logic of bjt regulator - Page 1
Forum section: electronics / projects

author: john23
author_rank: Frequent Contributor
timestamp: March 15, 2025, 08:52:52 am
content: Hello, I am trying to understand the dinamics of the circuit taken from the youtube video.if we apply 2V DC the node 2 on the photo will be 1.4 (Vbe=0.6)The node 1 will expirience some voltage wich will be translated to the other PNP base.Could you describe how the regulating dinamics works here?Thanks.

author: awesomechapro
author_rank: Contributor
timestamp: March 15, 2025, 11:40:12 am
content: The 2 transistors on the left a forming are rudimentary op-amp. This forms a voltage feedback loop with the PNP transistor, which stabilizes the output voltage at point 3.In this configuration, when the output voltage is less than the 2V reference voltage, this rudimentary op-amp will pull more current from the base of the PNP transistor, turning it on and increasing the output voltage.Once the output voltage is above the 2V reference, the op-amp will pull less current from the base of the PNP transistor, decreasing the output voltage.Here's a simulation with an op-amp instead of the discrete transistors, but the same thing still applies.https://tinyurl.com/298vm4zw

author: john23
author_rank: Frequent Contributor
timestamp: March 15, 2025, 02:02:58 pm
content: Hello ,I was trying to build the regulator in LTspice.2N3906 PNP is being used however I get barely 2mA in Q3 which is very low as shown in the plots.I know it needs to be linear.How do you reccomend to BIAS the PMOS so I'll get higher collector current from this PMOShttps://www.onsemi.com/pdf/datasheet/2n3906-d.pdf

author: awesomechapro
author_rank: Contributor
timestamp: March 15, 2025, 02:13:17 pm
content: R3 and R5 make no sense in this circuit, try removing them.The feedback loop in this circuit should only operate the transistor in the linear range (which is why this topology is called a Linear Power Supply), and shouldn't require biasing.

author: john23
author_rank: Frequent Contributor
timestamp: March 15, 2025, 03:11:32 pm
content: Hello, Linear mode is high IC current in bjt I cant get that in my PNP, where did I go wrong?Thanks.

author: john23
author_rank: Frequent Contributor
timestamp: March 15, 2025, 05:33:44 pm
content: UPDATE:Hello,I have implemented the varying resistor however I want to improve the PNP biasing situation in order to improve the voltage regulation on em_p node.given the data sheet how canI improve the situation?Thanks.LTspice file attached.https://www.onsemi.com/pdf/datasheet/2n3906-d.pdf

author: awesomechapro
author_rank: Contributor
timestamp: March 15, 2025, 11:44:46 pm
content: Remove R3 and R5

author: PCB.Wiz
author_rank: Super Contributor
timestamp: March 16, 2025, 03:43:56 am
content: Quote from: john23 on March 15, 2025, 05:33:44 pmWhat exactly are you trying to achieve here?What is youractual use case, and what voltage and current does thatneed, and what regulation?A simple 3 transistor regulator will always have limits, as VBE changes with IC and IC changes with load.

author: MathWizard
author_rank: Super Contributor
timestamp: March 16, 2025, 05:15:23 am
content: IDK what currents and voltages you want, but for small signals, I'd choose much larger resistors for the diff-amp, a max current of few milli-amps is pretty good for most small-signal diff-amp circuits. Or just lower Vcc. But yeah it depends on the goals, if you want high currents, that's can be fine too.You can choose and set currents and voltages, like the tail current of the diff pair. You might choose 2mA, and then choose some voltage, and resistor. And a VBE above that point, will be your base voltages. There's a lot of freedom to choice and set these values, in any order.And then you can consider where you want the collector voltages to be, and the min/max, and maybe the saturation point. And what sort of voltages you want or are possible when connecting to the next stage. And so you can choose some Rc's based on that. Depending on the circuit, you still might add the 2nd diff-amp collector resistor, the Vce will affect the transistors beta. Likewise with the PNP emitter resistor, you still want it, in some cases, it depends on the the circuits, and your goals.

author: john23
author_rank: Frequent Contributor
timestamp: March 18, 2025, 06:13:37 pm
content: Hello ,given the curcuit below(with the resistors I need to remove)I saw that with increasing resistance the output gets less and less regulated as shown below.How can I improve tthe regulation?what is the circuit intuition?Thanks.

