Maker Pro
Maker Pro

Boost converter 12V to 36V (25 mA load) using 555

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
Take a look at the dedicated smps ICs.

The intent of that link was to show how complex it can be to design a "simple" DC-DC converter.

Sure, I could offer improvements that could boost its performance, but all that would do is add more complexity. And that's fine if your aim is to learn more about techniques and pitfalls, but less so if you want something that just works.

The dedicated solutions will give you a guaranteed performance and their datasheets will contain equations or tables to guide you to appropriate value components. There are some that require in the order of 3 additional components.

I do understand that using something that just works, won't give me any knowledge. Really, I am eager to learn but, you would agree that a hobbyist's IQ would be much lower than that of an electronics graduate/engineer. Therefore, sometimes its is not so easy for me to understand the configurations and terminology etc. But never give up, I keep on searching for further explanations on the net so that I can catch the exact idea.

Believe me, I am really grateful to guys like you, Kris, Adam and others from whom I have learned a lot. I shall be happy to receive your improvement tips, of course, to learn more.

Lastly, again for the sake of learning, may I ask about the 555 design which I posted at #18? Would appreciate to have some comments, if it is somehow workable or it is just nonsense.

Thanks.
 
Last edited:

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
I've never tried simulating that circuit. I did my tests in real life.

As per my normal practice, first I do my best to understand the schematic and if required I usually take help of one or two simulators to see the performance/behavior. And then I assemble it on breadboard before final soldering.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I've got no problems if you want to experiment and try stuff out, just let us know. Most people are looking for a simple, hassle free solution, but I understand that you may have different priorities.

This humble hobbyist will take a look at the 555 circuit as soon as he gets in front of a computer with a real screen and keyboard :)
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Ok, a quick first impression of the 555 circuit.

This is another circuit without a feedback path to enable regulation. This means that the voltage will be set by the weakest link in your system.

At any particular time this might be the breakdown voltage of the transistor, diode, filter capacitor, or even the insulation on your wires.

It's hard to argue that this is other than a bad thing. It will give you an output voltage that may be unpredictable and your circuit may gradually fail with no obvious signs of damage, or it suddenly fail with a bang or a flame or a puff of smoke.

With very few exceptions we normally try to avoid designs that place so many components under so much stress.

If you go back to the thread about my experimental DC-DC converter you'll see that I ran it without feedback at first, but then a lot of effort was put into the feedback network to get a useful circuit.

An earlier circuit had a zener diode across the output to provide regulation. That would work, but in the same way as you could control the speed of a car by having your foot flat on the throttle and using the brake to determine the speed.

A feedback mechanism is more a control of the throttle than the brake. As you would appreciate from the analogy, controlling speed with the throttle is more efficient than controlling speed with the brake.
 

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
I've got no problems if you want to experiment and try stuff out, just let us know. Most people are looking for a simple, hassle free solution, but I understand that you may have different priorities.

This humble hobbyist will take a look at the 555 circuit as soon as he gets in front of a computer with a real screen and keyboard :)

Thanks for the positive response. :)
 

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
Ok, a quick first impression of the 555 circuit.

This is another circuit without a feedback path to enable regulation. This means that the voltage will be set by the weakest link in your system.

I assume you are talking about the 555 circuit posted at #18 and not at #13.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Excellent timing. I was just coming back to look at this again.

My comments were about the circuit posted at #13.

The one posted at post #18 has feedback and is thus a better circuit.

You do need to ensure that the components are not overly stressed. Let's take a look :)

The 2N7002 is a pretty small part, I'd probably use something a little more rugged.

I would also be careful about any voltage transients placed on the power rail by the switching of current through the inductor. You would be wise to decouple the connection to the 555 part of the circuit from the current fed to the inductor. Likewise the ground connections should be separate. For such low currents it's not going to be a huge issue, but it's always best to have these things in the back of your mind.

It would be interesting to see it in practice.
 

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
Actually I wanted to use 2N7000 but LTspice does not have that model. But I think 2N7000 would also be not suitable, have to use something which can bear about 2A. (The LTSpice shows that the mosfet and the diode are drawing about 1.5A). If this is the case, does this means that the input supply must be at least 12V 2A to give an output of 35V at 50mA from this circuit?

Can you please elaborate where to place the decoupling capacitors and their values/types?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Assuming 100% efficiency, for 36V 50mA output you require an average input current of 150mA at 12V (12 * .15 = 36 * 0.05).

