Control current from 90V PS with RC PWM signal?

[landru]

Hi All,

Let me admit up front that I'm semi-literate in electronics at best. However, given a good push in the right direction, I can find my way around a data sheet and a breadboard and come up with usable results. So here I am looking for some electronics expertise.

The project is to build a circuit to provide current control of a 90 VDC floating power supply using a standard RC 1-2mS PWM signal as the control signal. At 1mS the ps voltage output should be zero; at 2mS the ps output should be its maximum. Response between the endpoints should be linear. The object of the whole exercise is to provide PC control of a 90V hysteresis brake from existing hardware/software that already outputs a standard RC PWM signal.

The solution I've been looking at is an opto-isolated circuit that converts the PWM signal to an analog voltage and then uses the amplified result to regulate the current from the power supply:

pwm signal -> opto-coupler -> RC filter -> op-amp -> power transistor -> power supply

The RC filter isn't fast but it's fine for my purposes. Moreover, I've breadboarded a version of circuit that provides just the control I need -- but with a 12V voltage supply standing in for the 90V supply.

Here's the hitch: modifying the circuit for the 90V supply isn't as straightforward as I'd anticipated. I have identified a high voltage opto-coupler (PS2513), op-amp (OPA454), and transistor (TIP102) which should do the trick. However none of the components are available locally (a small problem). Plus the op-amp is only available in a surface mount 'Power Pad' package as far as I can see (a bigger problem). SMD I can handle, but is there any way to solder Power Pad other than reflow?

Before I bumble onward, it occurred to me to ask whether I'm on the wrong path altogether. Am I overlooking a simpler and better solution?

Thanks,
Andrew

[Bruce Abbott]

The project is to build a circuit to provide current control of a 90 VDC floating power supply using a standard RC 1-2mS PWM signal as the control signal. At 1mS the ps voltage output should be zero; at 2mS the ps output should be its maximum.As I understand it, a hysteresis brake needs a current regulated supply, to minimize torque drift as its coil heats up.

Requiring a high voltage opto-isolator and opamp suggests that you are making a constant voltage power supply. If you re-configure the circuit to provide constant current then the output transistor will be the only high voltage component required.

[landru]

Hi Bruce,

Thank you for the very helpful reply and circuit diagram. Much appreciated.

In your circuit I see that the opto-coupler, op-amps, 12V supply, and 90V supply share a common ground. Will that approach be problematic considering that the 90V supply in question is a floating supply?

Isolating the floating supply was my reason for including the high voltage op-amp. In effect, my own circuit is very similar to yours -- with the exception that all the components upstream from the opto-coupler are high voltage and share the same floating ground as the PS.

I popped in to the small, local electronics supplier yesterday and the owner gave me an interesting suggestion: treat the project as a high voltage version of a brushed ESC. That approach seems like a possibility -- albeit one that introduces more complication. I wonder, too, if I won't have to solve the same problem of isolating the 90V supply from the rest of the circuit.

Regards,
Andrew

[landru]

I wonder if I can come at the problem from a different direction.

Looking at Bruce's circuit above, it seems that the question is how to provide a low voltage supply (for the opto-coupler and op-amps) which shares the same floating ground as the 90V supply. If I understand correctly, that floating ground is at significantly lower potential than mains ground.

Battery power, or a separate mains-referenced DC supply of appropriate negative voltage don't seem like practical solutions. The switching DC-DC converters I've looked at are too complicated and/or too expensive.

So how about a sequence of linear voltage regulators which reduce the 90V supply to 12V in two or three steps? Not very elegant perhaps, but would it work?

I see that the humble LM317 is happy to operate at up to several hundred volts as long as the differential between input and output voltage isn't greater than 40V. Something like 100V -> 65V -> 30V -> 12V perhaps?

Feedback appreciated.

Andrew

[Bruce Abbott]

The control circuit normally only draws about 3mA, so you could just use a simple shunt regulator (eg. 15k 2W resistor, 12V 1W Zener, 10uF smoothing cap). Alternatively you could build a conventional 12V regulated supply (mains transformer, rectifier, filter cap, 3 terminal regulator etc.) or just use a 12V regulated 'wallwart'.

