Doubling up a regulator for more current ??

[unclejed613]

a zener only works as a regulator when you have a relatively stable current draw, and with limited currents. not an option here. it's a very poor choice for a regulator, especially with the type of circuit you're driving. think of a zener in terms of a passive regulator. it can only clamp the voltage at a certain level. if the load draws higher currents and the "regulated" voltage drops below the zener voltage, the zener can't supply more current to make up the difference like a real regulator can. zeners also waste a lot of power as heat and tend to burn themselves up often.

you've "switched gears" here several times between running your device from an AC supply and a battery. which is it? or is the battery a backup supply if the power goes out? is there a problem with the circuit running on 12V or is it running on 10V because it's maxing out a 12V wall wart? if it's running on 10V because it's drawing too much current from the wall wart, then build a 12V regulated supply for it that can supply 3A, and have the battery backup available directly(after the regulator) if the power goes out. i don't see any logical reason the circuit can't run off of 12V.

if you have the capacitor AT the point of load, then it can "dump" current more effectively to help maintain the voltage at a constant level when the motor is driven.

just how often is the motor driver circuit stepping the motor?

i still recommend that if you really want the circuit to run at 10V, use a "low dropout" regulator. this will give you 10V regulated output until the battery gets down to 10.6V.

also, the coil voltage of the stepper motor is most likely 12V. stepper motors usually have coil voltages of 12, 24, 36 or 48V. those are pretty much the standard operating voltages of most industrial electronics. there may be a rare occasion where you find a 6V stepper (which usually gets run from a 5V supply), but the stepper motor business is well established and standardized, and exceptions are as rare as Dunsel diodes.

 

[madhippiescientist]

you've "switched gears" here several times between running your device from an AC supply and a battery. which is it? or is the battery a backup supply if the power goes out? is there a problem with the circuit running on 12V or is it running on 10V because it's maxing out a 12V wall wart? if it's running on 10V because it's drawing too much current from the wall wart, then build a 12V regulated supply for it that can supply 3A, and have the battery backup available directly(after the regulator) if the power goes out. i don't see any logical reason the circuit can't run off of 12V.

The stepper motor runs ALL the time turning the helicorder drum, and I need a stable voltage so I don't lose/gain any phases if the voltage changes. I have been using a 10V 1A wall wart and the device runs, stepper and all, with VERY reliable timing, but the wall wart gets really hot. I want to build my own AC to DC supply which feeds into a relay that will also connect to the backup battery, for when the AC goes out. Between the power supply and the battery charger, I want ONLY one cord from the AC outlet (no wall wart). Then I'll regulate the voltage (though I previously inadvertently wrote of regulating each supply individually) from the relay to the circuitry.

Between regulator cut out and the amount of juice the battery has remaining, is why I originally designed the circuit for 10V (it will run MUCH longer under power outage conditions). I could do it at 12V, even with a low drop out regulator, but it will cut down on the amount of time it will work before the lower voltage causes the helicorder drum to lose time due to the stepper motor stopping and/or phases being lost (if it misses one single phase every 30 seconds, the drum will lose about a minute for every hour on the graph chart). The motor is only drawing 600ma, but I have other circuits, as well as another circuit that senses "motion" activity and drives a 9.6V servo.

As I stated before, I know I could separate the two, but I initially inquired about "doubling up a regulator for more current" (the thread's title), w/o having to buy an expensive regulator <read some of the previous threads!> (a 7810 in parallel?, which I know now is not an option!!). So I'm now convinced I need to utilize a variable regulator (LM338 / 5A, LT1528 / 3A or maybe even the LM318 / 1.5A), but I MOST IMPORTANTLY need for the output to STAY at a stable level (circa 10V), hence the zener diode inquiry.

Please, this isn't about my design. I'm trying to attain regulation at 10V for more than an amp of current, but keep the voltage at as constant of a level as possible (while the juice is there and the circuit IS always drawing, with a little more draw than normal when the sensor makes the servo go crazy). Once I tune the stepper's astable timer, I want it to "keep" the time I set (if it were to vary through it's 7805, which I doubt), but I want the motor not to miss any phases if the voltage supply varies under any condition EXCEPT the inevitable battery "below cutout" (taking into account operation for as long as possible under backup conditions).

As the battery dwindles, if it does drop too low, I would rather have it completely stop than spend time losing phases before it stops (accuracy of time on graph sheet). If I did the whole thing at 12V, then even with a 0.4V cutout as "comes" with the LT1528, and the battery is at 13.8V, I would only have about 1.4V of "availability" as compared to 3.4V (240% longer). I see your suggestion for the 10V, but I'm asking (since I have never built a power supply with a variable regulator) how I can keep it's output (voltage) as stable as possible (with a zener between the regulator and the circuits?), and if so, at what wattage and where I can get a zener to do it ???).

