# Voltage doubler circuit with 555 and current 500mA?

Discussion in 'Digital Circuit Design' started by abhay12, Jul 13, 2016.

1. ### abhay12 Thread Starter New Member

Jul 13, 2016
2
0
I am fairly new to Electronics, and creating a dc-dc voltage doubler circuit for my project. I have researched online and decided to follow this voltage doubler circuit but the problem is I am only getting around 45mA output current and I need atleast 500mA current. I also tried other circuits but output is more or less same, one constraint here is that I need to use 555 to double the voltage. So is that possible with 555? Should I increase the Battery power? Just guide me in right direction.
Thanks

2. ### TheButtonThief Active Member

Feb 26, 2011
219
38
Getting 500mA from a 555 is full of nope

Why must you use a 555? there's hundreds of better solutions for stepping up DC voltage. Note though that it will always be at the cost of current. You can't increase voltage and increase current from the same supply, or even maintain the same current, voltage goes up and so current must come down.

3. ### Alec_t AAC Fanatic!

Sep 17, 2013
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Welcome to AAC!
The output voltage of the standard 555 is always a volt or so less than the supply voltage, so even with an ideal voltage doubler you can never get double the supply voltage. Also, the 555 can't source or sink more than about 200mA.

4. ### AlbertHall Well-Known Member

Jun 4, 2014
2,261
447
You could disconnect C3 from U2 and add in this circuit:
(Note the transistors need to be able to handle 1A)

But as others have said you will get less than 10V output.

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5. ### AnalogKid Distinguished Member

Aug 1, 2013
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1. The 555 output square wave is not symmetrical. That is, it does not have a 50% duty cycle. This is impossible to achieve by design with the standard astable circuit, but is a natural fallout of an alternate astable design that has only one timing resistor instead of two.

2. The 22 uF capacitors are too small for a 1 A output current. At an operating frequency of 1.9 kHz and 50% duty cycle, the caps have over 500 us in each charge and discharge phase. But a 22 uF cap charged up to 10 V is discharged to less than 1 V in less than 200 us. Both the operating frequency and capacitor size need to increase.

3. Between the 555 output headroom loss and the two diodes, the voltage swing into the output capacitor is much less than Vcc x 2. Adding the current boosting stage in post #4 increases the losses. You probably should add another stage to the charge pump, making the overall circuit a voltage tripler. The output will be much closer to 10 V.

4. A fundamental flaw when applying a charge pump topology to a relatively high current output is that the two pump capacitors are effectively in series, and form a voltage divider whose impedance is based on the switching frequency. For the output voltage not to sag under load, the impedances of the caps have to be *very* low. Note that commercial charge pump chips generally quit at about 100 mA output current - for a reason.

ak

Last edited: Jul 13, 2016
6. ### Sensacell Well-Known Member

Jun 19, 2012
1,183
276
For a doubler with 500 ma output, you need a switching boost converter with an inductor.
It's just not going to happen with a charge pump.

7. ### DickCappels Moderator

Aug 21, 2008
2,758
667

This circuit is a little bit dangerous because if you unplug the NE555...ZAP!

You would need heavier duty components to get 500 ma. Remember, you will loose about a volt per half-wave doubler stage.

8. ### abhay12 Thread Starter New Member

Jul 13, 2016
2
0
Thanks for all the replies.
Although I am not able to understand all the answers completely, as I am new to electronics, but I will research and try to understand the terms used in Replies.
Further I shall try the circuits, suggested by AlbertHall and DickCappels.
As per the above discussion, we can not achieve 500mA from 555 IC, so if we remove the constraint of 555 IC, what are the other ICs which can be used to get 500mA and can double the voltage at the same time. Its better if someone can show me some circuit of this kind.

9. ### Bernard AAC Fanatic!

Aug 7, 2008
4,233
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Might find a boost converter on ebay like XL6009 by Banggod, 10 pc for \$ 16.79, Ouch!, or sim. with Vm., \$ 5.98.

10. ### ian field Distinguished Member

Oct 27, 2012
4,447
791
That will add more Vbe drop to an output that already doesn't go fully rail to rail.

If actual doubling is important, the final stage needs to be complementary pair common emitter - and that requires extra driver circuitry.

CMOS outputs tend to be more like rail to rail, but the 7555 is even less output current than the 555.

An old favourite was the 4007 uncommitted inverter pairs, that won't handle the required current but there was a RCA/Harris/etc buffer to boost the power of that device - I don't know if that can handle the current either, but you can probably gang a few to get the rating.

11. ### Bernard AAC Fanatic!

Aug 7, 2008
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414
Re post # 9, also available from amazon, OSOYOO, XL6009, 2 for \$ 7.50. 4A output, 3.5 V min input, adjust
output to 40 V.

12. ### crutschow Expert

Mar 14, 2008
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3,362
At that current level I would go with a boost switching regulator.

13. ### Bordodynov Active Member

May 20, 2015
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See

#12 likes this.
14. ### ian field Distinguished Member

Oct 27, 2012
4,447
791
It can be done with a couple of transistors and a suitable inductor.

You can pretty much copy the blocking oscillator style inverter used in cheap flyback SMPSUs. Just scale the relevant component values to suit the intended input voltage.

Anything over 1.5V Vcc and you need to capacitor couple the feedback to the base and provide it with a startup resistor.

Regulation is little more than a second transistor shunting the B/E of the chopper transistor.

I have seen MOSFET versions of that style SMPSU - but they're more complex with an extra winding to provide a standing negative rail for the regulation control circuit.

15. ### #12 Expert

Nov 30, 2010
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and Bordodynov comes through with the cash...again.

16. ### dannyf Well-Known Member

Sep 13, 2015
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Use a boost converter.