Two transistor voltage regulator analysis - Page 1
Forum section: electronics / beginners

author: Chris.M
author_rank: Contributor
timestamp: May 09, 2013, 01:31:49 am
content: Hi,I have extended the previous circuit I was trying to understand with an additional transistor to form a two transistor voltage regulator.https://www.eevblog.com/forum/beginners/relationship-between-transistor-base-current-and-emitter-resistor-voltage/msg227682/#msg227682Would someone mind taking a look at my analysis and giving me a little guidance on how to proceed.If we start with the path through R1, Q1, R2+R3:VCC = (IR1*R1) + Vbe + ((Hfe+1)*IbQ1*(R2+R3))And for the path through R1, Q2 and D1, D2, D3:VCC = (IR1*R1) + VQ2+ V(D1+D2+D3)We know that V(D1+D2+D3) will be approx 2.1v as they are forward biased through R4.If I equate the equations above, I would get (IR1*R1) + VQ2+ V(D1+D2+D3) = (IR1*R1) + Vbe + ((Hfe+1)*IbQ1*(R2+R3)).We can cancel out (IR1*R1) from both sides, and use 2.1V for V(D1+D2+D3) and 0.7V for Vbe.This leaves me with: VQ2+ 2.1 = 0.7 + ((Hfe+1)*IbQ1*(R2+R3)).Now is where I'm a little stuck. I need to get some extra information to replace VQ2, but I'm not sure what.Perhaps I've gone about the analysis the wrong way. Any hints on how to proceed without giving away everything?Thanks.

author: qno
author_rank: Frequent Contributor
timestamp: May 09, 2013, 08:46:05 pm
content: The junction between R2 And R3 has 3x 0.7 +0.7 V=2.8VThis means the current in R3 is 2.8/500 = 5.6 mAThis current als runs in R2 so output voltage is 5.6 V

author: Chris.M
author_rank: Contributor
timestamp: May 10, 2013, 12:19:03 pm
content: Quote from: qno on May 09, 2013, 08:46:05 pmIndeed qno you are right, and I was able to do a full analysis. I'm going to put the analysis into another post with a slightly modified circuit.However, it looks like tiny assumption was made, which allows for the remaining circuit voltages and currents to be calculated. I'll talk about this in the next post.

author: Chris.M
author_rank: Contributor
timestamp: May 10, 2013, 01:10:08 pm
content: Sorry about the long post. There are some calculations but their easy to follow.The circuit has been slightly updated, R4 now comes from the transistor emitter voltage rather than the collector.So to start the analysis we know that the voltage drop across the 3 diodes will be 2.1V (3 * 0.7V).VeQ2= 2.1VIf we add the 0.7V drop for Q2's base emitter junction, we get 2.8V.VbQ2= 2.8VThat same 2.8V is across R3, so using ohms law we can get the current through R3.VR3= 2.8VIR3= 2.8/500= 0.0056 = 5.6 mANow here is where we disobey Kirchhoff's current law. We assume that the same current through R3 must be flowing through R2.Therefore, because R2 and R3 have the same resistance, the voltage was equally divided, so we can double the voltage across R3 to get the voltage across R2.VR2= 2 * VR3= 5.6VThe voltage across R4 is therefore that same 5.6V minus the 2.1V across the diodes.VR4= 5.6V - 2.1V= 3.5VTherefore from ohms law we can calculate the current through R4.IR4= 3.5V/2200= 0.001569 A= 1.57 mAGoing back to the current through R2 and R3 which comes from Q1's emitter, the current through the transistor must therefore also be 5.6 mA.IcQ1= 5.6 mAFrom the current gain equation (Ib = 100*Ic), we can calculate IbQ1.IbQ1= IcQ1/100= 5.6 mA/100= 0.000056 A= 56 uAContinuing on from VR2, if we add 0.7V for Q1's base emitter junction voltage, we get 6.3V for Q1's base voltage.VbQ1= 6.3VBecause Q2's collector and Q2's base are at the same potential, Q2's collector voltage equals Q1's base voltage.VcQ2= 6.3VIf 6.3V is dropped across Q2, then VCC - 6.3V gives use the voltage across R1.VR1= VCC - VR1= 28 - 6.3V= 21.7VNow the current through R1 can be calculated using ohms law.IR1= VR1/R1= 22.3/1000= 0.0217 A= 21.7 mAFrom the 21.7 mA, 56uA will flow through Q1's base, and therefore the remaining must go through Q2.IcQ2= 21.7 mA - 56 uA= 21.644 mAKnowing IcQ2we can work out the current through Q2's base.IbQ2= IcQ2/100= 216.44 uAFinally the voltage across Q1 and Q2 can be calculated.VceQ1= VCC - 5.6V= 28 - 5.6V= 22.4VVceQ2= VcQ2- Vdiodes= 6.3V - 2.1V= 20.2VAnalysis Complete.Now going back to the where we break Kirchhoff's current law, in which we say that the current through R2 must be the same as R3 and the voltage across R2 is twice that of R3, because R2 and R3 are in series, aren't we forgetting the small current will flow out of R2 and into the base of Q2.Therefore the current through R2 is slightly higher than R3, and therefore the voltage of R2 is slightly higher than R3.Without breaking Kirchhoff's current law, we can no longer double the 2.8V across R3 to get R2's voltage, and we no longer know the true current through R2 and therefore IeQ2.So I understand that the base current will be considered negligible and we can ignore it, but for the sake of knowing how to accurately determine currents and voltages in such circuits, is there a way to do the analysis taking into account Q2's base current?

