How to build a variable boost converter?

Question

Hi,

I am trying to build a variable DC power supply. Let me explain my setup:

1. Using a 9V battery, I am powering an LM555 timer to generate a variable frequency (~680 to ~6800 Hz) square wave which has a duty cycle of roughly 50%. One of the resistors in my circuit is a potentiometer which allows to change the frequency between 680 and 6800 Hz while preserving (mostly) the 50% duty cycle.

2. The varying frequency output from the timer will be used as the switching signal to a transistor. This will be implemented as a Boost Converter. Check wikipedia for a quick lesson on how the Boost Converter works if you don't know (may also be known as a "joule thief"). Controlling the frequency controls the output DC voltage.

First, understand this is a project I am taking up in my spare time. The values for my resistors with the timer may not provide an adequate range of frequency to give me the boosted voltage I want (ideally I want to reach 25-30 V and ~50 mA).

Now my questions are these.

1. Is this elementary idea feasible? I think it is but I am very open to suggestions/constructive criticism.

2. While I haven't implemented the actual Boost Converter yet, I am getting a mixed result on the PWM from the timer. My frequency range is indeed correct, but if the timer works away (and gets hot), the frequency of the output steadily decreases, which will affect the output on the Boost Converter. Why is this so? My suspicion is the heating up of the 555 timer. Any suggestions as far as using a heatsink (a penny?) or using a recommended chip other than the 555? Or should I follow another way to make a variable voltage supply? Please help.

In case you were interested, the resistors in my timer circuit are 1 KOhm (resistor) and 100 KOhm (pot). I didn't include an image to the diagram because it isn't necessary.

Many thanks in advance!

Answer 1

Roger seems to have provided all the calculations. I assume they are correct. So to add to that & try and answer your specific Q's --



1) seems simple enough & feasible. Though you *might* mess-up your USB port at some time :D



2) the 555 is a very stable & simple IC to use. It generally does NOT get hot. However you DO have to watch the currents flowing through your resistors - which, if underated, will get hot and increase in value. I suggest you connect a small resistance in series with the pot to take care of minimum resistance in that leg. I assume the pot is connected between the 1K and the cap ? The cap discharges through this, and if its resistance goes really low - you can have heating in the pot !! Which might explain the freq drift downward - especially when you set the freq at high end (pot is at its lowest resist)



Lastly - I believe the actual voltage of the booster depends on the duty cycle of your PWM. However you have mentioned this to be steady at 50% ?!?

So here is an alternate circuit which might work better for you http://www.dprg.org/tutorials/2005-11a/index.html

Answer 2

1. To get an output of 25 to 30 volts at 50 ma ( 1.25 to 1.5 watts) requires the 9 volt battery to provide over 140 ma. This will quickly kill the battery. The duty cycle D will have to be greater than 50%, since Vo/Vi = 1/(1-D). D must be 0.64 to get 25 volts out, and D must be 0.70 to get 30 volts out. I suggest you use the 555 timer to drive a NPN transistor like the 2N2222A, use a 1k ohm resistor from pin 3 of the 555 to the base of the 2N222A. The transistor's emitter connects to the negative side of the battery and the collector connects to the inductor. It is easier to make the 555 "discharge" (when the 555 output is low) time constant (0.693*R1*C) and vary the "charge" ( 555 hi) time (0.693*(R1 +R2)*C).

For the 25 volt output D is 0. 64, the current I in the inductor must rise from 0 ma to 430ma.

Using ΔI = D*T*Vi/L where ΔI is 440 ma, Vi = 9 volts and T is the ramp up time (555 charge time)

and L is the inductor value T/L = 0.076, if you make L = 2 millihenries, T = 150 microseconds, this is the charge time of the 555. the discharge time would be 54 microseconds

For the 30 volt output, Duty cycle D is 0.7 the input current must average 170 ma, the current in the inductor will start at 0 and ramp to 485 ma, T/L is also 0.076, the charge time T = 150 microseconds, the discharge time would decrease to 45 microseconds.

Answer 3

555 Boost Converter

Answer 4

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