expected Hfe for it.What you should always do is chose the minimum so that you can always cover the worst case.So for 2N6133 the minimum Hfe is 20, and this is your number.The last step is to assign Vbe a value.More modern power transistors have graphs with detailed info on just about anything you need, and ideally you should find the expected Vbe at the IC needed.Unfortunately 2N6133's pdf does not clarify this either, they just say it to be at maximum -1,4V ( minus sign meaning is a PNP transistor ) which to me seams to much for a single silicone BJT, but since we do not have any other info and because as i said one should always consider the worst case, this is what you have to work with.Only now you have what you need for those calculations.So:Quote from: khatus on October 09, 2022, 12:28:30 pmThe proper way to name those currents is to say R1 current ( namely the current that flows through some part ) and regulator current, and no, they are not the same, they cannot be, just because Ireg is the sum of Ib and IR1. Thus IR1=Ireg-Ib.You now know IC to be 1,5A, you need to know Ib, and you also know ( or should know ) that Ib=Ic/Hfe.Thus 1,5/20=0.075A ( 75 mA ).IR1=Ireg-Ib=1.5-0.075=1.425A.R1=VR1/IR1, and knowing that VR1 is also Vbe of the series pass PNP transistor:R1=1,4/1.425=0.98Ohms, round it up to 1 Ohm ( 1R ).

author: gbaddeley
author_rank: Regular Contributor
timestamp: October 10, 2022, 09:09:08 am
content: Set R1 to 1 Ohm for a 1A regulator. As the load current ramps up, the transistor will turn on at about 0.6 - 0.7 A, and pass virtually all the current above this value. Whatever I b-e is required will pass through the regulator and load and does not affect regulation, if the total I regulator does not go above 1 A.HTH

author: Terry Bites
author_rank: Super Contributor
timestamp: October 10, 2022, 10:09:11 am
content: Ireg is your choice. Ireg is Ibase + what you'd like to set for the regulator. So for a 1 amp regulator you might want to choose an Ireg of say 0.75A. That gives you 25% margin. You might want to be more or less conservative. 50%? When Ireg*R1= Vbe the power transistor will wake from its slumber. You want it to start conducting well before the regulator maxes out and starts self limiting. A plain old PNP is not a good choice. A Darlington or Sziklai pair are a good alternnative. You need to add a Vbe or 2x Vbe for a Darlington or a Vgs ( to suit MOSFET Id) to the regulator's dropout voltage.If you use a Darlington you can ignore the base current and beta R=2Vbe/Ireg you don't have to be precise about it.A Sziklai pair only needs one Vbe(ish) to turn on.The wanderings of beta and Vbe within the transistor(s) SOA(s) won't have a dramatic effect.Most of the outboard transistor schemes lack short circuit protection, an invitation to thermal run awayThe regulator can only protect its self. See my replies atwww.eevblog.com/forum/projects/design-help-reference-voltage-with-increased-current/msg4429831/#msg4429831Simple short circuit protection for the outboard transistor is easy to implement. I will work with any three terminal reg or ref.I'd put a fuse or polyfuse on the raw input and maybe a crowbar if failure is going to be expensive, tearful or dangerous.The calulations shown in previous posts are for a minimum resistor value. This may be thermally efficient but might not the best solution.Its worth thinking about the balance between regulator current vs transistor current to keep both within their SOA.Good heat sinking is a must.

author: khatus
author_rank: Regular Contributor
timestamp: October 10, 2022, 02:07:12 pm
content: Which book is it? I want to know the name of the book.

author: khatus
author_rank: Regular Contributor
timestamp: October 10, 2022, 02:23:20 pm
content: I want to clarify one thing, the input current of voltage regulator and the output current of voltage regulator will be the same for a particular load?

author: ledtester
author_rank: Super Contributor
timestamp: October 10, 2022, 02:48:07 pm
content: To a first approximation, yes, but consult the datasheet for the quiescent current (I_Q) or in the case of the LM317 I_Adj.

