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Electronics Question
By ChrisW, Section Controls Posted on Tue Aug 03, 2004 at 06:51:03 PM MST
Guru needed, please...

Greetings all -- I'm working on a circuit that switches a moderate amount of power (about 10 amps) through a MOSFET, an IRF520N in this case. I've seen a number of schematics where a power resistor with a really low value (like .1 or .05 at 5 watts or higher) is used between the Source pin and the load (an SLA battery in this case), and similarly, on the emitter of NPN power transistors when used in voltage regulator circuits. (In case that's not clear, think of it this way: current source is connected to the Drain pin, Source pin connects to the power resistor, other end of the power resistor goes to the battery). Given the number of times I've seen this configuration, I was wondering 1) what's the resistor's function considering its low resistance and high wattage value, and 2) if it's really necessary since the circuit I'm working on seems to get along just fine without it.
-Chris-

Electronics Question | 12 comments (12 topical)

Re: Electronics Question (3.00 / 0) (#1)
by RobD on Tue Aug 03, 2004 at 07:08:38 PM MST

I'd have to see it but it might be a current sensing resistor. Are there leads running to an OPamp input from it? Can you display a circuit?

Re: Electronics Question (3.00 / 0) (#2)
by ChrisW on Tue Aug 03, 2004 at 07:23:15 PM MST

No, it's not a current sensing resistor. Take a look at the schematic on this page:

http://www.mitedu.freeserve.co.uk/Circuits/Power/1230psu.htm

In this case, he's using TIP2955s (PNP), but I've seen other versions of this regulator that use 2N3055s (NPN) wired up the same way, i.e., with .1 ohm, 5W resistors on the emitter.

In similar current switching circuits that use a power MOSFET, the resistor is connected to the Source pin -- apparently for the same reason it's used in the voltage regulator on the aforementioned page.

-Chris-

[ Parent ]

Re: Electronics Question (3.00 / 0) (#4)
by drdongle on Tue Aug 03, 2004 at 07:36:38 PM MST

Current limiting. .1 Ohms at 12 volts limits current to 120 amps

Carpe Vigor

Dr.D
Carpe Vigor, Dr.D
[ Parent ]

Re: Electronics Question (3.00 / 0) (#5)
by Ungrounded Lightning Rod on Tue Aug 03, 2004 at 07:50:39 PM MST

It's a balancing resistor.

If you didn't have it, minor variations between the transistors would cause the ones with a lower junction voltage or higher gain to try to carry more current. (I think that even if you had perfectly matched transistors, the ones that got warmer would try to pull more current and thus you'd get thermal runaway, with the current concentrating in the hotter transistors.)

With the resistors each transistor's section is acting as an emitter follower, trying to keep the voltage at the emitter at a fixed point, one junction-drop down from the base voltage. Because of the gain of the transistor the bulk of the voltage across the resistor is from the output current. The resistors thus surve as current-distribution sensors, keeping the emitter currents, and thus the collector currents, about equal in all the transistors. You pick a very small resistor because the base current slope is very steep and you thus don't need a lot of feedback signal, and you don't want to burn a lot of power in the resistors.

You need these if you're paralleling junction transistors, but not if you only use a single transistor per switched leg.

You probably don't need them for FETs: FETs are swtiched resistors, and the switching gate is an electrode that doesn't pull current (except to charge its capacitance). Where one junction transistor would hold down the gate voltage of all of its partners, the FETs don't do that. So the gate voltage rises beyond the point where the transistor is fully turned on and the current is limited by the FET's on-resistance - which is non-trivial and shouldn't vary greatly between different instances of the device. (I think the thermal coefficient is also positive, i.e. more heat more resistance, which should also tend to keep things balanced if they start out unbalanced or heating is uneven.)

[ Parent ]

Re: Electronics Question (3.00 / 0) (#3)
by ChrisW on Tue Aug 03, 2004 at 07:34:15 PM MST

Okay, guess I should have read the WHOLE page! Down at the bottom he says: "Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor."

So it sounds to me like a balancing resistor. Since I'm only using one transistor, e.g., not in a ganged configuration like in the voltage regulator schematic, I'm beginning to think I can do without it.

-Chris-

[ Parent ]

Re: Electronics Question (3.00 / 0) (#6)
by Ungrounded Lightning Rod on Tue Aug 03, 2004 at 07:58:47 PM MST

Since I'm only using one transistor, e.g., not in a ganged configuration like in the voltage regulator schematic, I'm beginning to think I can do without it.

Count on it.

Also, since it's a MOSFET you wouldn't need it even if you DID have multiple units paralleled.

[ Parent ]

Re: Electronics Question (3.00 / 0) (#7)
by Ungrounded Lightning Rod on Tue Aug 03, 2004 at 08:02:08 PM MST

Also, since it's a MOSFET you wouldn't need it even if you DID have multiple units paralleled.

