In this post we learn to make 3.3V, 5V voltage regulator circuits from higher voltage sources, such as 12V or a 24V source without ICs.
Linear ICs
Normally a step down voltage from a higher voltage source is obtained by using a linear IC such as a 78XX series voltage regulator IC or a buck converter.
Both the above options can be costly and/or complex options for getting a particular desired voltage quickly for a particular application.
Zener Diodes
Zener diodes also become useful when it comes to achieving a lower voltage from a higher source, however you cannot get sufficient current from a zener diode voltage clamp. This happens because zener diodes normally involve a high value resistor for protecting itself from high currents, which restricts the passage of higher current to the output to just milliamps, which mostly becomes insufficient for an associated load.
A quick and a clean way to derive a 3.3V or 5V regulation or any other desired value from a given higher voltage source is to use series diodes as shown in the following diagram.
Using Rectifier Diodes for Dropping Voltage
In the above diagram we can see about 10 diodes being used for acquiring a 3V output at the extreme end, while other corresponding values can also be seen in the form of 4.2v, 5v and 6V levels across the relevant dropping diodes.
We know that typically a rectifier diode is characterized to drop around 0.6V across itself, meaning any potential fed at a diode's anode will generate an output at its cathode which would be normally approximately 0.6V less than the input at its anode.
We take the advantage of the above feature in order to achieve the indicated lower voltage potentials from a given higher supply.
Using 1N4007 Diode for 1 Amp Current
In the diagram 1N4007 diodes are shown which might yield not more than 100mA, although 1N4007 diodes are rated to handle upto 1amp, it needs to be ensured that the diodes do not begin warming up, otherwise that would result in higher voltages being allowed to pass.
Because as the diode heats up the rated drop across it begins receding towards zero, that's why not more than a 100mA max should be expected from the above design for preventing over heating and enabling an optimal response from the design.
For higher currents one may opt for higher rated diodes such as 1N5408 (0.5amp max) or 6A4(2amp max) etc.
The drawback of the above design is that it does not produce accurate potential values at the output and therefore might not be suitable for applications where customized voltage references may be needed or for applications where the load parameter could be crucial in terms of its voltage specs.
For such applications the following configuration could become very desirable and useful:
Using an Emitter Follower BJT
The diagram above shows a simple emitter follower configuration using a BJT and a few resistors.
The idea is self explanatory, here the pot is used for adjusting the output to any desired level right from 3V or lower to the maximum fed input level, although the maximum available output would be always less than 0.6V than the applied input voltage.
The advantage of incorporating a BJT for making 3.3V or 5V regulator circuit is that it enables you to achieve any desired voltage using minimum number of components.
It also allows higher current loads to be used at the outputs, moreover the input voltage has no restrictions and may be increased as per the BJT's handling capacity and by some minor tweaks in the resistor values.
In the given example, an input of 12V to 24V can be seen, which can be tailored to any desired level such as to 3.3V, 6V, 9V, 12V, 15V, 18V, 20V or to any other intermediate value simply by flicking the knob of the included potentiometer.
5V Stabilized Regulator
The working of the 5V transistor regulator circuit is as follows.
The 1k resistor attached between the collector/base of transistor BD131 typically means that the BD131 always stays in the conducting mode.
However the 4.3 V zener diode ZD1 clamps the base voltage of the transistor 2N697 to approximately 4.3 volts lower than the emitter voltage of the BJT BD131. The 2N697 begins turning ON as soon as its base voltage reaches around 0.6 V positive with respect to its emitter voltage, and at this point the BD131 emitter reaches a potential of around 4.9 V positive.
An elevated voltage at this point allows higher amount of current to flow through the 2N697 (every 80 mV rise at the base of a conducting transistor enables the collector current to increase by 10 times value), causing the transistor to pull more current by means of the lk resistor which consequently decreases the base/emitter voltage of the BD131.
By doing this the circuit ensures proper stabilization at approximately 4.9 V - 5.0 V. Another zener diode, ZD2 has a value of 5.6 V positioned for the safety in case the BD131 malfunctions causing a short circuit. In that situation, ZD2 may possibly absorb the excess current until the attched fuse blows. If the circuit is operated with a battery, a 500 mA fuse should be just fine.
I’m confused by your mention of a 1N5408 and saying it has a .5A max. The 1N5400 series are rated for 3A. And any of these from the 1N5400, 5401, etc. will all work for voltages starting at 35V max. The datasheet shows derating based on lead length. At 3A and a .9V drop, that is 2.7W. The datasheet seems to clearly show that it can handle at least 2A as long as it can dissipate enough heat to keep temperature below 110 Centigrade. But going further, the thermal resistance is 20 degrees centigrade per watt. Even at 3A, using ambient temp of 40 degrees + 20 x 3 (bumping it to 3 watts) that is 100 degrees. That is well under 110. Am I missing some other factor? Thanks!
Yes I know the datasheet rating for 1N54XX series is 3 amp, but if you check practically you will find they can get pretty hot even at 2 amp current. I mentioned 0.5 amp as the optimal current since beyond this value the diode can start getting warmer.
At 100 degrees the characteristics of the diode can change drastically, affecting its working ability and performance.
