In this post we discuss how an effective and efficient, yet very cheap, and stabilized bench power supply can be designed by any electronic hobbyist for safely testing all types electronic projects and prototypes.
The main features that a bench power supply must have are:
- Should be built with cheap and easily available components
- Should be flexible with its voltage and current ranges, or simply must include the facility of a variable voltage and variable current outputs.
- Should be over-current and over-load protected.
- Should be easily reparable, in case a problem arises.
- Should be reasonably efficient with its power output.
- Should facilitate easily customization as per a desired specification.
General Description
The majority of the power supply designs thus far incorporate a linear series stabilizer. This bench power supply design uses a pass transistor which works like a variable resistor, regulated by a Zener diode.
The series power supply system is the more popular, possibly due to the fact that it is a lot more efficient. Except of some minor loss in the Zener and feed resistor, noticeable loss only happens in the series pass transistor during the period it is supplying current to the load.
However, one disadvantage of the series power supply system is that these do not provide any kind of output load short-circuit. Meaning, during output fault conditions the pass transistor may allow a large current to flow through it, eventually destroying itself and possibly the connected load also.
That said, adding a short circuit protection to a series pass bench power supply can be quickly implemented through another transistors configured as a current controller stage.
The variable voltage controller is achieved through a simple transistor, potentiometer feedback.
The above two additions enables a series pass power supply to be highly versatile, rugged, cheap, universal, and virtually indestructible.
In the following paragraphs we will briefly learn the designing of the various stages involved in a standard stabilized bench power supply.
Easiest Transistor Voltage Regulator
A quick way of getting an adjustable output voltage is to hook up the base of the pass transistor with a potentiometer and Zener diode as shown in the figure below.
In this circuit the T1 is rigged as an emitter-follower BJT, where its base voltage VB decides its emitter side voltage VE. Both VE and VB will precisely correspond with each other, and will be almost equal, deducting its forward drop.
The forward drop voltage of any BJT is typically 0.7 V, which implies that the emitter side voltage will be:
VE = VB - 0.7
Using a Feedback Loop
Although the above design is easy to build and very cheap, this type of approach doesn't offer great regulation of power at the the lower voltage levels.
This is exactly why a feedback type control is normally employed to get an improved regulation throughout the entire voltage range, as demonstrated in the figure below.
In this configuration, the base voltage of T1, and therefore output voltage, is controlled by the voltage drop across R1, mainly due to the current pulled by T2.
When the slider arm of the pot VR1 is at the ground side extreme end, T2 becomes cut off since now its base becomes grounded, allowing the only only voltage drop across R1 caused by the base current of T1. In this situation the output voltage at the T1 emitter is going to be almost same as the collector voltage, and can be given as:
VE = Vin - 0.7, here VE is the emitter side voltage of T1, and 0.7 is the standard forward voltage drop value for BJT T1 base/emitter leads.
So if the input supply is 15 V, the output can be expected to be:
VE = 15 - 0.7 = 14.3 V
Now, when the pot VR1 slider arm is moved to the upper positive end, will cause T2 to access the whole emitter side voltage of T1, which will cause T2 to conduct very hard. This action will directly connect the zener diode D1 with R1. Meaning, now the base voltage VB of the T1 will be simply equal to the zener voltage Vz. So the output will be:
VE = Vz - 0.7
Therefore, if the D1 value is 6 V, the output voltage can be expected to be just:
VE = 6 - 0.7 = 5.3 V, so the zener voltage decides the minimum possible output voltage that could be obtained from this series pass power supply when the pot is rotated at its lowest setting.
Although the above is easy and effective for making a bench power supply, it has a major disadvantage of not being short circuit proof. That means, if the output terminals of the circuit is accidentally short circuited, or an over load current is applied, the T1 will quickly heat up and burn.
To avoid this situation, this circuit could be a simply upgraded by adding a current control feature as explained in the following section.
Adding Overload Short Circuit Protection
A simple inclusion of T3 and R2 enables the bench power supply circuit design to be 100% short-circuit proof and current controlled. With this design even an intentional shorting at the output will not cause any harm to the T1.