However the peak current will be much higher. if the duty cycle is 10% (10% of the time it is on and 90% of the time it is off) then the current rises to a peak during the on time and falls again during the off time. If we assume that it's symmetrical, the current is actually averaged over 20% of the time period, and it rises to a peak of approximately double this average. So the peak current is approximately .150 * 100/20 * 2, which is about 1.5A.

The peak current for these mosfets (even for pulsed loads) is under that so yeah, you do want a meatier mosfet.

decoupled.jpg

That's what I'm talking about. Decoupling the high current and the low current paths. The diode and the cap (10uF should be OK) will help isolate the 555 from any noise created by the "high" current switching.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
I've had a bit of a play with the 555-based design from post #18 in LTSpice.

I don't like the fact that there's no monitoring of the current in L1. This is likely to lead to saturation, and loss of the magic smoke from the MOSFET.

Also with the original design, the ON time of Q1 is too long for a 100 µH inductor at 12V; with C1 = 2n2 the ON time is around 16 µs, and the increase in inductor current can be calculated from:

dI = dT V / L
= 16×10-6 × 12 / 100×10-6
= 1.9A

That's far more peak current than you need; for a 25 mA load and a step-up ratio of 3, your peak inductor current should be about 300 mA.

For a 100 µH inductor, that corresponds to an ON time of 1.667 µs, which is pretty fast for a 555. So I tried increasing the inductor to 470 µH and recalculated the required MOSFET ON time.

dT = dI L / V
= 0.2 × 470×10-6 / 12
= 7.8333 µs

Next I changed the operating mode of the 555. You really want it to operate more as a pulse generator, generating high output pulses with a fixed width of about 8 µs. So it should operate a bit like a gated oscillator: when the feedback voltage from the zeners drops low enough, it should produce an 8 µs high pulse to the MOSFET.

This can be done by changing the charge path for C1 so that instead of being charged directly from the 12V rail, it is charged only when the OUT pin is high. This is done using a diode and resistor from OUT to pins 2 and 6.

The fundamental problem is inductor saturation at startup. During the first 5~10 switching cycles, when the output voltage is only slightly higher than the input voltage, there is very little voltage difference across the inductor when the MOSFET is OFF, and the magnetic field, and the inductor current, only drop slowly. You can see this from the formula dT = dI L / V; as V approaches zero, dT approaches infinity. Since the inductor current doesn't fall much between pulses of energy from the MOSFET, the inductor current climbs quickly at startup, like a staircase, at the start of switching.

Inductor saturation at startup can be avoided by monitoring the actual inductor current, and simply locking out pulse generation while it's over a certain point. This is a general way to design a boost converter - turn the MOSFET OFF when the inductor current reaches a certain value; this is how current mode controllers work (the inductor current is sensed in the path between the MOSFET's source and the 0V rail, for convenience). Most of the boost converters I linked to in my earlier post use current mode control, which is a good technique because it detects and prevents inductor saturation problems directly at the source (pun intended).

I modified the circuit again by adding a 2.2Ω current shunt resistor (marked RS) in series with the inductor, and a PNP transistor, Q2, monitoring the voltage across that shunt resistor. As soon as the voltage reaches about 0.7V (which corresponds to an inductor current of about 320 mA), the transistor conducts and charges the timing capacitor very quickly. The timing capacitor in fact has almost no effect on the ON time of the MOSFET, and I removed the path from the 555's output through the diode to the timing capacitor with little effect on the circuit.

The timing capacitor and the resistor from it to 0V (which I've called RT) now determine the minimum OFF time, which determines the maximum voltage step-up ratio. That small change (adding Q2) has simplified the design quite a bit. It also prevents the 555 from starting up until the initial power-on current surge has finished; during that time, Q2 is saturated, the CT voltage is held near to VCC, and the 555 does nothing.

The RT value is not critical. With CT = 1 nF and L = 470 µH, RT = 3k3 gives continuous conduction until the regulation kicks in. With RT = 10k there is discontinuous conduction before regulation kicks in, so the startup time is a bit longer - about 6.5 ms. So I chose 5k6 as a compromise :)

So here's the schematic that I've ended up with:

epoint 269429 schematic.png

I've changed Q1 to another smallish MOSFET with a lower ON-resistance. I've used a 1N5819 for D1 because my LTSpice installation doesn't have the UF400x series; you should continue to use a UF400x in your circuits.

And here are some graphs from the simulation:

epoint 269429 graph 1 overview.png

This graph shows the first 5 ms of operation from power-up. The green trace shows the output voltage, using the legend at the left side. At point B, around 4.2 ms, the output voltage has reached around 36V and stabilises due to the zener regulation.