author: blackdog
author_rank: Frequent Contributor
timestamp: October 10, 2022, 05:21:07 pm
content: Hi khatus,Look at the picture i edited.Kind regards,Bram

author: MrAl
author_rank: Super Contributor
timestamp: October 11, 2022, 05:34:45 am
content: Hello,The whole idea with these current boost schemes is than the LM317 handles some of the current and the transistor handles some of the current, and when the LM317 goes into current limit that limits the base emitter current in the transistor and thus limits the collector current. So the LM317 current limit still limits the total output current although now it is higher.One thing to think about here though and that is that there are other regulators that are the same as the LM317 but they handle more current. One of them handles 5 amps for example, and that is with no extra parts or worries of what the Beta is doing or whatever. Im sure you will find several on the web with a little search. I used a 5 amp device in a power supply project several years ago. I added some extra parts anyway though for better temperature characteristics and more precise regulation and fine output voltage adjustment.One thing to keep in mind with all these single chip linear regulators and that is that the voltage reference is on the SAME chip as the pass transistor. What that means is that as the pass transistor heats up, the voltage reference also heats up, and there is no way to prevent that because they are on the same die.What this in turn means is that the temperature characteristics of the voltage regulation is not as the data sheet would have you believe. That is because the temperature characteristic of the voltage reference is given and since that is heated as the current increases, the voltage variation is more than you might expect. In fact, it could be worst case for a design that goes up to near full current of the LM317 (or similar IC).This could affect some applications such as Li-ion chargers that rely on a very stable voltage output. The voltage could vary more than the Li-ion specification would suggest.Let me qive a quick example.Say you have an LM317 that outputs 4.200 volts at no load and ambient temperature of say 25 degrees C.As the device is loaded and heats up (even with a heat sink) if the voltage reference was external it would still be close to 25 C, but because the die heats up it could be much hotter now. With that change in temperature the voltage reference voltage changes and then so does the output of the LM317. To find out how much, just check the data sheet for the variation of the voltage reference and then calculate the new output voltage of the LM317. it's not a huge amount, but it is still significant in applications that need a very stable output voltage. There were other reasons, but that's one reason why i added external voltage regulation to my past project.

author: khatus
author_rank: Regular Contributor
timestamp: October 14, 2022, 02:48:02 pm
content: I want to know how the above equation arrives.I Have tried but I can't make itFor transistor Q2,VBE(Q2) = 0.8 = ISC × RSCFrom output side of the regulator we get,ILOAD = IREG + IC1From input side of the regulator we get,IREG = IB1 +IC2 +IR1IB2 =( β+1)× IE2

author: ledtester
author_rank: Super Contributor
timestamp: October 14, 2022, 03:11:20 pm
content: In my previous post I wrote:Quote from: ledtester on October 09, 2022, 04:08:17 pmThe derivation is:$$\begin{align}I_O &= I_R+I_C \\&= I_R + \beta I_B \\&= I_R + \beta(I_R-I_1) \\&= (\beta+1)I_R - \beta I_1 \\&= (\beta+1)I_R - \beta \frac{V_{BE}}{R_1} \\\beta \frac{V_{BE}}{R_1} &= (\beta+1)I_R - I_O \\\frac{1}{R_1} &= \frac{(\beta+1)I_R - I_O}{\beta V_{BE}} \\R_1 &= \frac{\beta V_{BE}}{(\beta+1)I_R - I_O} \\\end{align}$$

author: khatus
author_rank: Regular Contributor
timestamp: October 14, 2022, 04:30:57 pm
content: The derivation of the first circuit and second circuit is same?

