Doubling up a regulator for more current ??

[madhippiescientist]

If I wire in two voltage regulators (in parallel), will I get more current available?

Might there be such a formula to tell me how much it would increase?

I need just a little more than an amp, at 10v, and I know with a LM7810 I can get
a little more than an amp (intermittently) with a heat sink, but I just want to take it
easy on the regulator since it runs 24/7. The circuit works w/o an additional regulator
but it gets pretty hot. Without going for a variable TO-3, or even a 10v TO-3 (if there
is such a thing because I can't find one, plus they're expensive), I'm thinking maybe
a few TO-220 in parallel would do the trick. I already utilize a large buffer capacitor, but
does anybody have any ideas if my inquiry might lighten the load of the single TO-220?

 

[dougy83]

I take it you're using a heatsink?

To boost the power, have a look at the 7810 datasheet; there should be a section on just that. They recommend using a PNP pass transistor to provide the extra current. I don't think using 2 in parallel is a very good way to go as they won't necessarily share current that well.

 

[madhippiescientist]

I am using a heatsink, but the single 7810 still gets a little warm (can only touch it for less than
a second!). Normally w/ the 78XX regulator series', with a heatsink, I believe they give about 110%
(on a draw of about 90% of the time). My problem is, when the servo does become active, it
draws a significantly higher amount of current from the rest of the board, though I yet have to
blow a 7410. I just want to make it more reliable. With the servo active, I draw circa 1.3A with
it initially peaking up to 1.8A. I didn't think the parallel would make much difference either (the
shortest route law). So what is this "PNP pass transistor for extra current" thing ??

 

[dougy83]

Sorry, it's not in every datasheet. Have a look here: **broken link removed**

 

[madhippiescientist]

That's FANTASTIC "dougy" !!

Just (one?) "Q" though, how does the addition of the transistor allow the current to
pass, without allowing the higher voltage by it? Do you know ?? Is it just because
of Rsource and transistor "cutout"??? Does "R" do the regulating via the transistor
(by turning it on/off) ????

Also, the link you provided indicates a user's inquiry re calculating the "Rsource" and
the "R", and "admindata" answers the inquiry w/ example (no formula), while advising
to look at the datasheet (pg.14) for the formulas (which aren't there!). He seems to
have utilized the formula (and it's example) for the diagram (current regulator), that
is above the current boost regulator circuit (the one I want to use). Where could
I get the formulas for the Rsource & R, so I know what resistors to use? If anybody
knows the formulas, pass them on and I'll calculate it. If someone wants to do the
math for me, I'm supplying circa 13.8v (battery and/or wall adapter) and want 10V/2A.
I'll apparently also need aside of the resistance, the tolerance and the wattage as well.

 

[dougy83]

The regulator passes current from its input pin to its output pin so long as the output pin has a voltage <=10V. That is how it functions as a regulator.

With the resistor and PNP pass transistor in place, under light loads the transistor is off and the current is provided (through the resistor - the PNP is off) to the output via the regulator. When sufficient current is sunk at the output, the current through the resistor creates a voltage drop high enough to turn on the PNP transistor - which provides extra current to the output. If the output voltage gets a bit high, the regulator will reduce its output, which will in turn reduce the voltage across the resistor and the PNP will slow down (i.e. reduce its current) or switch off. So the output is still regulated.

The resistor should drop ~0.7V when you want the PNP to start pitching in. I guess at ~0.5A might be a good point? So 0.7V/0.5A ~= 1.5 ohms. Sounds a bit small, I'll see if I can find an example somewhere.

 

[Mr RB]

I might get jumped on for this, but you could just use two 7810 regulators in parallel, and use an output resistor on each one.
An output resistor of (say) 0.15 ohms would drop about 0.1v at 600mA, so they would share the load pretty well.

 

[madhippiescientist]

Anyone with any other idears, or the formulas for R & Rsource ??