I don't want the zener to actually regulate the entire power supply, and I am well aware it won't supply more voltage in "lower than" conditions. I DO want to try and clamp the output at 10V <circa a constant value>, even if I set up the variable regulator for 10.25V. Then if the supply dwindles a pinch it won't matter, because the zener will still only be letting (circa) 10V through.

 

[unclejed613]

when i said a zener wouldn't regulate if the voltage dropped, i meant on the load side. it's obvious that it would drop if the supply side went too low. but if a current peak were to occur on the load side that brings the regulated voltage down, a zener won't compensate, but an active regulator will, and keep the voltage constant. all active regulators (including the 3 terminal regulators) have a zener internally, but not connected to the load. they are used as voltage references only, so they run with a constant current. the voltage reference is then compared with the regulated voltage output of the regulator, and the difference is amplified and sent to the pass transistor. to increase or decrease it's conduction to maintain a constant voltage output. this all happens in real time, so the small variations that cause all of this to happen aren't noticeable.

what i was getting at with running the circuit from 12V, was that you already know it will work down to 10V, so take advantage of the extra "cushion" between 12V and 10V (at least with the battery). running a power supply at a raw 14V into the battery, while the regulator running off of 14V gives you a regulated 12V. the battery and regulator are "diode OR'ed" to the circuit. the diodes do two things. first they automatically switch from regulated DC to battery DC in a power failure without the short delay and cutout you would get from a relay. second, the diodes isolate the battery from the regulator, and keep the battery from discharging through the unpowered regulator. you then have about 2 volts of battery discharge "cushion" to work with before the battery drops down too far to operate the circuit. the 14V raw dc from the supply trickle charges the battery through a large value resistor. if you use schottky diodes, you only get a forward drop across them of 0.1-0.3V.

 

[MrDEB]

Here is a little different spin on increasing current
High Current Voltage Regulation - Electric Circuit

 

[mneary]

Here is a little different spin on increasing current
High Current Voltage Regulation - Electric Circuit

They look a lot like the suggestions in posts #2, #4, and #7. The regulation needs to be good with an input voltage of 11.5V, so those circuits won't work here.

The first circuit needs the overhead of the 7810, plus the Vbe of the transistor. It is a good circuit, but it won't work with only 11.5V.

The second circuit mistakenly assumes that 1N4007 has the same voltage drop at 1A as it does at 12mA. It also mistakenly assumes that the two diodes will somehow assist in the current sharing (If anything they make it worse due to their temperature coefficient.) This circuit also has the overhead of the 7810, plus the 1N4007 diodes.

The third circuit has poor regulation, and it does not resolve the overhead issue of the 7810.

The first and second circuits in the "comment area" use the MJE3055 as an emitter follower. This uses even more overhead, and the regulation is the poorest of them all. (He warns that the pins could be wrong, and they are wrong, yet he didn't bother to fix them.)

 

[unclejed613]

this should work just fine, it's a 2A regulator, and a LDO regulator with a dropout voltage of 0.45V. the closest voltage to what you need is 9V, but if you put a forward biased diode between the "ground" pin of the regulator and ground, it will bump up the regulated voltage by 0.7V, so it will be a 9.7V regulator. so your circuit will operate from a fully charged battery until the battery voltage drops to 10.15V, which gives you a lot more battery headroom.

**broken link removed**


edit: just thought of a painfully obvious solution to the problem. use a 7810 and pass transistor (there are plenty of TO-220 transistors you can use for this), get a second battery, wire the batteries in series for 24V, get a 24V open frame adjustable power supply and adjust it for 28V output (or build one from two 12V ones wired in series). this way you can run the circuit until the batteries are each discharged to 6V, which gives you a 400% increase in battery operation time.

 

[mneary]

get a second battery, wire the batteries in series for 24V

Problems:

1) Once the battery is below 90% of its nominal value (10V for a "12V" lead-acid), almost no more energy is available.
2) Discharging almost any battery to 50% of its voltage will destroy it.
3) A 24-28V input voltage would require a regulator, transistor, and heat sink capable of 36 watts at 2A, instead of 9W.