author: AnalogTodd
author_rank: Regular Contributor
timestamp: March 18, 2025, 07:32:49 pm
content: Quote from: john23 on March 18, 2025, 06:13:37 pmThis is where you actually need to review the operating point for the circuit overall. R5 is in there for a reason--if you run R2 with 8.75mA through it, that means the base of Q3 sits 8.75V below Vdd. Q3 wants to have a VBEof roughly 0.6V, so R5 drops about 8.15V across it. The issue is that if you want to increase the current in R4, you now need to increase the voltage across R5. You can't double the current in R5 before running out of headroom from Vdd. As your simulation goes to higher load currents, have it plot thecol_pnode and watch how far it drops. Thebase_pnode will be dropping at a slightly higher rate, but it can't get to where it goes below theemiternode. At that point, you can't drive R5/Q3 any more andem_pwill continue to crash down.It's a low gain circuit, optimized to run at a given bias point. Trying to get large variations in load current are not going to provide great regulation in the first place, but once you get past ~4.3mA it's going to crash a lot faster.

author: john23
author_rank: Frequent Contributor
timestamp: March 18, 2025, 07:49:54 pm
content: Hello AnalogTodd, I have noticed that when I remove R5 the PNP gets is linear mode with much higher current running threw it.Its very very clear that when Q3 has large Ic current and linear state that the regulation is best.I understood you intuition regarding the differential pair but Q3 is another story.As you can see the regulator can handle much higher loads with higher currents.Could you please give an intuition regarding why at higher load we need higher current threw Q3 so it will function properly?Thanks.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: March 18, 2025, 08:33:47 pm
content: Quote from: john23 on March 18, 2025, 06:13:37 pmYou have still not said what the actual use and design objectives are?Whatload currentdo you need to supply and whatvoltage changeis tolerable? What is theactual load?You have it working in Spice, you should be able to derive your own intuition by change of values.You can easily reduce the voltage change, by throwing in more gain, but that needs care with stability.You already found that reducing R5 helps, another change is to replace R2 and R3 with a current mirror (eg dual matched PNPs in one package)Change of Q3 to a PFET, further increases gain by removal of base current change with load but introduces other compromises.If you wanted this to work over a wide set-voltage range, you could change R1 for a current sink. If it only ever supplies 2.00V, a fixed R1 is tolerable.Note that testing with a slow variable load is also going to entirely miss stability issues.Smarter is to use an impulse load of 10%-100% of your target, then you can check overshoot and settling times.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 19, 2025, 11:58:28 am
content: Hi,Let me try a quick design of this circuit.Using the original reference designations.This is the design process that I followed: define the load current. I chose 0-20mA estimate the base current in Q3. Assume Hfe = 100. Ibase = Ic / Beta = 0-200uA the voltage across R2 is the same as Q3 Vbe = 0.6V. Chose R2 so the current is 600uA (3x Ibase Q3) the desired collector current in Q2 is 600uA. The desire is to have Q1 and Q2 operating at the same collector current. R1 sets the current IC_Q1 + IC_Q2. The target is 600uA x2 = 1.2mA. The voltage across R1 = 2V - Vbe = 1.4. R1 = 1.4 / 1.2 mA = 1.2kThe result is:As shown the load regulation is about 1.4%Jay_Diddy_B

author: AnalogTodd
author_rank: Regular Contributor
timestamp: March 19, 2025, 12:20:36 pm
content: Quote from: john23 on March 18, 2025, 07:49:54 pmQ3 is a bipolar transistor. It requires current out the base to provide higher collector current. The higher collector current of Q3 is required to maintain the voltage atem_pas the resistor value of R4 drops.

author: john23
author_rank: Frequent Contributor
timestamp: March 21, 2025, 01:05:08 pm
content: Hello , As a side issue my PNP is not acting like in wikipedia.photos 1 to 5 represent a state where E>B>C which acctording to wikipedia its supposed to be a high current state called forward active.but instead I get the opposite very weak current as if I am at saturation.where did I go wrong?

author: john23
author_rank: Frequent Contributor
timestamp: March 21, 2025, 02:46:05 pm
content: Hello Jay_Diddy_B,Very good explanation.I hav some logical question.You wanted the cuurent on R2 to be 600uA and also you said that Q1 and Q2 will have 600uA currents threw each of them.but 200uA will flow to Q3 so Q1 will have only 400uA,so Q1 will not have the same current.Where did i go wrong?Thanks.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 21, 2025, 04:23:43 pm
content: john23 and the group,The base of Q3 will vary with the load current. It will Iload/Beta. As the load is increased from 0 - 20mA, the base current will vary from 0-200uA, assuming beta is 100.The sum of the Q1 and Q2 collector currents will always be equal to the current in R1. Ignoring the Base currents of Q1 and Q2.As the load current increases, Ib3 increases. Q1 collector current increases. Which means Q2 collector current decreases, since IcQ1 + IcQ2 = IR1.You can check this in LTspice.Jay_Diddy_B