 

[dougy83]

The formula is as I said above, sorry, I should have confirmed it. R = VBE/It, where VBE is the base-emitter (or emitter-base in this case) turn on voltage of the transistor and It is the output current at which you want the transistor to start conducting.

Ref: Various Schematics and Diagrams, you'll see a 5 ohm resistor used here which equates to a turn on at around 0.12A output.

 

[madhippiescientist]

Dougy83...how about resistor tolerance and wattage? The formula for Rsource ??

 

[dougy83]

Dougy83...how about resistor tolerance and wattage? The formula for Rsource ??

Tolerance for the resistor is unimportant. Use anything from about 5 - 10 ohms I guess. Resistor dissipation is Vbe*Vbe/R, where Vbe is around 0.7V - 1.5V; check the datasheet for your transistor's Vbe.

Rsource? What do you mean?

 

[madhippiescientist]

From the schematic for the "78xx series Current Boost Regulator Circuit" within the 78xx Motorola spec sheets (see attachment, page 14 figure 9). I didn't even see your circuit link. Rsource is a series resisitor inline from VCC to the balance of a circuit similar to the one you presented. If I'm up to 14.5v in, and utilizing the circuit with a 7810, in order for the pass transistor to still do it's duty but take it easy on the entire circuit, what Rsource would be best? I know it's only going to be a few ohms, but how about it's tolerance/wattage? I know ohms law, but don't know how to figure this out (though I am aware the lower the tolerance% the better/more stable the reduction will be). How big should it be, both physically (wattage) and electronically (resistance) ? I want about 2 amps of power...

(open the "MC7800 Series Datasheet" link w/in this link, NOT the "MC7800 Series" link!)...

**broken link removed**

 

[mneary]

Rsource is the internal resistance of the input power source, which hopefully with only 14.5Vin, is near zero. Some designs with very high input voltages ((Vin -Vout) > at least 6V) might prefer to dissipate power in a resistor instead of the transistor. Since most resistors aren't easily attached to a heat sink, I usually use the transistor. In your case, with Vout =10V, you really don't want to add to Rsource with an additional resistor.

 

[crutschow]

I might get jumped on for this, but you could just use two 7810 regulators in parallel, and use an output resistor on each one.
An output resistor of (say) 0.15 ohms would drop about 0.1v at 600mA, so they would share the load pretty well.

Consider yourself slightly jumped.

The output voltage tolerance of a 7810 is 10±0.4V so, worst-case, you would need a resistor to drop at least 0.8V to insure that they reasonably share the current. That would be 0.8Ω at 1A.

If you measured the output voltage of the two units, then you could tailor the resistance value to the difference between the two and only use one resistor at the unit with the highest voltage.

 

[Mr RB]

Yep there is that risk and it was a long shot at a easy solution. 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.

Either way you could just wire up 2 regulators and 2 resistors to the load and measure the volts drop on the resistors and go from there, ie maybe tweak a resistor value or use larger resistance resistors, it's not likely the 10v needs to be perfect regulation anyway.

I've never tried paralleling 2 regulators like that but I have used a parallel resistor from Vin-Vout on a few occasions when the regulator was driving a known load like a light.

 

[madhippiescientist]

