MY goal is to design a regulator which could receive (12 V to 60 volts DC as an input) and 5 volt with 60 to 70 mA current as an output. The reason I used three BJT's NPN (Q1, Q2, Q3), so I thought I could get decrease the heat dissipation by dividing the current through 3 different BJTs (60 mA/3)= 20 mA. Power dissipation= (Vin At the collector -Emitter)*current= (50.30_5.33)*20/1000= 0.9 W and I test them on bread board and it gets very hot, and it makes sense because 1 watts will almost = to 50 degree Celcius, plus the ambient temperature which makes it 75 degree Celsius.

Any expert there in designing voltage regulator, with a smart solution, or what would you recommend? If I Increase the RX value to 800 ohm, the collector voltage will decrease from 50.5 to 5.33 volts and that is perfect, but the problem is when I change the voltage back to 12 volt, then I won't get 5 volt as an output. If I miss any informationenter image description here to add please let me know, and I will get back to you as soon as possible.

       --- Operating Point ---

V(n004):     11.0833     voltage
V(a):            6.19012     voltage
V(n002):     50.4175     voltage
V(n003):     6.18692     voltage
V(ve):           05.3231     voltage
V(n005):     5.16179     voltage
V(n001):     60  voltage
V(n006):     5.3231  voltage
Ic(Q2):  0.0319418   device_current
Ib(Q2):  0.000319418     device_current
Ie(Q2):  -0.0322612  device_current
Ic(Q1):  0.0319418   device_current
Ib(Q1):  0.000319418     device_current
Ie(Q1):  -0.0322612  device_current
I(C1):   6.19012e-018    device_current
I(D2):   -0.242137   device_current
I(D1):   -0.00180777     device_current
I(R6):   -0.000638836    device_current
I(R5):   -0.0645224  device_current
I(R3):   -0.0322612  device_current
I(Rx):   -0.0638836  device_current
I(R1):   -0.244583   device_current
I(R4):   -0.0322612  device_current
I(R2):   0.0024466   device_current
I(V2):   -0.308467   device_current
  • 6
    \$\begingroup\$ This just begs for a DC to DC switcher. Any reason why not? If not, that's fine. But provide some good reasoning about it. \$\endgroup\$
    – jonk
    Feb 11, 2020 at 16:18
  • \$\begingroup\$ Exactly what kind of transistors are you using, in what package? What kind of heatsinks? \$\endgroup\$ Feb 11, 2020 at 16:27
  • \$\begingroup\$ @ElliotAlderson digikey.ca/product-detail/en/diodes-incorporated/ZXTN25100DGTA/… \$\endgroup\$
    – user242388
    Feb 11, 2020 at 16:34
  • 1
    \$\begingroup\$ @user242388 Yes. Your situation just begs for an IC for that purpose. It's a lot of fun doing it discretely, without an IC. But an IC will be lots easier and lots better in the end. And probably not a lot more expensive. I prefer discrete because I'm anachronistic that way and I enjoy tweaking. I'm a hobbyist. But an IC is probably your better bet if you aren't like me. \$\endgroup\$
    – jonk
    Feb 11, 2020 at 16:43
  • 1
    \$\begingroup\$ @user242388 Besides, ICs have better, more robust protection schemes that are hard to add with discrete parts. \$\endgroup\$
    – jonk
    Feb 11, 2020 at 16:49

1 Answer 1


The thermal resistance of your transistors, if you only use a small copper area for the tab, is 104°C/W. If you ask them to dissipate 0.9W you should expect that the die temperature will increase by about 90°C above the ambient temperature. With a large copper area (50mm by 50mm of 2oz copper) you can get the thermal resistance down to 42°C/W, but you would still see a temperature increase of about 30°C.

If you want to use this linear regulator you should look for a transistor in a package that is better able to dissipate heat, such as a TO220 package, and look for a compatible heatsink.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.