 

[madhippiescientist]

MrDEB, I tend to agree w/ mneary re your suggestion looking similar to the pass transistor scheme and it also having a very low "overhead" for backup duration. In re to a pass transistor or the LDO, I would still be relying on discreets (which is what I was originally trying to avoid, for accuracy). If I'm going to use discreet components, I may as well go with a variable regulator at a higher current (3A or 5A) which only have about 600mv D/O (for the LM338) at the current I need, and I already had a suggestion for an LDO (LT1528) from mneary at 3A. If I can find one of the LDOs here in town (Phoenix), I will do it (I have yet to search for it, but I don't want to get any through mail order because I'm not going to pay $10 to have a $2 component shipped, and it's really hard to find the LDO's that everybody is listing within the couple of catalogs I have). I also don't have any credit or debit cards. This is why I'm likely going to do the LM338. Getting a second battery would also be a big waste of money. I also saw online a "78S10", which is a fixed 2A regulator, which I will look for. The big reason I'm aiming at the LM338, is because of the stable 10V circuit that National supplies in it's spec sheet (see link, page 9 under "High Stability 10V regulator"). It utilizes a voltage reference chip as part of the Vref (LM329B), along with discreet components that are of specific values and good tolerances. Besides, the manufacturer is worth trusting the most, normally...

**broken link removed**


In re the LDO suggested by mneary (LT1528), I stated an inquiry with my math and haven't heard anybody tell me if I have the values correct. See post#40 in this thread for my math & link.


Anybody interested in the 78S10, see this one...though it's D/O isn't pretty ! I probably won't be able to find this in town either.

**broken link removed**

 

[unclejed613]

mneary, you're probably right about draining the batteries too far down. it's probably better to improve the discharge time by paralleling two batteries.



ok, i looked at the LT data sheet. Isense is the current through the sense pin, and you need to add the additional voltage drop due to the sense current to the voltage drop across R2. 130uA isn't a whole lot, but if you're looking for high accuracy in your voltage setting, you will need to take it into account. for R1, R2, you can use a 5k linear potentiometer, and just connect a meter across the output and set it for 10V, and you won't have to worry about the math.

 

[mneary]

madhippiescientist, where are you located? Maybe one of us can help you find things locally (to you) at lower cost.

For example, here's a 5A LDO for $3 plus $2 shipping at an ebay advertiser in Torrance, CA, USA. Micrel MIC29502BT High Current 5A LDO Regulator TO220-5 (#290366682844). A glance at the data sheet suggests to me that it deserves a look.

Searching ebay for "LDO regulator" brings up a lot of possibilities.

Here's another, (#380199460297) for $4 plus $1.50 shipping you get twenty of them. (This one is a 1.3 volt dropout @ 2A; maybe marginal for you). I've bought a lot of things from dpi4parts, very satisfied. And being in the US you don't have to wait for the end of Chinese New Year celebrations.

 

[madhippiescientist]

I wrote of being in Phoenix (AZ), and the market for distributors has taken a big beating here. I've now (as of yesterday) actually got Mouser sending me an LT1528 (LDO 3A) for free!, which is the LDO that you originally suggested (also why I'm so interested in my math). I know about checking the Vo with a meter, but if I can get the values pretty close, and use some smaller valued pots, I'll know more specifically what value of additional resisitors to add/parallel in. While searching for an LDO on the web, I was looking for your suggestion, and sooo many distributors didn't have them OR I needed a card to buy them with (I could just get the LM338 here in town). I'm going to try the LT1528 as soon as I get it, and see just how stable it's Vo remains after tuning it (before the LM338 thing with the 1.2v Vref IC as listed in the National spec sheet).


Again, the formula is Vout=(3.3v*(1+(R2/R1))+(Isense*R2), so I figured the needed R2 as 680 ohms utilizing the maximum suggested value of R1 at 330 ohms. Is this at least going to be close ?

 

[madhippiescientist]

I have confirmation of the LT1528 having been shipped, but I'm still waiting for her...

Meanwhile, If I'm making a 10VDC 3A power supply, I would think after the bridge rectifier's dropout, that a 14V transformer would be sufficient. Should I go for the 16V?

Either way, tell me if I'm wrong, but I can't have the transformer exceed 3A output (14V...42watt, 16V...48watt), or I risk damaging the rectifier if I HAPPEN to try to draw more than 3 amps, correct???

 

[Mike odom]

The output voltages of 3 pin regs are pretty good these days with the new manufacturing tech or laser trimming or whatever they do. It's trivial to measure the output voltage and with 7805 (even cheap ones) they are rarely out of the range 4.96 - 5.08 or about 0.12v variation, and often better when from the same batch often within 40mV or less.