The working of this stage could be understood as follows:
As soon as the output current tends to go beyond the set safe value, a proportional amount of potential difference across R2 is developed, enough to switch ON transistor T3 hard.
With T3 switched ON causes T1 base to be joined with its emitter line, which instantly disables the T1 conduction, and this situation is maintained until the output short or overload is removed. In this way T1 is safeguarded from any undesired output situation.
Adding a Variable Current Feature
In the above design, the current sensor resistor R2 can be a fixed value if the output is required to be a constant current output. However, a good bench power supply is supposed to have a variable range for both voltage and current. Considering this demand, the current limiter could be made adjustable simply by adding a variable resistor with the base of T3, as shown below:
VR2 divides the voltage drop across R2 and thus allows the T3 to switch ON at a specific desired output current.
Calculating the Part Values
Let's start with the resistors, R1 can be calculated with the following formula:
R1 = (Vin - MaxVE)hFE / Output Current
Here, since MaxVE = Vin - 0.7
Therefore, we an simplify the first equation as R1 = 0.7hFE / Output Current
VR1 can be a 10 k pot for voltages up to 60 V
Current limiter R2 can be calculated as given below:
R2 = 0.7 / Max Output Current
Max output current should be selected 5 times lower than T1 maximum Id, if T1 is required to work without an heatsink. With a large heatsink installed on T1, the output current can be 3/4th of T1 Id.
VR2 can be simply a 1k pot or preset.
T1 should be selected as per the output current requirement. T1 Id rating should be 5 times more than the required output current, if it is to be operated without a heatsink. With a large heatsink installed, the T1 Id rating should be at least 1.33 times more than the required output current.
The maximum collector/emitter or VCE for T1 should be ideally twice the value of the maximum output voltage specification.
The value of zener diode D1 can be selected depending on the lowest or the minimum voltage output requirement from the bench power supply.
The T2 rating will depend on the R1 value. Since the voltage at collector of T2 will be always equal to Vin, the VCE of T2 should be higher than the Vin or the input supply. The Id of T2 should be such it is able to handle the base current of T1, as determined by the value of R1
The same rules apply for T3 also.
In general T2, and T3 can be any small signal general purpose transistor such as BC547 or maybe a 2N2222.
Practical Design
Having understood all the parameters for designing a customized bench power supply, it's time to implement the data in a practical prototype, as shown below:
You may find a few additional components introduced in the design, which are simply for enhancing the regulation capability of the circuit.
C2 is introduced to clean any residual ripple at the T1, T2 bases.
The T2 along with T1 forms a Darlington pair to increase the current gain of the output.
R3 is added to improve the zener diode conduction and therefore to ensure better overall regulation.
R8 and R9 are added to enable the output voltage to be regulated across fixed range, which are not critical.
The R7 sets the maximum current that can accessed at the output, which is:
I = 0.7 / 0.47 = 1.5 amps, and this appears quite low compared to the rating of the 2N3055 transistor. Although this might keep the transistor super cool, it may possible to increase this value up to 8 amps if the 2N3055 is mounted over a large heatsink.
Decreasing Dissipation to Increase Efficiency
The biggest disadvantage with any series transistor based linear regulator is the high amount transistor dissipation. And this happens when the input/output differential is high.
Meaning, when the voltage is adjusted towards lower output voltage, the transistor has to work hard to control the excess voltage, which is then released as heat from the transistor.
For example if the load is a 3.3 V LED, and input supply to the bench power supply is 15 V, then the output voltage has to be lowered to 3.3 V which is 15 - 3.3 = 11.7 V less. And this difference is converted to heat by the transistor, which could mean a efficiency loss of more than 70%.
However, this problem can be simply solved by using a transformer with tapped voltage output winding.
For example the transformer may have taps of 5 V, 7.5 V, 10 V, 12 V, and so on.
Depending on the load the taps could be selected for feeding the regulator circuit. After this, the voltage adjustment pot of the circuit could be used for further adjusting the output level precisely to the desired value.
This technique would increase the efficiency to a very high level, allowing the heatsink to the transistor to be smaller and compact.