author: khatus
author_rank: Regular Contributor
timestamp: October 21, 2022, 05:13:50 pm
content: Here is another example of an Outboard Current Boost Circuit. I am trying to find out the values of Rsc and R1 and R2.But did not succeed. Can anybody remove the term IR1 in equation 5 so that only IREG, ILOAD, and IC2 remain in the equation??In case my solution is wrong then please solve this equation and find the values of Rsc, R1, R2, and R3 for me.

author: ledtester
author_rank: Super Contributor
timestamp: October 22, 2022, 03:03:11 am
content: I would just use a symbolic solver like Wolfram Alpha or Sympy.Here is an example of using the online Sympy shell athttps://live.sympy.org/to solve the original problem:In the solve() statement the list of variables to solve for is given as [R1, I1, Ib, Ic]. Even though you are only interested in R1 the other variables are listed because you don't want them to appear in the expression for R1.

author: ledtester
author_rank: Super Contributor
timestamp: October 22, 2022, 12:20:12 pm
content: Regarding the more complex circuit, the problem is that there are not enough constraints to determine the currents.I've annotated the schematic so that all currents and nodes are labelled.There are 9 nodes: A, B, C, D and (1), (2), (3), (4) and (5) which give 9 equations via KCL.Each transistor yields two equations: Ie = Ic+Ib and Ic = beta*Ib.So there are total of 15 equations among the currents.But there are 18 current variables:- 9 for the 3 transistors (Ie, Ic, Ib for each transistor)- Ix, Iy, Iz, Iw- Ir1, Ir2, I3- Ireg, IloadEven if you take Iload as a given there are still 17 variables to solve for.When I row-reduced the equations I got a rank of 14.

author: MrAl
author_rank: Super Contributor
timestamp: October 24, 2022, 01:05:00 am
content: Quote from: khatus on October 21, 2022, 05:13:50 pmHello,Not sure what you are trying to analyze there but there is an approximation we can use.Pay most attention to the resistors R1, R2, and Rsc, and the approximate Vbe voltage of 0.7 volts. R3 could be superfluous because it is in series with a collector.All three resistor values are chosen in about the same way, on the basis of the resistor value and 0.7 volts, so the turn on current threshold for any transistor is:Ith=0.7/RbeTo start, set R1 for the max current of the linear three terminal regulator, which is probably around 1 amp and that is the current the linear regulator will start to limit its own output current. Using the rule above and 1 amp we have:1=0.7/R1and clearly R1 should be 0.7 Ohms. This is only if the max current of the linear is 1 amp.Next set Rsc for the max current output of the circuit minus the max current of the linear regulator. Since the linear regulator puts out 1 amp (as above) if we want 5 amps current limit we set Rsc for 4 amps again using the rule above. This means:4=0.7/RscThis comes out to Rsc=0.7/4 Ohms which is 0.175 Ohms.R2 probably isnt too picky but would depend on the Beta of Q1 and the size of that transistor, and sometimes smaller transistors are used for that kind of function so the collector current would be more limited than the other transistors. If you want Q3 to turn on with 100ma for example, then we would have:0.1=0.7/R2and so R2 would be 7 Ohms. The catch is that the Beta of Q3 would have to be at least 40 if the current capacity of the input source could put out the entire output current plus 4 more amps, for example. This may mean you would want to set R2 for 200ma instead of 100ma so the gain of Q3 can be even lower. Although it is probably better than that we have to be aware that the collector emitter voltage probably has to go to somewhere close to the saturation voltage and so the gain may be much lower.As to R3 you may even be able to leave that one out, unless Q1 gets too hot. This would depend on the output voltage and the available supply voltage as well as the type of transistor.Note that if R1 is not set for the FULL current of the linear regulator, when the circuit starts to get into the current limit mode it will actually force more current through the linear and so it wont actually be regulating at 4 amps it would be higher.So that should get you started.In any circuit that uses a base emitter voltage as part of the current determining calculation the resistors may need a slight adjustment, and also you probably have to use standard values unless you are willing to parallel some values to get the value you really want.Good luck and maybe you can report back here on your results.