Calculating the currents in this circuit - Page 1
Forum section: electronics / projects

author: saaluc
author_rank: Contributor
timestamp: April 23, 2025, 07:11:27 am
content: Hi thereThis circuitis Series Transistor Voltage Regulator. In such a circuit there is a formula for calculating and adjusting the Maximum Collector Current or ICMAX:ICMAX=VS-0.7 ÷ RSₓBETA.Based on this formula, as much as we reduce the RS, the ICMAXwill increase.Now if R1 and D1 are added to the circuit in the following way1. will ICMAXdecrease? If yes, how much?2. In the second circuit, is RSthe sum of R1+RSitself?

author: PGPG
author_rank: Frequent Contributor
timestamp: April 23, 2025, 08:00:17 am
content: R1 and Rs can't simply be summed as different currents flow through each of them.At some Ie voltage drop at R1 will make transistor saturate and Ic will no longer be beta*Ib.

author: saaluc
author_rank: Contributor
timestamp: April 23, 2025, 08:38:56 am
content: Quote from: PGPG on April 23, 2025, 08:00:17 amLet us concentrate on my first question "will ICMAX decrease? If yes, how much?".I did not understand the relation of your answer "At some Ie voltage drop at R1 will make transistor saturate and Ic will no longer be beta*Ib" to my first question. Perhaps you are pointing to the fact that every transistor has a maximum IC beyond which we cannot increase it by increasing Ib (current of base). If you mean this, I agree. But my question is something else. My question is about the range in which we CAN adjust IC.In the first circuit, where there is no R1 and D1, ICMAX=(16-0.7)÷330×30 (BETA of TIP41C)=1.39 amps (this is in the range in which we CAN adjust IC of TIP41C).My question is whether in the second circuit where a 6.8Ω resistor and a 1N4001 Diode are added, will 1.39 amps (maximum current of collector) decrease because of the two added components? If yes, how much? If no, why?

author: antenna
author_rank: Frequent Contributor
timestamp: April 23, 2025, 12:41:30 pm
content: Yes, it will decrease. Assume infinite gain for a moment. Adding a voltage drop in series with RS, whether from a collector resistor or diode, will decrease the voltage across RS and therefore decrease the current through it. Less base current equals less collector current.Consider the load trying to draw that same 1.38A, but this time through 6.8ohm. That's 9.38v across R1. Add the diode and you'll have 10v across the pair. Now, instead of 16.7v-7.5v=9.2v across RS, you only have 2.5v across RS. Looking at the TIP41C datasheet, there are a few values for Hfe, but lets assume 15. With 9.2v across RS (330ohm), you have 28mA split between the base and the zener. Assume it all goes to the base, with a Hfe of 15, thats not even half an amp at the collector. In other words, before you even add the 6.8ohm resistor, that base resistor isn't big enough for 1.38A collector current (unless I am wrong about the gain value). With a current gain of 15, you need 92mA of base current. With 7.5v at the base and a 16.7v supply, you would need RS to be no higher than 102ohm. With R1 and D1 in circuit, its more like 27ohm (again, assuming Hfe of 15). If your transistor has sufficient gain, then its a matter of the added components eating up the headroom at that current. Transistors drop some voltage too, even when fully saturated.Someone else can do the math, that's just my 10cents.

author: saaluc
author_rank: Contributor
timestamp: April 23, 2025, 02:56:59 pm
content: Quote from: antenna on April 23, 2025, 12:41:30 pmThank you very much indeed.I want the current for the load be limited to 1.6 amps, that is, the transistor deliver only up to 1.6 amps.1. Is this possible with this kind of circuit?2. How can I make sure about the Beta (hfe) of TIP41C?

author: ledtester
author_rank: Super Contributor
timestamp: April 23, 2025, 03:56:46 pm
content: Quote from: saaluc on April 23, 2025, 02:56:59 pmYou don't have a lot of guarantees about the beta of a transistor except perhaps a minimum value.To limit the current I would use a separate circuit such as a foldback limiter.Here's a video:https://youtu.be/RFfRrbqM_-sThe basic idea is to steal base drive away from the pass transistor when the current limit is reached.

author: saaluc
author_rank: Contributor
timestamp: April 24, 2025, 08:26:32 am
content: Quote from: ledtester on April 23, 2025, 03:56:46 pmThe circuit you linked has several components. A circuit with less components is preferable because I want to attach it to another existing circuit on an emergency light.So let me ask your idea about this high-current shunt voltage regulator:I want the voltage across the BT1 (a 6-volt 5-AH SLA battery) never go beyond 6.8 volts. And I want the initial current that the empty battery draws never go beyond 1.6 amps.Input voltage to this circuit, when no load is connected (the battery is disconnected) is 15.5-16 volts; when there is a load (battery is connected), the input voltage drops; at this moment when the battery is full, input voltage is 7.8 volts.With these specifications, for limiting the current for battery to at most 1.6 amps, what the value of R3 should be?To answer the above question, we should know which input voltage must be divided by 1.6 amps, no-load input voltage (15.5-16 volts), or loaded input voltage (which changes with changing load resistance and load current)?