Op Amp Controlled Precision Bench Power Supply
The working procedure of the opamp controlled bench power supply circuit is fairly basic uncomplicated, since regulated power supplies can be simply as particular forms of feedback amplifier. In this concept, the R1 and R2 generate a refernce sample signal from the output supply, which is by another reference voltage created by D2. The resulting correction signal is supplied back through the 741 on the series pass transistor Q1.
Observe that the circuit stability has been enhanced by providing the reference source R3 -D2 through the stabilized output rather than from the unstabilised input as is usually done inother bench power supplies. To ensure that the circuit initiates as soon as switched on, a leakage resistance R4 is placed in parallel with the series pass device. This means that the feedback loop begins to run as soon as power is switched ON.
Absolutely no regulation is sacrificed because of R4, since it is the general output which is sampled by R1 -R2, therefore, the impact of the ripple current moving through R4 gets adjusted through the feedback.
Making the Output Adjustable
The output may well be made adjustable by changing R1-R2 with a potentiometer, however in its existing layout, the circuit can't be forced to regulate under the zener voltage value of D2. When uninterrupted adjustment of the output voltage is necessary, the reference source R3-D2 should be furnished through the unregulated input, with accompanying minor lack of stability. The quantity of power the circuit is able to offer will be limited primarily by the current handling capacity of Q1 and the maximum power capacity of the unregulated supply.
More Elaborate Bench Power Supply Circuit
A regulated bench Power supply are normally a useful gadget for any hobbyists or engineer. Despite the truth that IC based voltage regulators have become very easily accessible, a circuit only employing ordinary discrete components can be appealing. In order to save power, and to restrict the dissipation across the series regulator, the entire 0-30 V control range is further divided into 3 scaled-down voltage ranges.
All the 3 ranges matches with a suitable secondary supply voltage (determined by the position of S1a) and a proper reference voltage (determined by S1b). So that you can get a consistent control of the output voltage to a minimum of0 V, a negative auxiliary supply needs to be added.
In this bench power supply circuit, this is extracted (by means of D5 and C2) via a different 12 V winding over the mains transformer. A different option could be to incorporate an additional separate mains transformer.
The final results tested on the bench prototype are pretty decent: ±35 V mains voltage swing induced just ±25 mV swing of the output voltage, with full load 1 amp load attached to the output. The A.C. ripples of the output (hum) had been lower than 15 mV.
How it Works
The circuit functions in the following manner.
A reference voltage, extracted through the zener diode(s) D6 -D9 and fixed using pot P1, is directed to the transistor T2 base by means of D10 and TI.
T2 and T3 work like a differential amplifier; wherein the base of T3 receives the output voltage by means of DI2. This differential amplifier's output is applied, through D11, on the base of the combined series regulator made up of transistors T4, T5 and T6.
Although the configuration might look a bit complex, it works like a typical regulator circuit; it maintains the output voltage virtually fixed over an extensive range of output currents.
Transistors T7 and T8 along with connected parts form the current limiter stage. As soon as the voltage over R10 gets to a particular value (as fixed by P2) T7 begins conducting. This, consequently, causes transistor T8 to get biased and starts conducting; which reduces the base drive to transistor T4, and the situation decreases the output voltage hence the output current continues to be inside the predetermined boundary.
When S1 is selected at position 1, that compares to an output range of 0-10 V, setting at 2 allows 10-20 V and adjusting at position 3 provides an output range of 20-30 V. P1 is used for tweaking the range as fixed by S1.
The highest amount of output current can be established using pot P2. This P2 pot could be either pre-programmed to supply a highest output current of 1 A or employed like a variable output current control.
ATX Bench Power Supply
The following ATX bench power supply schematic image and prototype images were submitted by one of the avid readers of this blog, Mr. V.
Loved the circuit very nice classic schematic
Thanks Bob
Thanks, Glad you liked it!
Sir pls my question is a bit off,,, I want to convert a 19v laptop charger to a +19,Gnd,-19,,,,,,,,,,,, pls how do I do it,, help me
Hi MD, you can try the IC 741 based circuit or the IC LM324 based circuit from the following post, and modify them according to your specification:
https://www.homemade-circuits.com/dual-power-supply-3v5v6v9v1215v-with/
It seems excellent, I will try it!
