# Power Supplies Tutorial No.5 (Adjustable Regulated Linear Power Supply) 29.04.18

Adjustable Regulated Linear Power Supply

In previous articles we have been speaking about generating a stable DC regulated power supply where the output voltage is designed to always remain the same. What about if you want a variable voltage, say for a lab bench supply. Today we take a look at this scenario.

In the first example we will look at using a normal old LM78xx voltage regulator to generate a variable regulated supply. Lets consider the following circuit. For this example lets select R1 = 470 Ohm

We know from experience that the LM7805 is designed to regulate the voltage at its output terminal to be 5 V with respect to the End terminal.

Vfixed = 5 V

From this we can determine the current flow in resistor R1 will be

Ir1 = Vfixed / R1 = 5 / 470 = 11 mA

From the data sheet of the LM7805 we know the device quiescent current

Iq = 8 mA

So the current through R2 will be equal to

Ir2 = Iq + Ir1 = 11 + 8 = 19 mA

Now if we select R2 to be 100 Ohm

The voltage across R2 will be

Vr2 = Ir2 x 100 = 19 mA x 100 = 1.86 V

So the output voltage of the circuit will be

Vo = Vfixed + Vr2 = 5 + 1.864 = 6.86 V

if we were to consider the situation where R2 was chosen to be 510 Ohm

Vo= 14.51 V

So using the formula Vo = Vfixed + R2 x (Vfixed / R1 + Iq) and selecting appropriate resistors you can vary the output voltage of the circuit.

Now consider if you were to replace R2 with a 500 Ohm variable resistor. This would allow you to vary the resistance from 0 - 500 Ohms and in doing so the output voltage would be variable from the fixed regulator voltage, 5 V in out example up to approximately 14.5 V.

So there you have it an adjustable regulated power supply, with over temperature shutdown and current limiting with just the addition of two more components.

LM317 Adjustable Linear Regulator

Of course the issue with the circuit above is what do we do when we want to set a 3.3 V supply. Well thankfully there is a specialist regulator to do just that. If we now take look at the LM317 data sheets we find it can provide an adjustable output from 1.25 to 37 V.

It may help to consider the LM317 to be like a LM78xx regulator but with a Vfixed of 1.25 V and now we call it Vref. The other main difference in working out how to use the LM317 is the quiescent current Iq is very low and is normally not considered when calculating the output voltage.

So looking at our previous formula

Vo = Vfixed + R2 x (Vfixed / R1 + Iq)

we drop the Iq and change the Vfixed term to become

Vo = Vref + R2 x (Vref / R1)

which if we simply becomes

Vo = 1.25 x ( 1 + R2 / R1)

Of course to make th epower supply adjustable we would select R2 to be a variable resistor so we can set the power supply voltage as we desire. The next question is how do we select the appropriate resistor values?

Now this is where it gets a little interesting. If we go back to the data sheet we find the regulator requires a worst case scenario of 10 mA load current to ensure good load regulation. Now if the load of the power supply is always guaranteed to draw more than 10 mA then all is well, however if not then we should select R1 and R2 to provide this minimum load current.

So lets set the current flow through R1 as we know it all flows through R2 as well.

R1 = Vref / 10 mA = 1.25 / 10 mA = 125 Ohm

So to guarantee the minimum load current we need to make sure R1 is not higher than 125 Ohm. Lets select 120 Ohm.

Ok so now we have R1 selected, we can calculate R2 for the desired output voltage.

R2 = (Vo / 1.25 - 1) x R1

I see a lot of circuit designs that recommend R1 values like 240 Ohm, if you use the typical minimum load current of 3.5 mA then this resistor will give 5.2 mA which should work fine but when designing circuits they should cater for the device worst case specifications. In this case the 5.2 mA is obviously way short of the 10 mA worst case specification.

This is something to look out for!!

Other considerations

Ok so I guess the other things we should recap are the dropout voltage, this device is just another linear regulator and has minimum input voltage requirements to guarantee it regulates correctly. Check the data sheet to ensure you have things correct.

The other main factor is power dissipation, as the device is a linear regulator, its job is to drop the additional voltage across itself. So depending on the load current and the voltage it is dropping the power can be quite high. Always try to keep the input voltage close to the output voltage whilst still ensuring the dropout requirements are met.

And just one last thing the circuit above is a simplified version, if you check the data sheets there are recommendations for output capacitors to improve transient response, an input bypass capacitor, decoupling capacitor on the adjust terminal and protection diodes where applicable. An example is below. Note, this must be designed with a load that will draw more than the 10mA requirement. Ok so that pretty much covers adjustable regulated linear power supplies. There are other chips, discrete circuits etc however I hope this provides you with enough information to research the rest for yourself.