Introduction: Easy Project - Bench Power Supply (0 .. 30V, 2 Amp)
POWER SUPPLY CONTEST ENTRY
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One of the most useful pieces of equipment for the electronic enthusiast is a good bench power supply. Buying one can be expensive, so most of us search the internet for circuits to copy, and just build our own.
But there is a disadvantage going this method. As the build will be just another "copy and paste" exercise, the builder does not understand how it works, and does not gain any knowledge.
With this Instructable, I hope I will be able to teach at least one person the basic operation of a variable power supply. I will break down the circuit, go through each step, and show how the components work, and how to calculate their values.
So what is a Bench Power Supply?
Well, it is simple, it is a power supply with a variable output voltage, and has adjustable current limiting. No more sparks and burned out components when used correctly on your projects.
It consists of two parts:
1 - Unregulated part
This section converts the utility AC voltage to the required DC voltage for our power supply. The transformer perform two tasks:
- It converts the utility voltage from a high voltage to a safe working voltage of your power supply
- It gives electrical isolation between the utility network and your power supply output.
Rectifier DB1 converts the AC voltage to a DC voltage.
Lastly, capacitor C1 is used to filter out the 50/60Hz components present on the DC output.
See Figure 1.
Refer to this Instructable:
2 - Regulated part
Two things happen in this section. The ripple factor is reduced to as low as 1%, and the output voltage will be adjustable. Both facilities employ negative feedback. See Figure 2.
Care should be taken when designing a power supply. With the output set at it's maximum output voltage, the output voltage must still be lower than the lowest voltage dips in the unregulated part of the supply. A good principal is to allow for at least 3 volt play. The idea is illustrated in Figure 3.
Step 1: Unregulated Design
Lets design a 30V, 2A power supply.
In a moderate power supply like this, we can allow for a 10% peak-to-peak ripple voltage across the output. The peak voltage at the output will then be:
Vpeak = ( Vout + 3V margin + ( 0.1 x Vpeak ) ) x 1.2 safety margin
Vpeak = ( 30V + 3V + ( 0.1 x Vp) ) x 1.2*
Vpeak = 45V
* 20% was added for transformer losses and diode forward voltage losses.
Capacitor C1 can now be calculated
Vripple x C = I x t, where t = 1 / (2 x line frequency)
( 45 x 10/100 ) x C = 2A x 1/( 2 x 50Hz)
4.5 x C = 2 x 0.01
C = 0.044 Farad, or 4 400uF
C = 4 700uF, 63V
The transformer output should be:
Vrms = Vpeak / 1.414 (square root of 2)
Vrms = 45V / 1.414
Vrms = 32V
The maximum VA delivered by the transformer is:
VA = Vpeak x Iout
VA = 45V x 2A
VA = 90VA
The diodes should be rated at 2 x Iload to make provision for charging the capacitor.
Idiode = Iload x 2
Idiode = 2A x 2
Idiode = 4A
The minimum reverse voltage of the diodes should be 2 x Vpeak-peak
Vd_reverse = 2 x Vpeak
Vd_reverse = 2 x 45V
Vd_reverse = 90V
Vd_reverse > 100V
The fuse is specified to withstand double the rated current:
Ifuse = 2 x Iload x (Vlv / Vhv )
Ifuse = 2 x 2 x ( 32V / 220V )
iFuse = 0.58A
Use 220V, 0.64A slow-blow fuse
When an electrical earthing point is available, it is always a good idea to connect the power supply case, as well as the core of the transformer to earth.
Depending on your requirements, you can leave the output of the power supply floating, or connect the 0V output to earth. I prefer to connet the 0V to earth via a 1Mohm resistor.
Step 2: Voltage Regulation Section
This section of the design is what controls the output of the power supply.
Looking at the transistors, we need to understand that the voltage input to the regulator will be fixed, and the output voltage can be varied by the user. The output voltage is determined by the bias current to the transistors. But there is one drawback. The input current will be the same as the output current, and the voltage across the transistors will be Vin - Vout. Thus, Ptransistors = ( Vin - Vout ) x I. Looking at Figure 3, this means that the area above the maximum output voltage, and below the unregulated voltage, is the energy that will have to be dissipated by the regulator, which is converted into heat.