Although it can easily be done, I don't recommend leaving the entire circuit 'floating', as seriously high voltages could build up and cause circuit malfunction or even insulation breakdown. If the power supplies are derived from the mains then you should tie their common negative rails to the mains Ground.

[gigelus2k3]

Plus the op-amp is only available in a surface mount 'Power Pad' package as far as I can see (a bigger problem). SMD I can handle, but is there any way to solder Power Pad other than reflow?


Andrew,

At my place of work, where we use conventional soldering (no reflow oven), all devices with exposed bottom paddle (QFN, DFN, SOs with exposed pad, all sort of power mosfets etc.) have a footprint with a large plated hole in the middle of the exposed pad.

In this way, the technician can slip in from the other side the tip of the soldering iron and, with the help of a quality flux, make a nice and shiny solder joint.

Serban

[landru]

Hi Serban,

That's a very clever solution! Thank you for sharing it. Even if I don't use the technique in the current project, I'm sure it will be useful in the future.

* * *

... Although it can easily be done, I don't recommend leaving the entire circuit 'floating', as seriously high voltages could build up and cause circuit malfunction or even insulation breakdown. If the power supplies are derived from the mains then you should tie their common negative rails to the mains Ground.

Bruce,

Thanks for the continued help.

The 90V supply I have on hand is a small solid state unit which provides DC output from full-wave rectification of the mains supply. It is without isolation as far as I can see. The documentation for the power supply specifically warns against connecting its negative output to mains ground. Immediate destruction, etc.

Given that restriction, and the perils of floating the circuit as you mentioned, perhaps I should look for a different way to supply the 90Vdc.

I didn't anticipate the floating ground and lack of isolation when I picked up the supply on eBay. The supply is an industrial unit intended for simple on/off switching of hysteresis brakes and clutches with a mechanical switch. I had hoped to add current control without much difficulty. Perhaps not. At least the investment was small.

Andrew

[Bruce Abbott]

The 90V supply I have on hand is a small solid state unit which provides DC output from full-wave rectification of the mains supply. It is without isolation as far as I can see. The documentation for the power supply specifically warns against connecting its negative output to mains ground. In that case it will be 'floating' at 0~-110V or -55Vrms (mains voltage, half-wave rectified). Provided that your opto-coupler and 12V supply transformer can handle the peak voltage it should be OK. Of course you must ensure that the entire mains side of the circuit is suitably insulated, to avoid electric shock.

Alternatively, you could run the 90V supply through a small isolating transformer.

[landru]

Given the fact that the 90V supply is not isolated, would it be safe to use a shunt regulator to provide the low voltage supply?

My thinking is that, although floating, voltage within the circuit would still be limited by mains potential because of the lack of isolation. Would that prevent the possible high voltage hazard you pointed out above?

Thanks again for the very helpful advice.

Andrew

[Bruce Abbott]

Yes. The circuit is 'floating' relative to the mains ground, but it's still connected to it via the rectifier, so there's no danger of it building up an excessively high voltage. You just have to treat it like any other device that's powered directly from the mains - ie. no exposed wiring, and all exterior metalwork must be grounded (including your brake).

[landru]

Many thanks Bruce. Now I can put together the circuit safely.

:) Andrew

[landru]

Hi Bruce,

I have your posted circuit built and running on a breadboard. For the low voltage supply I used a shunt regulator with a series transistor.

Since I'm using a brake with 1.5K resistance, I increased R10 to 20 Ohms to compensate. And that brings me to my question. At 2mS PWM input, I get a max of current of about 65mA, which is entirely appropriate. However, at 1mS input, I get a non-zero current output (~15mA), and that's a problem. How can I zero the current at 1mS?