Here in Cuba everything has to be DIY, I have for years one based on the old honorable 723.
I will comment on you. Here, from Cuba, we also follow you 😉
Thank you so much! I wish you all the best with the project!
Sir I wanted to make a 0-50 v dc & 10 amp variable power supply. Can you send me a ckt diagram for this.
Hello Ratnadip, you can use the theory explained above to design and build the required power supply easily.
For the main pass transistor T1, you can use a TIP35 with BD139 as a Darlington, with a large heatsink.
In order to keep the transistors cool, use tapped transformer, at 9 V, 18 V, 24 V, 35 V, 50 V. Depending on the output voltage you can select the appropriate range on the transformer, this will ensure that the T1 is much cooler always.
what’s S2 component call?
Hi sir, may I know what’s the resistor watt that you use and 400uf and 100uf what is the rated voltage you recommend?
Hi Desmond, power is equal to voltage drop across the resistor multiplied by the load current through it. The voltage rating of the filter capacitors should be at least 1.5 times the DC from the transformer
Thanks for sharing the design. Would it work with a TIP142 replacing TR1 and TR2 and BC546 instead of BC547?
yes, it will work!
Just have a few questions regarding your practical example:
1) I’m using TIP142 as T2 and T3 where only min. Hfe = 1000 is available . How would you calculate R1? Using your formula with Output current 2A gives R1= 350 ohm, does that sound correct?
2) you are saying that the voltage drop over R1 is always 0,7V, why is that?.
3) In your practical example you have introduced R2=470 ohm, but what is the reason behind?
Hope you can help me
1) the formula is (supply input – supply output)hFE / 2, if this is giving you 350 ohms then it is correct.
2) where did I say voltage drop across R1 is 0.7V? 0.7V is the drop across base/emitter of the pass transistor.
3) Electronics is not critical, there can be little bit difference in part values, the final diagram was taken from a magazine, i have not calculated the parts values for it, but I assume they are as per my given formulas.
Thanks for your answers, just some clarifications:
1) Could you just use the forward voltage drop (2×0.7V for Darlington) instead of (Vin-Vout)? I was more wondering if using the min. hfe value is correct?. The real Hfe value is properly much higher.
2) In the section regarding calculating the parts:
“Since the voltage across R1 will be always 0.7 V, the VCE of T2 becomes immaterial, and can be any minimum value. The Id of T2 should be such it is able to handle the base current of T1, as determined by the value of R1”. I’m asking because I want to calculate the power rating of R1.
3) Just don’t know what the purpose is of R2 🙂
Thank you so much of sharing the great design and spending time answering my questions.
Br, Michael
You are welcome!
1) For Darlington the Vin-Vout will be 1.4 V or 1.2 V
2) The statement is wrongly mentioned, the voltage across R1 will depend on the load, and the value of R2.
3) R2 is the current limiting resistor for the load
Hi Swagatam,
I have built your power supply cct “Practical Design”. I would like to know if I can modify the cct to also adjust voltages between 0V & 5V. At the moment the lowest I can go is 5V.
Your assistance will be much appreciated.
Regards Jan
Hi Jan, first you will have remove D5 and R9 and replace them with direct links which will allow you reach the minimum 0V level, then you may have to experiment with the value of R8 until you get the maximum level of 5V output
Swagatam,
Does that mean that I will get the full range from 0V – 24V?
Yes, if your input is 24V then the maximum output will be around 23V, and the minimum can be reduced to 0V. For getting 5V maximum you may have to adjust the R8 value accordingly.
Hi Swagatam, I just want to confirm with you. I have removed D5 & R9, now R3 will go directly to 0V. Also if I connect a 20W load with the voltage set to 24V then the output voltage will drop to 20V and TR1 will get extremely hot within about 2-3 sec. I am suspicious about R3, but when I remove it then I can not adjust the voltage.
Your assistance will be much appreciated.
Regards
Jan
Hi Jan, the resistor R3 had been used for biasing the D5, since D5 is removed R3 becomes meaningless now, so it can be removed entirely from the circuit. But make sure to connect the emitter of TR3 to ground line after removing D5.
Also, you can connect a 1K resistor in series with the base of TR3, and then remove R8, R9 both from the circuit.