author: c4757p
author_rank: Super Contributor
timestamp: May 10, 2013, 01:12:41 pm
content: Quote from: Chris.M on May 10, 2013, 01:10:08 pmYes, stop calling it an assumption and call itKirchhoff's current law.

author: Chris.M
author_rank: Contributor
timestamp: May 10, 2013, 01:18:20 pm
content: Quote from: c4757p on May 10, 2013, 01:12:41 pmThanks c4757p, have modified the post.So by "Yes", are you able to elaborate on the how?

author: mikes
author_rank: Regular Contributor
timestamp: May 10, 2013, 05:02:22 pm
content: You're focusing on the wrong things. This is a voltage regulator, you want to analyze how it regulates.0.7 V isn't exact, so the voltage at R2/R3 isn't either. It's only close to 2.8V, making the output close to 5.6V.Current through R4 will be in a range. Since the output is ~5.6V, the base of Q1 will be at ~6.3+. It depends on the load. For 150 mA Ic, Q1's Vbe(sat) is about 1.3 V; for 500 mA Ic, it's ~2.6 V (from the datasheet). You don't need full saturation to maintain 5.6 V from a 28 V supply, so the actual Vbe is less. But let's go with that. The point is that it will change significantly with load.So, Q2 has to pull the base of Q1 through a range of 5.6+0.7=6.3V (no load) to 5.6+2.6= 8.2V (500 mA).To do so, it has to pull down R1. So the current through R1 will be in the range (28-6.3)/1000 = 21.7 mA to (28-8.2)/1000= 19.8 mA.(edit: I forgot to consider the Q1 base current. At 500 mA Ic, Bc will be about 17 mA (min 30 Beta/HFE), so Q2 only has to pull about 3 mA - note that you're about to lose regulation here. The range through Q2 is then 21.7 - 3 mA. This affects the math below, and also causes a wider swing of Vf on the rectifier string, which will vary from ~.675-.750, ignoring R4.)Q1 has a minimum Beta (actually, HFE) of 75 @ 10 mA Ic, so the maximum current into base of Q2 would be around 21.7/75 = 0.29 mA through 19.8/75 = 0.26. You use 5% resistors, so the ~5% current going into Q2 compared to the ~5.6 mA through R3 is basically inconsequential.More importantly the change over load of 0.03 mA means it should provide good regulation by not upsetting the R2/R3 balance.Since ~20 mA is flowing through Q2, a similar amount will flow through the rectifier string. Vf for 1N4848 @ 20 mA is about 0.750 V, so R2/R3 is now minimum (.75x3)+0.7 ~= 2.95 and output is ~5.9 V. You can take that back to the top and iterate again, but things won't change much.Point is, if you care about what the exact voltage is, make R2 or R3 a trimmer, and be done with it. It's more important to understand why/how it provides regulation.

author: c4757p
author_rank: Super Contributor
timestamp: May 10, 2013, 06:35:32 pm
content: mikes is right. You're not "disobeying" Kirchhoff's law, you're just approximating it. Kirchhoff says that I(R2) = I(R3) + Ib(Q2). You're just saying that Ib(Q2) is tiny, so it might as well be zero and I(R2) approx= I(R3).

author: Chris.M
author_rank: Contributor
timestamp: May 10, 2013, 11:02:27 pm
content: Quote from: mikes on May 10, 2013, 05:02:22 pmWell said.I was concentrating on the exact numbers, when really what I was interested in was the possible range that each component would have, so I could determine how well it regulates and if any component would be close to their powering rating before bread boarding.Your analysis was clear and I understood all of it except for this little bit below.Quote from: mikes on May 10, 2013, 05:02:22 pmJust wondering if you meant Q2 here, otherwise could you explain how Q1's HFE is linked with Q2's base current.If its Q2, then it makes sense, the 0.29 mA base current is just the proportion from the collector current.Thanks mikes,R2/R3 is supposed to be a potentiometer, so now l can do the analysis again with the Q2's base connected to the top of R2 and then connected below R3 (gnd) can compare how well it regulates. Then I can bread board it see how the real circuit compares.

author: Chris.M
author_rank: Contributor
timestamp: May 10, 2013, 11:04:17 pm
content: Quote from: c4757p on May 10, 2013, 06:35:32 pmApproximating Kirchhoff's law. Yes, that is the best way to say what I'm doing.Thanks c4757p.

author: mikes
author_rank: Regular Contributor
timestamp: May 11, 2013, 11:42:15 am
content: Quote from: Chris.M on May 10, 2013, 11:02:27 pmYeah, that should be Q2 where I said Q1.