Pardon me if I sound confused, but all replies to my answer of the "Rsource" inquiry from dougy83 seems to be pointed towards the regulators output. This is a "drop resistor" from the power source, before regulation and the pass transistor scheme. If I'm drawing 2A from this circuit, I believe I WILL need to lower the voltage being supplied to the circuit, in order to prevent overheating of the passQ and the 7810, because that would be some kind of choppin' (14.5v down to 10.0v) for a 2A draw. I'm looking for the RIGHT resistor (instead of "tailoring the value" like crutschow states), because if I utilize say a 25W resistor, that would cost a few bucks for every "shot in the dark". The regulated voltage truly needs to be quite accurate, since it drives an astable timer that operates a stepper motor which turns a drum that needs to rotate as close as possible to 1RPH (yes, per hour) through a circa 10389.6:1 gearbox. The timer has 1% resistors and 2% capacitors to also expedite accuracy. Over the course of 168 hours (1 week), each pulse I'm off on the timer after a divide-by-100 still yields a variation of about 8.7 degrees rotation. If I can get my hands on a meter that will display the entire frequency (I have a 3-1/2 digit meter and I'm reading a signal over 7,00_ hz, the "_" is unknown), I would be pretty accurate (0.87 degrees rotation), but I need the supply voltage to be accurate as well otherwise the timer will vary it's oscillation according to how much quicker/slower the capacitor "fires" (which is why I sway from Mr RB's dual reg / dual resistor scheme). The math on my gearbox, (in case someone decides to see how accurate my pulse per second needs to be) yields a needed PPS of 69.264 for a motor requiring 24 pulses per revolution, but with mechanical drag and my test timing of the drum's rotation, I've come to right around 7,73_hz (in order to turn the drum 360 degrees per hour). I know, who cares, but my point is, that I really do need the voltage to be closely regulated, even if I ever do get to use a frequency counter (to adjust through physical testing of the drum the final digit in the timer's f). Also in regards to the regulators output and measuring a voltage with a series resisitor, the voltage is going to be significantly(?) lower with the circuit on (even with the passQ?), and vary on the meter w/o draw. I don't have to have exactly 10v, I just need the c 10v being supplied, to stay at a specific level, so once I tune the timer, It will keep it's accuracy...Hmmm. Maybe I should just go with a TO3, do they make it in a 10v regulated?? (NO! NOT the adjustable thing!!). If timing truly wasn't CRUCIAL, I would just run the circuit from the battery, but battery voltages fluctuate as they dissipate / discharge / etc., and timing of the astable and it's supplied voltage for oscillating the "firing" capacitor, is of the MOST importance !

 

[dougy83]

bloody hell that was a long rant.

1% resistors and 2% capacitors will potentially give you a 3% error which is not accurate in the least. You can use a resonator or crystal if you actually care about any form of accuracy.

14.5V to 10V is not an unreasonable ask. The 7800 series requires a drop of a few volts anyway or it won't regulate. As someone said earlier, it's easier to mount a transistor on a heatsink than your average resistor. the total power dissipated by the regulator & transistor and resistor is 9W max (4.5V * 2A). To work out the maximum thermal resistance allowed and therefore the size of the heatsink you need, divide the difference between the max junction temperature of the transistor and the max ambient temperature by the power; e.g. 120^C max junction temperature, 30^C max ambient, 9W power gives 10^/W - this number, 10, is made up of the junction-case + case-ambient + heatsink rating + maybe 0.5^/W (due to bad mounting). This is the minimum rating for the heatsink: use a better heatsink where possible.

 

[mneary]

Now that your requirements are more clear, I will re-state my answer:

Architecture first... why are you regulating the entire 2A when the oscillator is (hopefully) only a small part of that? The oscillator could have a good regulator supplying its <5mA (even this will cause temperature rise in the oscillator = drift). Send the other 1.995A going unregulated (or less critically regulated) to other circuits, including the stepper. Steppers really don't care what their voltage is, within reason. The stepper voltage could be independently managed by a series resistor, if you like resistors so much.

In general, a resistor is a bad way to off-load power (heat) from a regulator chip. Specifically in your case using "Rsource" to reduce transistor dissipation is a really horrible idea, because your 4.5V overhead (14.5-10) is barely enough to properly run the regulator. At best a resistor should be allowed to take only a volt, else the 7810 will begin to vary its output. That's only 2W out of 9. And then you would need to find a way to keep the resistor(s) cool. No matter what you do, you'll have to dissipate about 9W.

It is much cheaper to heat sink a regulator and pass transistor than to assemble enough resistors that handle the same power. Even if there was such a value that you could calculate. If your max load is 2A, "Rsource" might be around 0.5 ohm.