Since this a linear regulator the main transistor will become hotter as the difference between input and output voltage increases. Moreover, the maximum input current should be 1/3rd of the collector current rating of the transistor. For example if the collector current rating of T1 is 15 amps, then the maximum current through T1 should not exceed 5 amps. Even at 5 amp the 2N3055 can get very hot so a heatsink will be required.
Hi Swagatam,
Referring to your Practical Design. Can you please give me R4-R7 wattage or how to calculate it.
Regards
Jan
Hi Jan, you can calculate it by multiplying 0.6 with the maximum current limit value associated with the specific resistors.
Hi,
I am new to this so bear with me 🙂
The D1 diode is on the emitter of T2 to ground, and is a lower voltage limit for when the VR1 wiper is at min, so it’s only serving as a minimum voltage limit.
Normally (in series regulators) when T2 does not exist, circuits have a zener diode between R1 and ground to stabilize the voltage going into the base of T1 when load is applied.
In the case of this feedback loop, that zener diode is no longer needed? so when the circuit is loaded and there is a drop in the voltage , wouldn’t there also be a drop at the base of T1?
Hi, your assumption regarding D1 diode is correct.
If suppose the input voltage is 12V, and the output is adjusted to 6V with the potentiometer, and if the input supply drops by 2 to 10V due to a load at the output, there will be no change in the 6V output value, because there is still a margin of 4V above the 6V
But this will need to tested through experimentation, as I am not entirely sure about it.
Hello dear Sir Swagatam
As an interested follower of your Site, I have a question which could be somehow related to this page of course. I am sure you are kind enough to help me who know nobody to get answer.
I have a 6 volt DC motor . it needs to 4 nos of medium size rechargeable batteries (which are out of order) to run with maximum speed. when I connect it to a 15 volt power supply, the voltage drops to 6 volt but, motor runs with maximum speed which means that the power supply possesses enough current.
Won’t this 15v power supply damage the motor? If it will damage, Can I use a few Nos. of 1N5408 diodes to reduce the voltage instead of using transistors or ICs? and my third question please: Am I obliged to do reduce 15 V to 6V?
With many thanks
Thank you Kevin, for being a dedicated follower of this blog. Yes definitely, a 15V supply can do a lot of harm to a 6V rated motor in the long run, therefore you must drop the voltage to the correct levels. If you use diodes, then you may have to use many of them to reach 6V from 15V….doesn’t appear to be a great idea. Instead you can try 7806 IC or the concept explained above.
Hello dear Sir Swagatam
I never imagined that you would respond to my letter and so soon. I am so astonished and happy of course.Thank you very much Sir. I did as you had ordered but The motor did nit run with IC7805 or 7806. I built your circuit under title “Connecting Voltage Regulators 78XX in Parallel for High Current”. 3 Nos. of 7805 ran it but the speed of motor was low, heat sink got to hot and voltage dropped to 1.5V . I did another test and used 9 Nos. of 1N5408 diodes in series with power supply. It was interesting that voltage drop without load after the ninth diode was only 2V (13 of 15) and when I connected to motor, the voltage dropped to 2V and diodes got very hot. Do you think that there may be a solution to run it safely with needed current? It seems that motor needs much current. I took the said motor out of a Black & Decker dust buster WV4815.
Thank you again and wish you the best
Best regards
Thank you Kevin, I am glad to help you!
The 78XX ICs can deliver a maximum of 1 amp current when mounted on large heatsinks.
So maybe your motor requires current a lot higher than 1 amp.
In that case it is better to employ a transformer based power supply rated at 3 amp or higher current, and preferably rated at 6 V, so that no voltage regulator is required for the conversions.
Hello dear Sir Swagatam
I believe that this is the best way but I will search your blog to find a simple variable circuit based on 2N3055 which can deliver up to 4A. my present 15V transformer delivers 3A and seems good.
Thank you very much for your response, dear Sawgatam.