Hi Swagatam,
For this circuit, my input is tapped from secondary transformer of 20-0 AC.
If I use rectifying it will produce some 28Vdc. How can I modify to suit this voltage please.
I need to drive a 5vdc relay in delay mode.
Thank you
Hi Raj, no need of using a full bridge rectifier at the transformer output. You can use a single diode 1N4007, and use the output for feeding the transistor emitter follower circuit. At the emitter side of the transistor you can connect the relay, and connect a 1000uF/25V parallel to the relay coil.
You can replace the 10K preset with a 7V zener diode.
Sir, If I use 7805 and diodes to drop Voltage from 9v to 3v , what about heating problem and power consumption ?
Nitin, Heating of 7805 cannot be avoided, otherwise you may have to make a buck converter circuit.
Hi sir, I want to run small 3v motor . Can I use Voltage divider with 150ohm and 100ohm resistor to drop voltage upt 3.6v with 70mA current?
Or please suggest another good way to do it.
Thanks in advance.
Hi Nitin, what is the input voltage? Is it 12V?
Anyway you can drop the voltage by using the last concept explained above, or you can also use an LM338 IC variable voltage regulator to adjust the output to 3V.
Resistor divider will not work, because resistors will waste a lot of power and also will not be able to provide sufficient current to the motor.
Another simple idea would be to use a 7805 IC and 3 series diodes in series with its positive output
Sir, i wanted to use the voltage drop across a rectify diode to regulate my 12v source to about 10.6 or 10.8v at 3A current, but you just explain the drawback to this configuration, please which other configuration will be suitable for me, because my resultant voltage must be stable because of the sensitivity or my load.
Adams, you can try the transistorized circuit, and feed the supply from a 7812 IC, the output will be fairly accurate.
Yes sir, the Transistorize circuit will work better, but the current will be an issue. i am looking at a 3A current source but as explained, the transistor can handle 0.5A. What can i do, sir
Adams, TIP122 is actually rated to handle upto 5 amps, so you can expect it to work easily with 3 amp current just make sure to mount it on a good heatsink
Thank you sir.
gud am Sir! I’m I really love your projects, I pray that u will continue in experimenting and discovering new circuits that is really really helpful for many of your follower like me 🙂 Sir is heat sink necessary for the second circuit? for tip122 or tip142?
Thank you jindro, yes heatsink will be required if the load current rating is above 500mA
Can I use this schematic to step down from 24V (truck battery) to 19V at about 5A(the high current is needed to charge the laptop battery) or do I need to replace some components?
-hobby/newbie at electronics here
yes you can… try the transistor option, replace the TIP122 with a TIP142 for 5 amp limit
Thank you for the quick reply, I'll try it as soon as I get the components.
Sir do you have a 12v to 5v buck conv using 555 timer ckts, using my inverter to 12v SLA-battery consume a 10w of power even in no load situation. 🙁
You can try the second circuit from the following link:
https://www.homemade-circuits.com/2015/10/calculating-inductor-value-in-smps.html
Hi Swagatam,
What are the wattage of the resistor on the collector and base respectively Thanks
Regards
Kanta
Hi Kanta,
when wattage is not mentioned it's always 1/4 watt
Hi Sir,
I feel this may be a silly question to you. But I am new and just wanted to know the purpose of 1k & 10k resistors at the output in both the circuits?
Can you pl explain?
Hi Manoharan.
the 1K resistor is for biasing (triggering) the transistor into conduction, while the 10K resistor is just to tie-up the emitter of the transistor to a ground reference which actually is not essential and may be avoided.
Thanks much sir.
sir what is outboard transistor? and what is the current capacity of lm196
Mujahid,
please refer to the following article, the first diagram shows T1 connected as an outboard transistor while the second diagram shows how to use an LM196 transistor.
https://www.homemade-circuits.com/2012/05/make-this-voltage-stabilizer-circuit.html
LM196 will provide you with 10amp current
sir i want to handle 10 amp load with voltage regulator. so I want to know can I use two or three lm338 parallel controlling with single variable? is it possible? or any other solution. thank you
you can replace LM338 with a LM196, or you can use an outboard transistor with LM338 for increasing its current capacity.
can I replace tip122 with to-3 package transistor like the one 2n3055 or any for higher current
yes you can do it, but for optimal current you may have to reduce the base resistors value accordingly ( wattage may increase significantly)
I made a transformerless timer circuit with ic555. I used 0.1uf cap to drop voltage including 12v zener diode to get 12v supply. I did not use a resistor in series with zener diode. the circuit worked well for a few days but then zener got damaged. is it because that I did not use a resistor in series with zener or I used zener diode that was unable to support load.
as you have stated above we cannot get sufficient current from zener diode
I think you have used 1uF and not 0.1uF….anyway, yes it's because you did not use a limiting resistor in the circuit…but the resistor must be in series with the input supply not in series with the zener. use a 1watt zener diode, and if possible connect a 1N4007 in parallel to the zener diode.
sir I tried the second circuit and its very nice regulating voltage. sir what about the circuit with ic lm338
Mujahid, LM338 circuit is the standard, traditional way of regulating voltage, but it's much complex and expensive than the above concepts