Therefore, we need a decent power transistor, with a decent heat sink to dissipate this heat.
Transistor T1 will do the regulation, and must be able to handle the load current. Transistor T1 & T2 are connected as a darlington pair, and their combined gains will allow for a smaller biasing current. For this, lets use the old time favorite 2N3055 power transistor for T1.Transistor 2 can be a 2N3054.
The 2N3055 transistor has a gain of typical 50.
The 2N3054 transistor has a typical gain of 100.
Calculting Transistor Bias Current
Iload max = 2A
Ibase t1 = Ice t1 / gain t1
Ibase t1 = 2A / 50
Ibase t1 = 40mA
Ibase t2 = Ice t2 / gain t2
Ibase t2 = 40mA / 100
Ibase t2 = 0.4mA
So, to turn on T1 and T2 completely, we need at least 0.4mA in the base of T2. T3
Step 3: Constant Current Source
The bias current for transistor T1 and T2 need to be stable, irrespective of the input voltage to the regulator. To do this, a constant current source can be used.
R4 and zener diode ZD1 forms a constant current source. This current source must be high enough to cater for Ibase t2, as well as the voltage regulating circuit (not shown)..
Use BC179 transistors for T3.
The BC179 transistor has a typical gain of 100.
Our current source must deliver Ibase t2 as well as the regulation current. Lets limit the maximum regulation current to 1mA. The regulation current will be controlled by transistor T4, which we will discuss in the next step.
Ice t3 = Ibase t2 + Iregulation
Ice t3 = 0.4mA + 1mA
Ice t3 = 1.4mA
Take zener diode D1 as 1.8V. Now,
Vr4 = Vbe t3 - 0.7V
Vr4 = 1.8V - 0.7V
Vr4 = 1.1V
R4 = Vr4 / Ice t3
R4 = 1.1V / 1.4mA
R4 = 785 Ohm, so take next smaller value, then
R4 = 680 Ohm
Ice t3 = Vr4 / R4
Ice t3 i= 1.1V / 680
Ice t3 = 1.6mA
Take current through zener diode D1 as 2mA. This will ensure good regulation by the zener diode.
R5 = ( avarage unregulated voltage - Vzener ) / Izener
R5 = ( 45V - (4.5V/2) - 1.8V ) / 2mA
R5 = 20.47K
R5 = 22K
With the constant current source complete, transistors T1 & T2 can now be switched to deliver the 2A of the power supply.
Step 4: Voltage Feedback Circuit
To be able to control the output voltage, we need a voltage feedback circuit. This is done via resistor R6, R7 and transistor T4. The circuit is set up as a negative feedback loop.
As the output voltage rises, transistor T4 is turned on harder, thus more current flows through T4. As the current source is constant, thus will result in less current to bias transistor T1 & T2. This results in a lower output voltage.
The next step is to calculate the voltage feedback components:
Use BC109 transistors for T3.
The BC109 transistor has a typical gain of 100.
Ice t4 max = Ice t3, or our maximum biasing current available from the constant current source.
Ice t4 = 1.6mA
Ibase t4 = Ice t4 / gain t4
Ibase t4 = 1.6mA / 100
Ibase t4 = 0.016mA.
This is the minimum current into the base of T4 that will allow the 1.6mA to flow.
This current needs to be supplied by the resistor network R6, R7 & R8. To be safe, current through these 3 resistors should be 10 times the base current needed on transistor T4.
Ir6= 10 x Ibase_t4
Ir6 = 10 x 0.016mA
Ir6 = 0.16mA
THE DRAWBACK OF THIS DESIGN:
As can be seen, there must always be an output voltage present to keep transistor T4 biased, Therefore, this type of power supply MUST have a minimum output voltage. Lets make the minimum output voltage 2V.
The minimum output voltage is 2V. Therfore, we can calculate the combined resistance value of R6, R7 and R8 at the minumum output voltage.