Best regards
You are welcome Kevin, however a 2N3055 linear power supply will create a lot of heat, so a buck converter may be more suitable, for example the last design from the following article:
https://www.homemade-circuits.com/dc-to-dc-converter-circuits-using-sg3524-buck-boost-designs/
or the following design:
PWM Solar Battery Charger Circuit
Hello dear Sir Swagatam
Thank you very much for your friendly guidance. I should go out to buy the necessary components for “PWM Solar Battery Charger Circuit”
Best regards
No problem Kevin, I wish you all the best!
Hello, good afternoon, My name is Carlos. I am a faithful follower of his website. I have a problem with a variable power supply from 0v to 30v x 5A. The problem is when I connect a load of for example 500ma the voltage drops as 2v…. For example 12v drops to 10v. Only with 500mA… Will you have a circuit to connect to the output and correct the voltage? The source is built with 2 Lm358, one for current adjustment and the other for voltage where can I upload the circuit so you can see it. Thank you very much
Hello Carlos, if it has an emitter follower type linear regulator, then you will need to increase the rating of the output transistor, and also replace it with a Darlington transistor as explained in the above article. Also make sure that the input supply from the transformer is adequately rated, otherwise the output will drop due to lack of current.
By the way without seeing the schematic it can be difficult to suggest the right solution.
Hello sir Swagatam thanks for the quick reply. transformer is 7A the circuit is here
Hi Carlos, you design uses two 2N3055 and one BDX53 in Darlington mode which together can provide high power and high current output. It may be difficult for me to understand the fault in your circuit, unless it is checked practically be me which is not possible…however this circuit should be easily able to handle loads upto 4 amps. Please check if everything is hooked up correctly or not….otherwise you can try the simple bench power supplies explained in the above article, which will provide similar results.
Hi Mr Swagatam thank you for your quick response. I want to tell you that I finally found the problem of voltage drop with small loads. The problem was the R4 shunt. I put 3 resistors in parallel… Two of 0.22 and one of 0.1 the three are 5w. Now the power supply works very very well, the voltage only drops 200mv with 4 amps. Thank you very much in advance for all your help. Greetings from Argentina
That’s great Carlos, yes it seems we completely missed the current limiter resistor which was blocking the passage of higher current, and dropping the output voltage. Glad it is solved now.
Great step-by-step DIY !
This article should have been published earlier !
Congrats Prof.Swagatam.
All the best in assisting all learners from your blog.
Thank you Imsa, Glad you found the post useful….appreciate your feedback.
Hello Swagatam,
Thanks very much for your tutorial on the Op Amp Controlled Precision Bench Power Supply. Please can you specify the value of series pass resistor R4?
(and how to calculate it)
Kind regards,
Rob
Thank you Thomas,
The R4 can be simply a 1K 1 watt resistor. It doesn’t need to be calculated because any value between 1K and 10K can be used here. It is basically to activate the feedback link to the op amp circuit, when power is switched ON.
Hello Swagatam,
Thanks for the information.
Kind regards,
Rob
You are welcome Thomas!
Thank you so much for all of your circuits you have gone over on your websites. I have looked at many and follow the new and am particularly interested in your battery charger and power supply circuits. I have a broken ego lithium ion battery charger and from what I can tell the rectifier is placed before the large capacitors which then feeds a transistor before feeding a transformer. Battery is 14s so fully charged voltage should be 58.8v. Voltage after the capacitors is ~74v. I was wondering if a circuit similar to the 12v power supply from battery charger would work for a larger voltage power supply. Also if you have posted a similar circuit I would love to see it. Again, many thanks for all of your work and helping me understand how common circuits work.
Glad you liked them Kevin, thank you very much for visiting this site!
Yes a simple transistorized regulator can be used for charging your Li-Ion battery, as shown in the following diagram. Just be sure to keep the input charging current below 50% of the battery’s Ah rating. The zener can be a 59V zener diode customized using a few series connected zener diodes:
https://www.homemade-circuits.com/wp-content/uploads/2021/11/single-transistor-voltage-regulator.jpg
Thanks so much for the response Swagatam! That would allow me to charge the batteries in a pinch. I’m assuming the best spot would be to tap in after the output capacitors on the circuit. What about using the unit to make a low amp 60v power supply?
You are welcome Kevin! Yes you can extract DC after the output capacitor. A low current 60V unit should also work nicely, no issues with it.