The red trace shows the inductor current, using the legend at the right side. At the start, at point A, you can see that the inductor current was very high. With the values I used, it starts off at about 2.7A. This is way above its saturation current but that doesn't matter; no switching is occurring, and no current is flowing in the MOSFET. This is just the initial current required to charge COUT.

The rest of the time, you can see that the inductor current always peaks at around 320 mA. This is the point where Q2 starts to conduct. This is shown in more detail later.

The blue trace is the voltage on CT. It is not very useful in this view.

epoint 269429 graph 2 startup.png

This graph shows the very start of operation. Point A is marked again. Operation begins when the inductor current drops below about 300 mA and Q2 turns OFF, allowing the CT voltage (blue trace) to start to fall, as CT discharges through RT. Once this voltage reaches the 555's low threshold (1/3 of the VCC voltage), the 555 output goes high and the inductor current starts to ramp up again.

From that point onwards until regulation kicks in, the circuit operates as shown in the next graph.

epoint 269429 graph 3 startup mid-way.png

The inductor current is shown in red, using the right hand legend. The voltage on CT is shown in blue, using the left hand legend.

Once CT voltage goes below 1/3 VCC (the 555's falling threshold voltage), the 555 drives its output high, turning on Q1. This causes the inductor current to increase. At point C, the inductor current is approaching the level where the voltage drop across RS is enough to bias Q2 into conduction; you can see the CT voltage starts to level off. Between points C and D, conduction in Q2 is increasing - first gradually, then rapidly, pulling the CT voltage upwards towards VCC. At point D, with the inductor current at 320 mA, Q2 is conducting quite heavily and the blue trace (CT voltage) is nearly vertical.

As soon as the CT voltage hits the upper threshold of the 555 (2/3 of VCC), the 555's output goes low, and Q1 turns OFF, and the inductor voltage (drain of Q1) "flies back" as energy from the inductor is dumped into the load. This is shown in the next graph. The inductor current ramps down again, and CT begins to discharge again through RT until the 555's lower threshold voltage is reached again, when the cycle repeats.

epoint 269429 graph 4 change to discontinuous.png

This graph shows the point where regulation kicks in, i.e. the output voltage is high enough that current through D2 and D3 prevents CT from discharging all the way to the 555's falling voltage threshold.

The graph does not show the CT voltage; it shows the inductor current in red (right hand legend) and the Q1 drain voltage (green, left legend).

You can see that when Q1 is OFF, and the inductor current is falling, Q1's drain voltage is about 36V. This is due to the back EMF from the inductor as it dumps its stored energy into the load. The flat top corresponds to a steady transfer of energy through D1 into COUT and the load.

Around 4.2 ms you can see the first discontinuous cycle; that is, the first cycle where the inductor current reaches zero because the MOSFET OFF time is now longer than the time taken for the inductor to dump all of its energy into the load. Subsequent cycles show the inductor "ringing" with the parasitic capacitance (in the inductor, Q1, and D1) as the last bit of energy decays to zero. The ringing is interrupted by the next MOSFET ON pulse, which pulls Q1's drain hard to 0V.


This design seems to simulate fairly well. I would be interested to know how it performs in real life. I suspect it will be noisy - both electrically, and audibly. I suspect it will suffer from "subharmonic oscillation", which is where a pattern emerges over a period of 2, 3, 4 or more cycles, that causes an audible sound from the inductor. This can be quite irritating and is often made worse if the load current varies significantly, since reduced load current causes the switching rate to drop from the maximum of 50 kHz during startup and overload, to something much lower.

In the simulation, reducing the load current to 10 mA (40% of specified) causes the switching frequency to drop to about 12 kHz! This is well within the audible range; even at higher frequencies, the subharmonic oscillations can be very audible.

If you build this circuit up, please let me know how it goes :)
 
Last edited:

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
Really no words to thank both of you!

And Kris, I can't express...................your devotion is greatly appreciated. I assume that you might have been playing with the LTSpice for the satisfaction of your electronics passion but then sparing time to write technical explanations in teaching style; really deserves kudo.

I have gone thru the whole post once and got a rough idea. But, I need to read it a few more times before simulating it. I am very happy to have the improved design with its detailed working. I hope to learn many new ideas from your design. Although a simple 3-component LT1173 would be hassle-free solution for my current requirement, I would certainly build this circuit not only for my satisfaction but also to express my thanks to you.:)

EDIT: By the way, is there any relation between the zener voltage and the output voltage? In my design, as I reported earlier it is
Vout = Vzener + 8.
 