R6 + R7 + R8 = Vminimum / Ir6
R6 + R7 + R8 = 2V / 0.16mA
R6 + R7 + R8 = 12.5Kohm . . . . . . ( Eq 1 )
Now, with R7 set for maximum output voltage;
Vmax = ( ( R6 + R7 + R8 ) / R8 ) x Vbe_t4
30V = ( ( R6 + R7 + R8) ) / R8 ) x 0.7V
42.8 x R8 = R6 + R7 + R8
( 42.8 x R8 ) - R8 = R6 + R7
41.8 x R8 = R6 + R7
R6 + R7 + 42.8 R8 = 0 . . . . . . . . ( Eq 2 )
and with R7 set for minimum output voltage;
Vmin = ( ( R6 + R7 + R8 ) / R8 ) x Vbe_t4
2V = ( ( R6 + R7 + R8) ) / ( R7 + R8 ) ) x 0.7V
2.857 ( R7 + R8 ) = R6 + R7 + R8
R6 + R7 - 2.857 R7 + R8 - 2.857 R8 = 0
R6 - 1.857 R7 - 1.857 R8 = 0 . . . . . . ( Eq 3 )
Now, Eq 2 = Eq 3 is
2.857 R7 - 40.943 R8 = 0. . . . . . ( Eq 4 )
substitute Eq 1 in Eq 2
Vmax = ( ( R6 + R7 + R8 ) / R8 ) x Vbe_t4
30V = ( 14.2Kohm / R8 ) x 0.7V
42.85 R8 = 14.2Kohm
R8 = 331ohm
Substitute R8 into Eg 4
2.857 R7 - 40.943 R8 = 0
2.857 R7 = 40.943 R8
2.857 R7 = 40.943 x 331ohm
R7 = 4.748Kohm
Lastly, substitute R7 and R8 into Eq 1
R6 + R7 + R8 = 12.5Kohm
R6 + 4.748Kohm + 331ohm = 12.5Kohm
R6 = 12.5Kohm - 4.748Kohm - 331ohm
R6 = 7.42Kohm
To get realistic component values, take:
R6 as 10K preset
R7 as 4K7 linear potentiometer
R8 as 470R preset
When setting up the power supply, R6 and R8 can be adjusted to limit the minimum voltage to 2V, and maximum voltage to 30V.
Step 5: Current Limiting Circuit
The last pat of the design is a current limiting circuit, which can be adjusted up to 2 amps.
R1, R2, R3 and transistor T5 makes up this circuit.
Use BC109 transistors for T3.
The BC109 transistor has a typical gain of 100.
R1 must be a low value, high wattage resistor. A typical value will be:
R1 = 0.47ohm
Pr1 = Ir1 ^ 2 x R1
Pr1 = 2 ^ 2 x 0.47
Pr1 = 1.88 watt minimum.
Take rating of R1 at twice calculated rating, or 5W, then
R1 = 0.47ohm, 5W
Vr1 at maximum current is
Vr1_max = Ir1_max x R1
Vr1_max = 2A x 0.47
Vr1_max = 0.94V
For the current limiter to ensure zero voltage output in an overcurrent condition, transistor T5 must be able to direct all the current from the constant current source to 0V. Thus,
Ice_t5 = Ice_T3
Ice_t5 = 1.6mA
Ib_t5 = Ice_t5 / gain_t5
Ib-t5 = 1.6mA / 100
Ib_t5 = 0.016mA
But, Ir2 = Ir3, and we want the current through them 10 times higher than the base current for T5. Now,
Ir2 = Ir3 = 10 x Ib_t5
Ir2 = Ir3 = 10 x 0.016mA
Ir2 = Ir3 = 0.16mA
Now, with pot R3 set to wiper of R2, voltage across R2 will be the same as transistor T5 Vbe, or 0.7V
R2 = Vbe_t5 / Ir2
R2 = 0.7V / 0.16mA
R2 = 4.375Kohm
R2 = 4K7 preset
At maximum current of 2 amp, Vbe_t5 should still be 0.7V. Now
Vr1_max = 0.94V, and
Vr1_max = Vr2 + Vr3
Vr1_max - Vr2 = Vr3
0.94V - 0.7V = Vr3
Vr3 = 0.24V
R3 = Vr3 / Ir3
R3 = 0.24V / 0.16mA
R3 = 1500ohm
R3 = 1K linear potentiometer
Step 6: Output Filtering and Stabilization
Zener diodes are likely to develop audio noise. Capacitor C2 mainly serves as a noise filter across this zener. The value is not critical, with a typical value of 10uF acceptable. Voltage across capacitor C2 is the zener voltage, so it is quite low, thus
C2 = 10uf, 16V
Transistor T4 forms the negative feedback loop used to regulate the output voltage. As with any negative feedback circuit, the circuit can easily go into oscillation. We can prevent self-oscillation by adding capacitor C3. The exact value of C3 will be dependent on circuit board design, and specific components used. A good value will be anything between 10pf and 100pf. Thus,
C3 = 10pf .. 100pf
Capacitor C4 plays an important role, and it must cater for frequency variations in the output current. The output impedance of the power supply is very low for DC and low frequency current variations. However, if the power supply is say, connected at an audio amplifier, the amplifier might require high current peaks at around 10KHz. This will make the voltage regulator unstable, and cause a high frequency ripple on the output.
