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I am making a 12 volt regulator for a solar panel to power an audio amplifier. I already have a switching regulator for most application but it is very noisy for amplifier even with a lot of filtering. The amplifier draws over 3 amps on full load.

I want to build a regulator using 7812 with series pass transistor for higher current as shown in the datasheet, page 20.

http://www.mouser.com/ds/2/149/LM7812-461970.pdf

I made the same circuit as shown in the datasheet and used the 2SD1047 as Q1

https://www.st.com/resource/en/datasheet/2sd1047.pdf

schematic

simulate this circuit – Schematic created using CircuitLab

But as I connected it to my bench power supply set to 18V I was not able to get a regulated output. It just showed ~2V less then the input voltage.

At 18V no load I got 16V output, 0.5A load 15V and 3A load 12V. I tried different values of R1 calculated by the formula shown in the data sheet.

Formula for R1

Is the output voltage affected by the the value of R1? Please Help. Beta of 2SD1047 Vbe While calculating R1 I assumed the value for Ireg to be 0.1A, Beta of transistor to be ~95, and IQ1 to be 3A, when that didn't work I tried a bunch of different values of resistors with the same results.

The LM7812 works great when no series pass transistor is attached. And the transistor is good too.

I already have a switching regulator and I also have a higher current linear regulator so don't suggest them please. I want this circuit to work, and I also want to understand this circuit further so if you could point me to other learning material concerning this circuit then that would be great. The only thing I had to work with was the datasheet. Thanks

PS: SOLVED!! My mistake, I connected a NPN BJT instead of PNP.

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    \$\begingroup\$ Glad you could work out some issues yourself. Yes, with a 'boost' transistor a minimum load is needed so the Vbe is high enough. Major mistake though. The boost transistor MUST be a pnp, with the 12 volt current boost coming from the collector. Try a MJE172. \$\endgroup\$
    – user105652
    Commented Mar 5, 2019 at 5:02
  • \$\begingroup\$ Oh Sorry!! I made such an obvious mistake and Thanks for not criticizing. \$\endgroup\$
    – Hamza Khan
    Commented Mar 5, 2019 at 5:07
  • \$\begingroup\$ This is a really good question btw, quality-wise. \$\endgroup\$
    – pipe
    Commented Mar 5, 2019 at 9:28

1 Answer 1

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Here is the diagram from page 20 of the datasheet you linked:

Figure 13 from Fairchild LM78xx datasheet

Image source: Fairchild LM78xx datasheet, Figure 13

Notice that the pass transistor is PNP. The mistake in your design, is that you are trying to use a 2SD1047 NPN pass transistor - which won't work, as you have seen.

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  • \$\begingroup\$ Thank you for the fast reply and not making fun of me. \$\endgroup\$
    – Hamza Khan
    Commented Mar 5, 2019 at 5:12
  • \$\begingroup\$ @HamzaKhan - No problem, and well done for writing a clear question and trying to solve the problem yourself :-) I see that you got the same reply in a comment, while I was typing this answer... \$\endgroup\$
    – SamGibson ♦
    Commented Mar 5, 2019 at 5:20
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    \$\begingroup\$ Keep in mind Q1 needs a heat sink of ~ 2'C/W for 40'C rise @18Vin*3A so efficiency is 67% \$\endgroup\$
    – D.A.S.
    Commented Mar 5, 2019 at 5:23
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    \$\begingroup\$ @Hamza - Since you won't automatically be notified about the comment made by Sunnyskyguy EE75, I'm writing this comment, which will notify you. As he points out, one disadvantage of a linear regulator is that you have to dissipate all the heat from that series pass transistor. You say that your requirement is "over 3 amps max", so you will have to consider the heatsinking for that transistor and the regulator. If you need help with that topic, then feel free to ask a new question about it. \$\endgroup\$
    – SamGibson ♦
    Commented Mar 5, 2019 at 5:29

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