L78XX high current configuration - Page 1
Forum section: electronics / beginners

author: sy
author_rank: Regular Contributor
timestamp: September 18, 2024, 06:02:54 am
content: Hello guys, I want to clarify how we can calculate the value of R1 for this linear regulator L78XX.I am not sure if it is a typo where it says "Ireq" instead of "Ireg" because it is not mentioned anywhere on the diagram or notes of the datasheet.Additionally, I am assuming bq1 is referring to the hfe value of the transistor. Is this correct? Because I thought that value changes overtime and is a bit ambiguous for calculations.Edit: additionally, how do I determine the value of Ireg? I assume the current would flow through R1 until enough voltage is dropped to activate Q1, but where does it go from there?

author: Smokey
author_rank: Super Contributor
timestamp: September 18, 2024, 06:52:48 am
content: I know this exactly isn't an answer to your question but LTSpice should be able to simulate this no problem.

author: magic
author_rank: Super Contributor
timestamp: September 18, 2024, 07:12:53 am
content: Yes, Ireg makes sense here and bQ1 is beta. They basically tell you to subtract base current of the PNP from regulator current and select a resistor which provides that much current.Ireg should be selected to be no less than the minimum load current for correct operation, under all possible conditions. Besides that, I think just about any current will work here. If you keep it low enough, you may get away without heatsinking the regulator in some cases. (Note that if the transistor fails open, Ireg will increase to maximum and the regulator will start to cycle between full power output and thermal shutdown.)

author: sy
author_rank: Regular Contributor
timestamp: September 18, 2024, 07:26:25 am
content: Quote from: magic on September 18, 2024, 07:12:53 amIn terms of power considerations for this resistor, if I am understanding correctly, would be the minimum possible load current to the minimum current required to turn on the transistor Q1?So from like 0-700mA through R1 if R1=1Ohm and VBE is about 0.7V?

author: ledtester
author_rank: Super Contributor
timestamp: September 18, 2024, 07:37:28 am
content: In the formula for I_O, "Q1" should really be b_Q1 and there is a minus sign missing between I_REG and V_BE/R1, that is, it should read:$$I_O = I_{\text{REG}} + b_{Q1}\left( I_{\text{REG}} - \frac{V_{\text{BE}}}{R_1}\right)$$which is to say that the output current is the current through the regulator plus the collector current (which is beta times base current).

author: iMo
author_rank: Super Contributor
timestamp: September 18, 2024, 07:43:02 am
content: I would do it simply in following way (aprox):1. I want the max transistor current 5A to deliver into the load2. let the transistor's beta is 50 at that current3. I would need at least 5A/50 = 100mA emitter-base current in order to deliver I_collector 5A4. the resistor R1 is there because we want the transistor to start to conduct (the I_collector) from a specific current only (R1 is "closing" the Q1)5. let us use for example a 7812, its max I_O current is 1A, I will choose 200mA I_REG max output current for the 7812, that is the current the transistor shall start to conduct already, so the I_REG current should be larger than those 100mA (2-3x let say)..6. the transistor opens itself and boosts the 7812 at around 0.65V V_BE voltage on the R17. so 0.65V/200mA = 3.25 Ohm8. everything above aprox 200mA to the load will be redirected over the transistor9. mind this circuit does not limit the output current..

author: David Hess
author_rank: Super Contributor
timestamp: September 18, 2024, 07:53:14 am
content: Usually the current through R1 is high enough that the base current and hfe can be ignored.

author: floobydust
author_rank: Super Contributor
timestamp: September 18, 2024, 08:02:41 am
content: Example 5V 3A regulator schematic in a bench PSU, using 51Ω resistor, I think 3W is overkill.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 18, 2024, 11:10:08 am
content: With the added transistor the regulator may well need additional capacitance at the output, not just the 100 nF for the 78xx. The 2 x 4-7 µF electrolytic in the plan from floobydust look more reasonable.The stability against oscillation is still one of the tricky points and one should check for oscillation.Usually I would prefer a regulator made for the higher current (e.g. LT1085), or if really needed a regulator chip like the 723 that is made to work with an external transistor.