Looking at the circuit schematic, I conclude that there is always a positive, non-zero voltage at the '+' input (pin 5) of the second op amp. In order for current through the brake to be zero, I figure that the same voltage would need to be present at the '-' input (pin 6). However, I can't see how such a voltage can be created on the emitter side of Q2 when it's in the off state. What am I missing, I wonder.

As an alternative approach, I supplied the appropriate 'zeroing' voltage to the op amp '-' input with a trimmer and eliminated its connection to Q2 through R9. With a larger 10K resistor at R8, this approach works adequately, the hitch being that brake current doesn't switch on until about 1.2mS and reaches maximum at about 1.8mS. Am I correct in thinking that those dead zones represent the 'toe' and the 'knee' of the active range of the transistor?

This has been a rambling post. As you see, I'm interested in both the practice and the theory of solving this little problem. Any thoughts or suggestions you might have would be much appreciated.

Regards,
Andrew

[landru]

...In order for current through the brake to be zero, I figure that the same voltage would need to be present at the '-' input (pin 6)...

I think I misspoke last night when I posted the above.

Let me try again. For zero current flow through Q2, the output of the second op amp to the base of the transistor would need to be less than about 0.6V (i.e. to keep Q2 turned off). The gain of the second op amp isn't clear to me, so I won't attempt to theorize about voltages. Measurements showed about 1.3V at the '+' input with a 1mS signal; and that about 0.7V applied to '-' input is sufficient to prevent Q2 from turning on (with 1mS signal). Does that mean gain is one?

No matter, I'm still not sure how that voltage can be derived from the emitter side of the transistor. Can you guess that I struggled in the electronics portion of undergrad physics? And that was back then. :D

[Dan Baldwin]

You may have 2 different problems here, well actually, you do have one, and you may have another;

1) Bruce's schematic assumes that you have a fixed frame rate of 20 ms, and a standard 1-2 ms pulse. Since your signal is coming from a computer, not a known RC transmitter, we don't know that that's the case. Do you know if the software puts out a fixed frame rate? Do you know what it is. If it's a fixed frame rate, but not at 20 ms, we can probably juggle resistor values and make it work. Hook up the system and note the voltage at pin 3 of IC1 with the PC outputting 1 ms, and then at 2 ms.

2) The second problem is a bit harder to handle. The LM358 has an input and output range that includes ground, but it can't pull the output to ground. The resistor ladder formed by R5, R6, and R7 are trying to pull the output of IC1 to about 1.2 volts. We can get it close to ground by adding a pull down resistor to pin 1, but you can never get it all the way to ground. You can, however add a small offset to the feedback signal from the 20 ohm current sense resistor to allow your system to go to zero ma. You won't have this problem with the second op amp because there is nothing trying to pull it away from ground, and as long as it's output is below about 1 volt, you shouldn't have any current flow.

Dan

[landru]

Hi Dan,

Thanks for having a look at the issue. Let me try to answer your questions.

The software in question is the PC interface for the Medusa Power Analyzer. Unfortunately, I couldn't find any info on the frame rate it produces.

The following are voltages I measured in the circuit (with R10=17 Ohm):

Pin......PWM......Voltage
-----------------------
3........1mS.......0.635V
3........2mS.......1.245V

7........1mS.......1.35V
7........2mS.......6.30V

There is some slight fluctuation in the voltages, and a tendency for them to rise slightly with the output of the low voltage regulator. The regulation of the low voltage supply could be improved easily if necessary.

I follow your comments on pulling the output of the first op amp to ground. What would you suggest for resistor values?

Regards,
Andrew

[landru]

... You can, however add a small offset to the feedback signal from the 20 ohm current sense resistor to allow your system to go to zero ma. ...

Ah yes. Now I see what's meant to happen. The feedback signal itself has nothing to do with setting the current to zero at 1mS PWM input. I was working under a mistaken assumption.

As you suggested, I added an offset voltage at pin 6 using a 10K trimmer as a voltage divider. Now the current can be adjusted to zero. And with a smaller current sense resistor (5R), the full range of current adjustment is available across the PWM signal range.

Thanks Dan. And thanks again to Bruce for drafting the original schematic.

Andrew