3

I plan on using a 555 timer for a PWM signal (variable with a potentiometer) and a gate driver to use a MOSFET as a switch. Essentially, I am trying to take a +300V rail and "step" it down using the PWM. The total current through the MOSFET wouldn't likely be more than 25mA, and would be more likely under 10mA. Mostly, what I am wondering is if it is safe to leave the MOSFET on for an extended period of time- at most, a few hours. I see no problem, but I haven't exactly tried it either.

Schematic

I also understand that the greatest losses are switching losses, so I'll try to minimize the switching frequency while still providing a, relatively, clean output that capacitors can filter.

The alternative to using such a PWM scheme is to simply put the potentiometer on the voltage rail and generate a drop across that, but that would likely load the circuit more than the MOSFET would, which is why I prefer the MOSFET.

Anindo Ghosh
  • 50,188
  • 8
  • 103
  • 200
Hari
  • 656
  • 5
  • 14
  • 2
    It might be worth sketching up the circuit you had in mind and picking a MOSFET you think may be suitable for review. You can edit and press ctrl-M to launch the on-site editor or just add a link to an image and someone can include it. – PeterJ May 18 '13 at 04:57
  • I've used MOSFETs in numerous home light switching designs, usually handling around 12V DC up to 10A. As long as the component you choose can handle the voltage/current you plan to use with it, then it can easily handle extended on periods. As long as the FET is not overheating from the power dissipation, it should be fine. – Kurt E. Clothier May 18 '13 at 07:02
  • Kurt, that is good to know. It is as I expected, but I definitely wanted to make sure. Peter, I've come to realize something. This is not a trivial circuit to design at all. I've been playing around with simulations for a good couple hours now, and I am no closer to a good solution than I was when I first set out to design it. Ultimately, I am looking for a high efficiency DC-DC converter circuit that takes ~275V and converts it (adjustably) to 150-250V. – Hari May 18 '13 at 08:07
  • Even worse, any working circuit I make has a LOT of power dissipation... I am basically taking a 555 output to drive the gate of a FET. The FET has a drain resistor of ~50k, but it forms a voltage with another 25k resistor placed closer to the V+ rail. In the middle of the divider is a parallel RC circuit, where R is the load (~75k) and C is conveniently sized to smooth ripples. I can get really small voltage swings by changing the PWM on the 555, but they are not really productive. I can also change the drain resistor (the 50k one). This does a better job, but it dissipates lots of power... – Hari May 18 '13 at 08:41
  • Show us a circuit - there are many ways this can be done but what you describe as your simulation means nothing without a circuit. – Andy aka May 18 '13 at 10:43
  • I'll upload that soon. – Hari May 18 '13 at 20:00
  • The circuit diagram is linked. The battery is a 300V source. In essence, I want to make this with a MOSFET, but replacing the switch with a MOSFET doesn't seem to work exactly the same. I should also be able to modulate the switch frequency so that I can vary the range from V_min to V_max (~200 - 275V) [Circuit](http://s832.photobucket.com/user/Silntsoldier/media/Circuit.jpg.html) – Hari May 19 '13 at 03:32
  • Quick addendum, the circuit above is the test circuit. The MOSFET ones will be added shortly. – Hari May 19 '13 at 06:50
  • [MOSFET Circuit](http://i832.photobucket.com/albums/zz248/Silntsoldier/Circuit-1.jpg)- This one shows a similar circuit (the one I originally designed). The problem is that when the MOSFET is on, the current through it is less than anticipated, so the voltage drop is smaller than expected and the swing at the load is correspondingly smaller. I'm not entirely sure of the problem here. – Hari May 19 '13 at 07:13

3 Answers3

2

If I'm reading this correctly you are using the MOSFET (switch) to charge a capacitor to a certain voltage. The voltage across the capacitor is being drained by the (fixed) load connected across it. The turn ON/OFF times are to be controlled by a 555 used as a PWM. Something like this:

enter image description here

Essentially it is an open loop system. The ripple voltage across the capacitor will be a function of the chop rate and the M/S ratio of the PWM. My first reaction would be to make it a closed loop system by monitoring and comparing the voltage across the capacitor (too high/too low) and using this information to control the PWM. Instead of altering the PWM 'manually' you would simply change the reference voltage (variable resistor) and the circuit would self adjust. I would also think an inductor between the storage capacitor (C) and load (and possibly a second capacitor across the load - pi filter) would be helpful to smooth the ripples. You may have to add a resistor in series with the MOSFET to limit the charging current and a fuse (from 300V supply) would be helpful in case of a short circuit caused by a faulty capacitor or an accidental short circuit output.