Last edited:

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Thank you abuhafss. It's nice to be appreciated :) And you're almost exactly right. I enjoyed experimenting with the design, and I did write up the explanation for the benefit of you and anyone else reading the thread, but it also benefitted me - I had to check my understanding and try some experiments to confirm my ideas, and I learnt from that process as well.
 

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
Thank you abuhafss. It's nice to be appreciated :) And you're almost exactly right. I enjoyed experimenting with the design, and I did write up the explanation for the benefit of you and anyone else reading the thread, but it also benefitted me - I had to check my understanding and try some experiments to confirm my ideas, and I learnt from that process as well.

But I was wondering how would it help others when the title of this thread is irrelevant. How the title could be revised?
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
How the title could be revised?
By magic :) Are you happy with the new title?

BTW I was wondering about the efficiency of that circuit, because of the 2.2Ω series resistor, but it's actually about 85%, which is a lot better than I expected.
 

abuhafss

Aug 3, 2010
348
Joined
Aug 3, 2010
Messages
348
By magic :) Are you happy with the new title?

BTW I was wondering about the efficiency of that circuit, because of the 2.2Ω series resistor, but it's actually about 85%, which is a lot better than I expected.

Better to add "555" before this new title and then it would be perfect.

Can the efficiency report of this 555 based boost circuit be measured or printed in LTSpice? Actually I failed because the options in the VIEW menu are grayed. I assume it is to be used only for ICs like LT1170 or LT1173.

EDIT: By the way, is there any relation between the zener voltage and the output voltage? In my design, as I reported earlier it is
Vout = Vzener + 8.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
I just graphed the efficiency using the formula shown in red at the top of the graph. You'll see that it's low during startup but levels off around 85%.

I had to increase CIN from 100 µF to 1000 µF otherwise there was too much "wiggle" in the efficiency trace. This is also why it takes about 10 ms to stabilise.

efficiency.png

The output voltage in the simulation is about 36.8V. The zener voltage is 33V. So the output voltage is roughly VZ + 3.8V.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
EDIT: By the way, is there any relation between the zener voltage and the output voltage? In my design, as I reported earlier it is
Vout = Vzener + 8.

That one is easy :)

Remember that the 555 has three 5k resistors in series to give the two trigger points at 1/3 and 2/3 of the supply rail.

With a 12V rail, the upper trigger point is 2/3 * 12 = 8V

When the output voltage is sufficient to raise the trigger input to 8V or higher the 555 stops switching and leaves the mosfet off. The zener diode is connected between the output and the trigger , so the output has to be Vz higher than the trigger point. Note that this means the output gets to Vz + 2/3 Vcc. With the supply voltage you have, this means the zener voltage + 8V

Two other small wrinkles are that (a) the output voltage is dependent on the input voltage, and (b) the zener voltage is dependant on current. The current to the trigger pin is low, so (especially for low voltage zeners) the voltage you get may be less than you expect.

This voltage dependency is almost exactly like the original design I pointed you to with the schmitt trigger inverters. If you want a stable output voltage and your input voltage can change then this circuit is not the one for you :)
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
When the output voltage is sufficient to raise the trigger input to 8V or higher the 555 stops switching and leaves the mosfet off. The zener diode is connected between the output and the trigger , so the output has to be Vz higher than the trigger point. Note that this means the output gets to Vz + 2/3 Vcc. With the supply voltage you have, this means the zener voltage + 8V
The 555 won't switch back ON again until the voltage drops below the lower trigger voltage, which is about 4V. Also, there's a 1N914 in series with the zener, which would make the trigger voltage about 4.6V. The voltage in the simulation, 3.8V, is probably lower than 4.8V because of the relatively low zener current. But there could well be other factors. One thing I noticed is that Q2 doesn't turn OFF immediately when the rising voltage threshold is reached; there's a flat top on the CT voltage waveform. There are probably more imperfections in the circuit. It's pretty crude. I'm surprised that it simulates so well.
Two other small wrinkles are that (a) the output voltage is dependent on the input voltage, and (b) the zener voltage is dependant on current. The current to the trigger pin is low, so (especially for low voltage zeners) the voltage you get may be less than you expect.
The zeners are loaded by RT, which is 5k6. That means a zener current of about 700 µA. That is low, and there might be better ways to apply the feedback - especially to avoid generating subharmonic frequency components. Do you have any suggestions for improving the design? If not, it doesn't matter; I'm just messing around with it for fun.
This voltage dependency is almost exactly like the original design I pointed you to with the schmitt trigger inverters. If you want a stable output voltage and your input voltage can change then this circuit is not the one for you :)
Ah yes, I didn't think of that!

