# Can I use 2 LM317s cascaded?

I have a 20 - 0 - 20 transformer which is giving about +/-27 V after rectification. I need +20 VDC, +16 VDC, 5 VDC and -20 VDC. I thought of regulating from 27 to 20 then 20 to 16, and finally 16 to 5. Will it work using cascaded LM317s or LT1085s?

• yes, you can cascade, but going from 27V to 5V will be very inefficient. Apr 1, 2021 at 3:02
• Thanks for the answer, I guess the same goes for the LT1085, right? About the 5V, I need such voltage to supply the +VCC of a STPD and also as high state voltage on the control pin, which I assume it won't drain more than a couple mA. Apr 1, 2021 at 3:17
• Yes, it can work. But please note that when you draw current from 5V output, the current will flow through the other LM317s (i.e. 27-to-20 and 20-to-16) as well. So, if the current drawn high enough, there's a risk of breaking down the other regulators. Apr 1, 2021 at 3:19
• Right Rohat Kılıç, I guess it won't be a problem considering the first one, from 27 to 20, will be a LT1085 a 5A capable regulator, then, I'll use a LM317 to reduce from 20 to 16 and supply a couple BC549 transsistors, and then, another LM317 suppling 5V to a SPDT. Apr 1, 2021 at 3:24
• Yes, any linear regulator, whether or not it is Low Drop Out, will dissipate at least as much power as the voltage drop x the output current. So, it doesn't matter how many steps you take, if you use linear regulators to go from 27V to 5 volts you are going to lose 22V x the current at 5 volts. If the current is small, you will be OK, but some regulators have a minimum current. :-(. Apr 1, 2021 at 3:25

You can chain regulators as you suggest BUT the lower voltages will incur large inefficiencies & also large power dissipation in the regulators.
Efficiency can be improved with switching regulators.

This (10 year old!) SE answer of mine to "My linear voltage regulator is overheating very fast" should be useful.

Linear regulator dissipation can be greatly reduced at lower voltages by using a series input resistor sized to provide adequate voltage at max current.
eg providing the 5V regulator with a resistor feed from 27V, if you assume a headroom of say 3V is needed and the regulator takes 1.25v for itself then
R = Vdrop / Imax
= (27 - 3 - 1.25 - 5)/Imax
~= 17V/Imax.

eg for say 0.5A max at 5V R = 17/0.5 <= 34 Ohm.

Resistor power dissipation will be Imax^2 x R or in this case 8.W watts.
A 10 Watt resistor would be very marginally suitable and a 20 W one advisable.
Use of air cooled resistors allows use without heatsinks.
Use of series resistors also allows substantial reduction of regulator heatsinks.