JIm Dearden
  • 111
  • 3
  • I've tried this circuit, and it doesn't work quite right. R_load is about 100k. Connected across a 100uF filter/storage capacitor, only about 8.4V is maintained across the resistor. The load, would draw, at max, 3mA from V+ if it were connected directly, and requires at least 1mA to operate. In theory, if the MOSFET were acting as a simple switch, the average current through it should all flow through the resistor (if the capacitor is "fully" charged), so the capacitor voltage is IR. Unfortunately, I (current) is not the expected ~1-3mA. The problem appears to be that V_GS is incredibly low. – Hari May 19 '13 at 04:04
  • Could you post your actual circuit driving the MOSFET? At the moment I am simply guessing the circuit setup from your written description and its difficult to suggest a solution to the problem. – JIm Dearden May 19 '13 at 09:06
  • [Circuit](http://i832.photobucket.com/albums/zz248/Silntsoldier/Circuit-2.jpg)- The linked circuit only charges the cap to ~8.4V, which makes some, since V_GS has a threshold of 1.5V in the simulation. In any case, the voltage could never be greater than 10V across the cap, since the 555 output is 10V. – Hari May 20 '13 at 01:11
  • @user24050 I think you may have interpreted my circuit too literally. The dashed line from the PWM is simply some sort of suitable control signal/interface device to turn the MOSFET 'switch' on and off according to the output state of the PWM. As I couldn't be certain if you had used a P or N type device I left it in this general form. As the 'switch' is a series element to the load it could be put below the CR section and does match up nicely with the circuit below. – JIm Dearden May 22 '13 at 16:28
  • Ah, I understand. That's actually what I now have just below this. Thanks! – Hari May 22 '13 at 23:55
1

I may have happened on a better circuit that performs the same function. In fact, it is actually very similar to a normal MOSFET/load switch. The load has a capacitor around it to smooth ripples. Also, the gate takes an inverted output from the 555 (will be done with an inverting gate driver), since the 555 can never have a shorter mark time than space time.

improved circuit

While I think I figured it out, I am still very open to other answers.

Kortuk
  • 13,362
  • 8
  • 60
  • 85
Hari
  • 656
  • 5
  • 14
  • The last line makes it sound like you are asking about this circuit, not just stating that you are still open to other answers. This lead to a mistaken delete, sorry. – Kortuk May 21 '13 at 22:07
1

What are you trying to accomplish with PWM? Do you want to convert the voltage efficiently? You can't do that without an inductor:

Can a charge-pump be 100% efficient, given ideal components?

If you do add an inductor, then you have a buck converter. You can roll your own, or buy them as complete modules.

Or is efficiency not as much of a concern as simplicity? If your load won't require more than \$25mA\$, then we aren't talking about a whole lot of power. At worst:

\$25mA \cdot 300V = 7.5W \$

is dissipated, either in the load, or in something dropping the excess voltage. The share of that between the load and the something else is determined by the voltage required by your load. A TO-220 can dissipate \$7.5W\$ with a heatsink, and around \$2W\$ without.

If you can deal with the excess heat and reduced battery life, then what you want is a linear regulator, which will be simpler, cheaper, better regulated, and more reliable than any inductorless 555 PWM scheme, while not being any less efficient.

There are many ways to make linear regulators, enough to merit another question, but it would be hard to get simpler than this:

schematic

simulate this circuit – Schematic created using CircuitLab

Regulation is poor and could be improved with an error amplifier, but it's hard to get simpler. It will be just as efficient as your 555 circuit, and at \$2.5mA\$, how efficient do you need to be?

Community
  • 1
Phil Frost
  • 56,804
  • 17
  • 141
  • 262
  • Thanks. I actually looked at buck converters, and tried to design a simple circuit for one, but I was unable to make one that I could effectively control with a potentiometer. If I understand correctly, it works by "charging" an inductor, switching off the power supply, then allowing the inductor to try and maintain the current. According to this [link](http://www.ti.com/lit/an/slva477a/slva477a.pdf), it appears the buck converter is acting like a low pass filter on a square wave. Anyway, the actual current, after a little revising, is more likely 2.5mA. – Hari May 23 '13 at 00:04
  • Also, doesn't that mean that V_DS will be V_in - V_out? How would I build this with a MOSFET and a 555? – Hari May 23 '13 at 00:12
  • @user24050 yes, \$V_{DS}\$ will be the supply voltage less the output voltage. All the voltage has to go somewhere. I'm still not sure what you hope to gain with a 555 -- the switching gets you nothing if you don't include some inductance somewhere. – Phil Frost May 23 '13 at 02:30
  • @user24050 if you want to learn more about how buck converters work, ask another question. – Phil Frost May 23 '13 at 02:40
  • My point about V_DS being in - out was because then the MOSFET would have a negative V_GS, since the source voltage is greater than the 10 to 15V a 555 runs on. The point of the 555 is simply switching. It was one of my ideas for solving this problem, but I'm fine with not using it. Anyway, I still don't see why inductance (specifically) is necessary. Inductors still "charge" with some resistance, and should still lose half the stored energy, just like a capacitor. Unless I'm missing a huge chunk of theory, I still don't understand the **need** for an inductor. – Hari May 23 '13 at 03:02
  • @HariGanti if you put the load on the source, then yes, you will get a negative \$V_{GS}\$. Don't do that. Put the load on the drain, so the source is on the same ground as the 555. The explanation of why an inductor is necessary is in that other question I linked. With an inductor, you do still incur some resistive losses, but the current isn't infinite, so they are small. Consider that \$P=I^2R\$ (\$R\$ being the wire resistance), and consider what effect an infinite current has. If it still doesn't make sense, consider asking a new question. – Phil Frost May 23 '13 at 12:05
  • I wasn't aware that you could put the load on the drain side of a MOSFET for a buck converter. That would certainly solve that problem. I understand where you are going with the inductor having a finite current, whereas the transient through a capacitor can be infinite. By analogy, does this mean that the transient voltage across an inductor (when suddenly disconnected) is infinite? That would make sense given the law of induction. – Hari May 24 '13 at 00:34