Simulating at 10V, 12V and 14V input voltages gives output voltages of 36.0V, 36.8V and 37.55V. The differences in output voltages are pretty close to one third of the diffrence in input voltages, as you would expect. If that's a problem, add a 6.2V 0.5W zener from pin 5 to 0V:

epoint 269429 schematic with zener.png

Here's the simulation (output voltages for VIN = 10V, 12V, 14V):

epoint 269429 output with input voltage varied.png

As you can see, they reach roughly the same output voltage; the obvious difference is the startup time. The actual average output voltages are 35.7873 at VIN=10V and 35.8792 at VIN=14V; a variation of slightly under 0.1V.

And here's the graph with a constant 12V input voltage, with three different load currents: 5 mA, 15 mA and 25 mA. Again the startup time is affected, but once it hits regulation, the output voltage variation is less than 40 mV between 5 mA and 25 mA load.

epoint 269429 output with input voltage varied.png

The following graph shows a close-up of the output voltage at 2.5 mA (the green trace) and 25 mA load (the blue trace).

epoint 269429 ripple closeup.png

The difference between the average voltages is slightly less than 40 mV but the ripple is very obvious. The ripple amplitudes are the same, but the frequencies are quite different. This is because the converter uses the gated oscillator concept, with a fixed amount of energy stored in the inductor on each pulse. When less energy is needed at the output, pulses are "missed" (really, the oscillator is held inactive), so the ripple frequency varies quite a lot; at low loads, it can be quite low.

I also just realised that I didn't specify any ESR figure for COUT when I captured those graphs, so the ripple at COUT is going to be worse than in the graph above.

So I made some changes to the output components: COUT now has an ESR of 1.2Ω (fairly typical for a 10 µF, 50V electrolytic); there's a series inductor of 3.3 mH (must have a rated saturation current of at least 50 mA) with a DC resistance of 14 ohms (this one is suitable: http://www.digikey.com/product-detail/en/RLB0812-332KL/RLB0812-332KL-ND/2561320), then a second identical 10 µF capacitor to 0V. The ripple is now a bit under 5 mV p-p at 2.5 mA load current. The regulation suffers, of course, because of the DC resistance of the series inductor; it's 0.32V from 2.5 mA to 25 mA output current! You can optimise the output components further if you like.
 

Attachments

  • epoint 269429 output with output load varied.png
    epoint 269429 output with output load varied.png
    2.8 KB · Views: 167

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I'm surprised the output voltage is largely unaffected by the supply voltage.

Damn, I'm so used to seeing trigger, threshold and discharge drawn on the dame side that I missed that trigger and threshold were connected.

I'd prefer to see the feedback exercise control in a way that's unaffected by supply voltage. The classic way is using an optocoupler, but you can also switch on a transistor that has its emitter tied to ground.
 
Last edited:

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
I'm surprised the output voltage is largely unaffected by the supply voltage.
It's mostly affected by the zener. The supply voltage only comes into it because it affects the falling voltage threshold at the 555's input, which is 1/3 VCC. That's why for every 1V change in input voltage, you get a 0.333V change in the output voltage. Unless you add the zener on pin 5.
Damn, I'm so used to seeing trigger, threshold and discharge drawn on the dame side that I missed that trigger and threshold were connected.
Yeah, it's really just used as a Schmitt trigger. The timing capacitor doesn't even affect the MOSFET's ON time; the ON period is ended when the inductor current reaches about 320 mA and Q2 charges it up very quickly. It's a bit of a Frankenstein's monster circuit.
I'd prefer to see the feedback exercise control in a way that's unaffected by supply voltage. The classic way is using an optocoupler, but you can also switch on a transistor that has its emitter tied to ground.
Hmm. I don't think the existing design is ideal, but it seems to work pretty well and it's fairly simple. I don't like putting common emitter stages inside feedback loops - in my experience it leads to instability. I think that's because of the delay (or phase shift, if you want to look at it that way) and the extra gain, but stability is a subject I've never understood, so maybe I've been doing it wrong.

Setting the 555's threshold voltages (with a 6.2V zener from pin 5 to 0V) makes the voltage feedback independent of supply voltage, as long as it doesn't go below about 10V.

I'm slightly concerned about the possibility of damaging the 555 if the output voltage was to jump upwards, perhaps due to a sudden drop in load current. I'll have a look at its response to load current changes.

Edit: The main circuit seems to handle load changes very well. If the pi filter is present on the output, that causes a slow overshoot because of the inductor. But without that, as soon as the load is taken away, the oscillator stops. The most overshoot you would see would be if the load disappeared while the MOSFET was ON; in that case you would see overshoot equivalent to the ripple voltage.
 
Last edited:
Top