In this article we learn how to make a couple of simple DC to DC voltage doubler circuits using a single IC 4049 and IC 555 along with a few other passive components.
If you are wondering how a simple IC 555 can be used for making a powerful voltage doubler circuit, then this article will help you to understand the details and construct the design at home.
What's a Voltage Doubler
A voltage doubler is a circuit that uses only diodes and capacitors for raising an input voltage into a higher voltage output, twice the magnitude of the input.
If you are new to voltage doubler concept and desire to learn the concept in-depth, we have a good elaborate article in this website explaining different voltage multiplier circuits for your reference.
Voltage multiplier concept was first discovered and used practically by British and Irish physicists John Douglas Cockcroft and Ernest Thomas Sinton Walton, hence it is also called the Cockcroft–Walton (CW) generator.
A good example of a voltage multiplier design can be studied through this article which exploits the concept for generating ionized air for purifying air in homes.
A voltage doubler circuit is also a form of voltage multiplier where the diode/capacitor stage is restricted to a couple of stages only, so that the output is allowed to produce a voltage that may be twice of the supply voltage.
Since all voltage multiplier circuits mandatorily require an AC input or a pulsating input, an oscillator circuit becomes essential for accomplishing the results.
IC 555 Pinout Details
Circuit Diagram of Voltage Doubler using IC 555
Referring to the above example, we can see an IC 555 circuit configured as an astable multivibrator stage, which is actually a form of oscillator, and is designed to produce a pulsating DC (ON/OFF) at its output pin#3.
If you recall, we had discussed an LED torch circuit in this website, which quite identically uses a voltage doubler circuit, albeit the oscillator section is created using an IC 4049 gates.
Basically, you can replace the IC 555 stage with any other oscillator circuit and still get the voltage doubling effect.
However using IC 555 has a slight benefit since this IC is able to generate more current than any other IC based oscillator circuit without using any external current amplifier stage.
How the Voltage Doubler Stage Operates
As can be seen in the above diagram, the actual voltage multiplication is implemented by the D1, D2, C2, C3 stage, which are configured as a half-bridge 2-stage voltage multiplier network.
Simulating this stage in response to the IC 555's pin#3 situation can be a little difficult, and I am still struggling to get it running in my brain correctly.
As per my mind simulation, the working of the mentioned voltage doubler stage can be explained as given in the following points:
- When the IC output pin#3 is in its low logic or ground level, D1 is able to charge C2, since it is able to get forward biased through C2 and pin#3's negative potential, also simultaneously C3 is charged via D1, and D2.
- Now, in the next instant as soon as pin#3 becomes at high logic or at the positive supply potential, things get slightly confusing.
- Here C2 is unable to discharge via D1, so we have a supply level output from D1, from C2, and from C3 also.
- Many of the other online sites say that at this point the stored voltage inside C2, and the positive from D1 is supposed to combine with the output of C3 to produce a doubled voltage, however that does not make sense.
- Because, when voltages combine in parallel, the net voltage does not increase. The voltages must combine in series to cause the desired boosting or the doubling effect.
- The only logical explanation that can be derived is, when pin#3 becomes high, C2's negative being at the positive level and its positive end also held at the supply level, it is forced to produce a reverse charge pulse which adds up with the C3 charge, causing a instantaneous potential spike having a peak voltage twice that of the supply level.
If you have a better or technically more correctexplanation, please do feel fre to explain it through your comments.
How much Current?
Pin#3 of the IC is assigned to deliver a maximum of 200mA current, therefore the maximum peak current can be expected to be at this 200mA level, however the peaks will get narrower depending on the C2, C3 values. Higher value capacitors might enable fuller current transfer across the output, therefore make sure the C2, C3 values are optimally selected, around 100uF/25V will be just enough
A Practical Application
Although a voltage doubler circuit can be useful for many electronic circuit applications, a hobby based application could be to illuminate a high voltage LED from a low voltage source, as shown below:
In the above circuit diagram we can see how the circuit is used for illuminating a 9V LED bulb from a 5V supply source, which would normally be impossible if the 5V was directly applied on the LED.
Relation between Frequency, PWM and the Voltage Output Level
The frequency in any voltage doubler circuit is not crucial, however faster frequency will help you to get better results than slower frequencies.
Similarly for the PWM range, the duty cycle should be roughly 50%, narrower pulses will cause lower current at the output, whereas too wide pulses will not allow the relevant capacitors to discharge optimally, again resulting in an ineffective output power.
In the discussed IC 555 astable circuit, the R1 can be anywhere between 10K and 100K, this resistor along with the C1 decides the frequency. C1 consequently can be anywhere between 50nF to 0.5uF.
R2 will fundamentally enable you to control the PWM, therefore this can be made into a variable resistor through a 100K pot.
Using IC 4049 NOT gates
The following CMOS IC based circuit can be used for doubling any DC source voltage (up to 15 V DC). The presented design will double any voltage between 4 to 15 V DC and will be able to operate loads at current not more then 30 mA.
As can be seen in the diagram, this DC voltage doubler circuit employs just a single IC 4049 for achieving the proposed result.
IC 4049 Pinouts
The IC 4049 has six gates in all which are all effectively for generating the discussed voltage doubling actions. Two of the gates out of the six are configured as an oscillator.
The extreme left of the diagram shows the oscillator section.
The 100 K resistor and the 0.01 capacitor form the basic frequency determining components.
A frequency is imperatively required if a voltage stepping actions needs to be implemented, therefore here too the involvement of an oscillator becomes necessary.
These oscillation become useful for initialing the charging and discharging a set of capacitors at the output which amounts to the multiplying of the voltage across the set of capacitors in a such a way that the result becomes twice the applied supply voltage.
However the voltage from the oscillator cannot be preferably applied directly to the capacitors, rather its done through a group of gates of the IC arranged in a parallel way.
These parallel gates together produce a good buffering to the applied frequency from the generator gates so that the resultant frequency is stronger with respect to current and does not falter with relatively higher loads at the outputs.
But still keeping the specifications of a CMOS IC in mind the output current handling capacity cannot be expected to be larger than 40 mA.
Higher loads than this will result in the deterioration of the voltage level toward the supply level.
The output capacitor values can be increased to 100uF for getting reasonably higher efficiency levels from the circuit.
With 12 volts as the supply input to the IC, an output of around 22 volts may be acquired from this IC 4049 based voltage doubler circuit.
- R1 = 68K,
- C1 = 680pF,
- C2, C3 = 100 uF/ 25V,
- D1, D2 =1N4148,
- N1, N2, N3, N4 = IC 4049,
- LEDs White = 3 nos.