Now, moving on to using an analog timer: If I understand the task, you have an appx. 7000 Hz oscillator divided by 100. You're seeking a precision of 1 pulse per week on that 69.264 Hz, or 1 in 43 million (0.02ppm). You can't even expect that from a crystal. It's mad to try it with an astable. Maybe if the crystal oscillator is in a temperature controlled oven.

Maybe I should just go with a TO3, do they make it in a 10v regulated?? (NO! NOT the adjustable thing!!).

"Fixed" regulators are just adjustable regulators with the supplier providing the resistors inside.

 

[madhippiescientist]

I see you love the "Rant". Missing a pulse per SECOND (not per week) causes the indifference to be significant. The pulses are so high and divided in order to create accuracy, along with the incredibly geared down box FROM the stepper motor. Again, my 3.5 meter is only accurate to 10 pps until I get a hold of a true frequency counter, and believe it or not, once I set a frequency on my meter (with the astable), it stays there (within TEN pulses, ex.- I set it at 773 in the /10 scale on my meter and it DOES stay there, but again, that's 7730 to 7739 which could result in several minutes of inaccuracy after a week). I didn't try a crystal because in time the gearbox/etc will develop more/less drag, and I didn't know how much drag I would have to begin with (my math was originally circa 6926.xxx pps before gearbox reduction and division by 100). In regards to regulating only the astable (which it DOES have a 7805), the device (seismograph, which is a seismometer with a helicorder) has a battery backup which would be simpler to integrate into the entire implementation with everything else during a power failure (the divide by chips, the 4-bit bi-di shift register, the drum limitation indicator flasher chips/leds/buzzer, the cooling fan, the battery voltmeter, the PWM servo controller/chips, the stylus servo, the base horizontal level indicators, etc.). I also desire implementing the regulator due to some of these items being subject to damage in case of power surge or backup battery high voltage (up to 14.5v). Also, instead of regulating several devices (at 12v), I designed it all with the implementation of a single regulator to begin with, designed to be less than 12v in order to accommodate battery depletion. (some of these circuits/items DO get further regulation anyways, and everything works off 10v, and my wallwart is 10v). Thanks for the resistor and wattage values! Again, the frequency is pulses per second, not per week, and 1 week is circa 604k seconds, so if I miss ONE pulse EVERY second (even though it's divided by 100 and geared down), my math reflects 1.45 minutes per week <8.73 degrees on the drum> (with a frequency counter), or 14.5 minutes per week <87.3 degrees on the drum> with my cheap 3.5 digit multimeter. The 5v astable may not vary (much) with voltage, but as previously stated, the servo momentarily pulls the draw up to almost 1.7A when it is actively drawing on the drum (normally the entire device's circuitry is a couple hundred ma), and the 9.6v servo needs regulation. Another big point also (as already noted), is that I will have a much longer battery backup by having designed it all to work at 10v.

I would gladly do the TO3 variable reg, but it's so darned expensive ($23) compared to a few 7810s or a 7810 with the passQ. My "simple" question was, is increasing the 7810's capacity (available current) possible, while whatever voltage it sits at (it could be a few tenth's above or below), it would at least remain at that specific value so once I tune the astable, it definately won't be affected by higher draws (the servo) and will therefore keep an accurate timeline on the drum's weekly graph paper. So, anybody w/o the TO3? How 'bout anybody with the TO3 at 10v (the circuit) ??

My thread is getting old, and too complicated !!! Maybe I'll excite you all with a few pictures of this electro-mechanical behemuth...

 

[Hero999]

I've not read through every word of the long posts so forgive me if I've missed anything.

You shouldn't need to worry about Rsource because the regulator and transistor should be able to dissipate the power with no problem.

Will the motor always be connected to the LM7810?

Is there a minimum load current?

If the current never drops below a certain level you could connect a resistor between the input and output of the LM7810.

Suppose the regulator can continuously supply 1A but you need 1.8A and the minimum load is 1A. If you connected a resistor that always passes 1A from the input to output , the regulation won't suffer as long as the minimm load current is drawn.