C4 forms a bypass filter at high frequencies. Thus, C4 performs the same function at 10KHz, as that C1 does at 50Hz. A typical impedance for C4 can be somewhere between 1 ohm and 2 ohm at 10KHz. So lets make it 1.5 ohm @ 10KHz
Xc4 = 1 / ( 2 x pi x f x C)
1.5 = 1 / ( 2 x 3.14 x 10KHz x C4)
C4 = 10.61uF
Voltage across C4 is the maximum output voltage, or 30V, thus
C4 = 10uf, 50V
Step 7: Output Voltage Time Constant
We need a stable output voltage at low and high current frequencies, and calculated all capacitor values to make this happen. But, there is one last issue we need to calculate. It is the output voltage time constant. Ideal, this time constant should not exceed 0.25 seconds.
With the introduction of capacitor C4, changes made to the output via the resistor R7, will not appear at the output immediately. This is due to the time constant of the output circuit.
This can be calculated as follow:
R x C = time constant
R is the output resistance (R6 + R7 + R8) in parallel with R9
C i is the output capacitance (C4)
Wthout R9, the time constant = R x C
t = ( 10K + 4K7+ 330R ) x 10uf
t = 0.152 seconds
Our time constant is faster than 0.25 seconds, so resistor R9 is not required.
Step 8: Additional Power Supply Protection
Diode D1 protects the power transistor should a reverse voltage spike appear across transistor T1. This can be any diode rated at around 1 .. 4 amp, and can be the same as used in the diode bridge. Thus
D1 = 1 amp, 100V
This diode protects the power supply against a reverse voltage being applied to the output. For instance, when the power supply is being used to charge batteries. This diode must be mounted directly to the output terminal posts, and should not be rated mmore than twice the output voltage, or 4A. This will ensure that a small battery will be short-circuited, without the power supply being damaged.
But, if a large capacity battery is connected incorrectly, the diode must fail. Although this might damage the power supply, high currents will only be present for a short period of time.
Step 9: Voltage/Current Meters
This meter must be custom scaled. Make use of a standard 100uA meter, and add shunt resistors so that the meter gives full scale at 2 amp.
This meter must be custom scaled. Make use of a standard 100uA meter, and add series resistors so that the meter gives full scale at 30 volt.
Step 10: Final Design and Mounting
In the final step, we must place the components.
- Transformer TR1, diode bridge DB1 and capacitor C1 is mounted separately, not on the control board.
- Resistor R1 must be placed where there is good ventilation, as it will get hot at maximum current. Ensure no wires are running nearby this resistor.
- Transistor T1 must be mounted to as large as possible heat sink on the outside of the power supply enclosure, without any air flow restrictions. Diode D1 is soldered directly onto transistor T1.
- Diode D2 must be connected directly across the output terminals of the power supply.
- The volt meter is connected directly onto the output terminals.
- The ammeter is connected directly across resistor R1.
- The rest of the components inside the dotted area are all mounted on a PC Board or veroboard, leaving only 5 wires connecting the circuit to the high-current components.
Participated in the
Power Supply Contest