Power Supply & Voltage Regulation Notes

This area discusses different ways that voltage and currents from a power supply or battery are controlled. Power Supplies

The Current Source (Series Resistor)

Inside the LED The simplest way to supply power to a load is through a single series resistor. For example, if your load was an LED that conducted a steady 5 mA, the series resistor is simple to calculate.

 LED supplied by series resistor The schematic shown at right shows series resistor R1 acting as a current limiting device. The LED being a diode, will have a fixed voltage drop (LEDs often have higher drops like 1.2 volts).

But in this arrangement, we are not regulating voltage at all. R1 simply limits the current flow. In this case we know the maximum current flow will be 5 mA (see at right). I = V/R

Calculate R1 When determining the resistance to use for an LED circuit like this, you must:

  1. First determine the current that the LED should conduct (ILED)
  2. R1 = VT / ILED

Check Power Dissipation

P=VI After you compute the resistance needed, be sure to check the power dissipated in the resistor. Worst case for R1 in this example is 45 mW. Since a 1/8th watt resistor handles 0.125 Watts (125 mW) then a 1/8th watt resistor would safely handle this power requirement.

Brightness Tip

Optimal R1 If you are calculating resistance for maximum brightness (thus maximum safe current), then do:

  1. Compute your LED resistance as per normal (above)
  2. Test the LED with the resistor, measuring the voltage drop across the LED (VLED). The voltage drop measured may be anywhere from approximately 0.6 volts to 3.5 volts
  3. Subtract the measured LED voltage drop (VLED) from the power supply voltage (VT).
  4. Divide this voltage by the current of the LED (ILED).

The first calculation always results in less than your target current flow in the LED. This is because like all diodes, some voltage is always dropped across the LED (robbing voltage from the resistor R1). When this factor is subtracted out, your calculated resistance will more accurately result in your targeted current flow.

The Voltage Divider

Voltage Divider Voltage dividers are often used for low current situations like biasing a transistor. The main design constraint is that the pair of resistors dividing the voltage must carry a minimum current of approximately 10 times that used by the load.

I=V/R In the circuit shown at left we have a 9 Volt battery and divider resistors R1 and R2. Here the goal was to supply 5 Volts to the load. Without the load connected, we have a total resistance of R1+R2, which is 450 ohms. The voltage divider circuit conducts 9 volts / 450 ohms or 20 mA.

V=IR We also check that 20 mA of current through R2 (250 ohms) = 5 volts (250 x 0.020 ⇒ 5).

R=V/I In our hypothetical example, the load when supplied at 5 volts, conducts 2 mA of current. This means the load resistance represented by RL is represented by a resistance of 5 volts / 0.002 amps ⇒ 2500 ohms.

When we connect RL up to the voltage divider, the circuit currents change somewhat. However, since the divider is conducting 20mA and the load only 2mA, the difference is small. The divider conducts 10 times the amount the load is.

R parallel We can see that RL operates in parallel to R2, so let's compute the parallel resistance of R2||RL ⇒ ( R2 * RL ) / ( R2 + RL ) ⇒ 227 ohms.

I=V/R The effect of R2||RL now changes the total current flow from the battery: R1 + 227 ⇒ 200 + 227 ⇒ 427 ohms. Now the current flowing amounts to 9 volts / 427 ohms ⇒ 21 mA.

V(RL) With 21 mA flowing through R1, there is now a larger voltage drop: 200 ohms * 21 mA ⇒ 4.2 volts. This means the voltage across the load has dropped to 9 volts - 4.2 volts ⇒ 4.8 volts.

If the current through RL was reduced, then the voltage at the load would move closer to the original 5 volts as planned.

Disadvantage One

By now you probably realize the disadvantage of this design. If RL becomes lower (fewer ohms), i.e. conducts more current, the voltage for RL drops. If the current differential gets significantly less than ten times, the voltage drops considerably. Clearly this arrangement has poor voltage regulation.

Disadvantage Two

The main disadvantage of this approach is the amount of current wasted to arrive at the target load voltage. Recall that the divider must conduct approximately 10 times that of the load.

Consider the situation where you needed RL to conduct 10 mA at 5 volts. This means the voltage divider needs to conduct 100 mA (current of the load times 10). The 9 volt battery will now have to supply 110 mA, while the load only consumes 10 mA. That is not good for battery life.