QUOTE]

where do you buy your regulators? Fantasyland??? I just built 4 test fixtures and I had (4) brand new 78L05s with the same date code, bought at the same time, and they were all over the place. The only reason I checked was part of the test fixture had a thermocouple input calibration circuit on it, so I used a precision source for that circuit. On a production run of 1000 boards, you'll find the full range of the +/- .4V. We've built systems where we have 70000 units in the field. I once did a statistical plot on a run of a 1000 boards (too much time on my hands). You couldn't have drawn a better bell curve. They set the parametric specs for a very specific reason. If they could repeatably manufacture CHEAP parts with that tight a spec they would have them spec'd that way. You could even 'grade' them, say, buy 100 and measure the voltage under a standard load and pick two that had the same voltage output, but come back in an hour, and they could be off... they might not have the same temp. drift... you should NEVER parallel two voltage regulator outputs, 'cept through a diode or a FET, for dual redundant, not double the current.

Consider yourself thoroughly 'jumped'...

And.... a fully turned on transistor has a VBE of .7v, but the knee starts at .6v, that's why in the previous message 5 ohm yielded .12A. I always use .6v when I calculate current limiting resistors. Here is a circuit I use, the one on the left. I adapted it from the data sheet of, YES, the 7805. Look at the output stage of the 7805 and compare that to the circuit on the right.

 

[madhippiescientist]

Hey Mr. Odom, I'm glad to see you have plenty of energy available to jump MrRB, do you have a little energy left to answer my most recent inquiries?

 

[unclejed613]

I have confirmation of the LT1528 having been shipped, but I'm still waiting for her...

Meanwhile, If I'm making a 10VDC 3A power supply, I would think after the bridge rectifier's dropout, that a 14V transformer would be sufficient. Should I go for the 16V?

Either way, tell me if I'm wrong, but I can't have the transformer exceed 3A output (14V...42watt, 16V...48watt), or I risk damaging the rectifier if I HAPPEN to try to draw more than 3 amps, correct???

that's what fuses are for. put a 2.5A fuse in line with the load.

 

[Mike odom]

Hey Mr. Odom, I'm glad to see you have plenty of energy available to jump MrRB, do you have a little energy left to answer my most recent inquiries?

my apologies to Mr RB... just want to err on the side of safety, especially when recommending circuits to newbies that might not fully understand the implications of what they are doing.

if you have specific questions left unanswered, please feel free to ask them again or pm me with them. I am new to this forum and want to help but don't think I have the time to sift through the 50 or so messages to pick out what still needs to be answered. If it's about the proper resistor for exactly 10v out of the regulator, you've probably noticed not too many engineers are worried about getting it exactly right the first time. Pick a value that's close, see what it does. You can spend lots of time defining exactly what parts are expected to be needed, then have to adjust them anyway. If you're using a 14v secondary on a transformer and a full wave bridge, you're going to see something like 19V on the DC coming from the cap. That should give you plenty, especially if you are using an LDO.

Let me know what else you need...

 

[Mike odom]

one other recommendation I haven't seen yet, look into poly fuses. These are the best thing since melted butter... they are resettable fuses. I use them every chance I get...

 

[madhippiescientist]

I planned on a fuse, but it's also no use to buy a transformer with a higher current than necessary (cost, power, size), so, should I just limit it to a 3A secondary, or is the bridge going to do some majical thing and double it's total overall current ?? (see next question). Aside, there's nothing wrong with attaining such a knowledge (LMK for good measures).

If you're using a 14v secondary on a transformer and a full wave bridge, you're going to see something like 19V on the DC coming from the cap

The secondary transformer thing...Isn't the total voltage of it's output going to be 14v (minus diode cut outs) after the bridge? I always thought the secondary (before rectification) only passes a maximum full wave output, from which IT is measured for marketing. I thought a transformer identified as having a 14v secondary, would be about 14v from negative apex to positive apex of the wave. Is it just 180 degrees of the wave (14v+ and 14v- for a t of 28v) ??

 

[Mike odom]

transformers are funny things... they are rated by load and voltage, so a 14V 3A secondary will be around 22V no load, and the voltage will decrease until the rated secondary load is reached... and no, the diodes will not magically double your current, however, the diode bridge (full wave bridge) and cap will give you a bigger output than a full wave diode setup. This is easy enough to experiment with... just take a transformer, hook it up to AC, and measure the no load voltage. Then run it through a full wave bridge and cap. Your voltage will be much higher than rated. Now, put various loads on it and see what that voltage drops down to. I'll set one up tonight and make some measurements and post them for you... today I have a meeting (damn work keeps getting in the way)...

 

[madhippiescientist]

I look forward to your response. I just want to buy the transformer to best suit my needs w/o going "overboard".