Additionally be careful about the power dissipation of the resistors. For example R1 and R2 in the 100 mA case requires a 1/2 watt (minimum) resistor to be used.

The Diode Dropper

The diode dropper circuit An often overlooked solution for dropping a supply voltage is the use of one or more series diodes. Diodes are used all the time in integrated circuits for this purpose, so why not in discrete circuits?

Diode drop The circuit at left uses two 1N4148 diodes to drop the regulated 5 volt supply down to 3.4 volts for the load (RL). The circuit will maintain the 3.4 volts fairly well as long as the current doesn't exceed the diode maximum (200 mA in this case). Unlike the voltage divider, this circuit has decent voltage regulation.

Each 1N4148 diode in this example drops the voltage by approximately 0.8 volts. More or less of a voltage drop can be efficiently managed by changing the number of diodes in series.

The 1N4148 diode is limited to 200 mA maximum of continuous current. If your circuit requires more than that, then you must use higher powered rectifier diodes. Keep in mind that the voltage drop varies slightly from diode type to diode and even among diodes of the same type and manufacture.

Another point worth mentioning is that for small power loads, LEDs can be used. They have the advantage that they often have much larger voltage drops in the range of 1 to 3.5 volts each. See also this note about diode drops.

One major advantage of this arrangement is that very little power is wasted in the voltage drop.

The Zener Regulator

Zener regulator If your current requirements are modest, a Zener diode can provide a fairly well regulated supply voltage. The regulator circuit consists of:

  • A voltage source (V1)
  • A limiting resistor (R1)
  • A reverse biased zener diode (D1)
  • A load (represented here as RL)

First the Zener diode used must be chosen with the required voltage. Here D1 is a 1N750 zener diode, which operates at 4.7 volts. It is limited to an absolute maximum of (reverse biased):

  • 70 mA for current flow
  • 500 mW power dissipated

Additionally, the diode D1 will conduct more current as the temperature increases. In this example, we'll operate D1 at a maximum of approximately 50 mA, which is well below it's maximum ratings.

Max Zener Current The voltage across the Zener diode (D1) will be regulated at 4.7 volts as long as the load operates within the current limits of D1. D1 regulates the voltage by shunting the current away from the load, should the voltage rise above the Zener diode's voltage rating.

When the load (RL) is disconnected from the circuit, D1 will conduct (shunt) it's full current of 49.7 mA.

If the load (RL) should conduct the full 49.7 mA, the Zener's current would be reduced to zero. As long as the load conducts no more than this, the voltage will remain at 4.7 volts.

However, if the load current should increase beyond 49.7 mA in this example, the voltage across D1 (and hence the load) will start to drop. Current in D1 drops to zero and the voltage across R1 increases, which then reduces the load voltage.

Review: In the circuit shown, the load will remain well regulated between 0 and approximately 49.7 mA of load current (through RL). Due to manufacturing variations in the zener (D1), for best regulation it is probably wise not to operate the load current too close to the circuit's limit of 49.7 mA.

Zener Power

P=VI Let's check that we haven't exceeded the zener's power rating. It's worst case is with no load attached. Recall that the Zener diode is used here as a shunt regulator so it must shunt the configured amount of current (when necessary) to maintain the Zener voltage. As shown at left, the maximum power shunted by the zener is 234mW, which is well within the 500mW absolute maximum rating.

R1 Power

R1 Power Don't forget to check the power level required for R1. At right the calculation shows 350 mW flows through R1. It will be more, if the load breaks the rules and conducts more than 49.7 mA. If the load is well behaved, R1 can be a 1/2 watt part.

Engineering Your Own Zener Regulator

Calculate R1 The basic steps required to engineer your own zener regulated circuit are:

  1. Establish your load's voltage requirement (VRL or VD1)
  2. Establish your load's maximum operating current (IRL or ID1)
  3. Choose a Zener diode (D1) supporting more than your load's maximum current (ID1max), and matching voltage (VD1)
  4. Establish what your unregulated supply voltage is (VT)
  5. Compute R1 and choose a standard value for the resistor
  6. Check the circuit without the load attached to make sure the current and power requirements don't exceed the absolute maximum ratings of your parts (R1 and D1).

If your load may exceed its normal load current, then factor that into your R1 power calculations.

power_supply_notes.txt · Last modified: 2011/06/09 09:10 by ve3wwg
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