Let me take that back partially. If you're using the MOSFETs as controlled voltage drops rather than switches you might want to put the resistors in there. In addition to balance they'd provide part of the power dissipation, so your transistors wouldn't have to work as hard.

[ Parent ]

Re: Electronics Question (3.00 / 0) (#8)
by ChrisW on Tue Aug 03, 2004 at 08:06:24 PM MST

Wow -- now that's what I call a GREAT explanation! Thanks for your help, that definitely cleared things up.

-Chris-

[ Parent ]

Re: Electronics Question (3.00 / 0) (#12)
by elektronix on Thu Aug 05, 2004 at 01:30:46 AM MST

Hi,
The explanation to the schematics has a small error - it says:
"Resistor R7 is 100 ohms and develops 4 Volts with maximun load."
It is impossible to get that voltage with all six power transistors
running total of 30 A. One transistor delivers ~5 A, and that gives voltage drop
of ~0.5 V across one of the serial resistors 0.1 Ohm, R1 through R6. With
additional voltage drop on base-emiter PN junction, lets say 0.9 V, total
is 1.4 V, and never can go to 4 V!

If the Rload voltage would go above 12 V (for example, a battery is fully
charged (a battery placed where Rload is shown)), the LM7812 would stop
delivering any current on the OUT, and that would stop current on the IN
side (except quiescent current ~8 mA), and consequently it would stop
any significant Base currents on 6 x TIP2955, limiting (or shutting) current
on Collectors, and finally at the Rload. Some voltage will show up
on an open output (high impedance Rload), as the quiescen current gives some
very tiny current through the Bases of transistors.

So, that circuit behaves as a self regulating high current voltage source at
about 12 V, with a current limitting function. The maximum Rload current is
defined through the maximum OUT current of the LM7812, that is about 1.5 A, giving
about 1.5 A on the IN side, shared by the Bases of six transistors.
1.5 A / 6 =~ 0.25 A per Base. With "beta" =~ 35, it could give maximum of
about 0.25 A * 35 = 8.75 A per transistor. The design specifies about 5 A
per transitor and total 30 A, however, my calculation shows higher total
current, at about 6 x 8.75 A = 52.5 A. If the "beta" is higher, the total
current will go up as well.

Resistor R7 = 100 Ohm plays no role with this example currents.

To get ~4 V across R7, the voltage drop across 0.1 Ohm would be 4 V - 1.0 V = 3.0 V,
meaning 30 A per a single transistor!, and not the total current.

Something is wrong here: the web site explanation, or my understanding of
that schematic.

Thanks,
E-x

[ Parent ]

Re: Electronics Question: alternative (3.00 / 0) (#9)
by VermontMaple on Wed Aug 04, 2004 at 09:50:07 AM MST

ChrisW,
I agree that these are "Swamping" resistors. I discovered this technique many years ago while working for Data General Corp. They used many paralleled very large transistors to power their power hungry systems. However one of the design engineers explained to me that their designs called for .05 and .1 ohm precision resistors but they found that a better appraoch for very small value resistor is to use high amperage fuses..from the same manufacturing batches. The fuses are really very very precise resistors and they also provide overload protection if a transistor fails. They also used the very same approach to paralleling lead acid batteries in their battery backup units. This is really good because if a battery shorts (or someone drops a wrench in there) the current is limited to the fused value and not the total sum of the bank. It works very well. I work currently as an engineer for the local power company (and yes I am off the grid, BTW) and we use the same technique substituting fuses for swamping resistors in our battery banks at our communications facilities. Just pick the amp value to protect the device (ie. if it is a 5 amp transistor use a good 5 amp fuse).
Also the total length of the wiring to the units is important be sure to use equal lentgh wires to(and from) the paralleled devices.
Good Luck

Re: Electronics Question (3.00 / 0) (#10)
by Nando on Wed Aug 04, 2004 at 11:10:10 AM MST

Chris:

I will be short since I need to go out.
The emitter resistors ( low value) when paralleling several transistors are used to equalized the emitter current.
The Base Emitter voltage is not precise. I. E. 0.65 -- it varies some millivolts during manufacturing.

The power dissipation is the voltage drop across it times the emitter current plus any additional wattage we desired to insure that the resistor does not glow during operation.
Sometimes some circuits have the resistors in the base and some times resistors in both places: emitter and base.

The IRF520N should not be used at currents higher than 10 amps because the curves display a negative behavior above 20 amps.

I you care to send to me the circuit you are working on, I will give you an analysis of it.

Regards

Nando

Re: Electronics Question (3.00 / 0) (#11)
by RobD on Wed Aug 04, 2004 at 02:51:24 PM MST

The 2955 PNP transistors are used as a closed loop pass transistors. I wouldn't use a mosfet in this application. Furthermore the 520 is an N type mosfet which would not be a closed loop and you would need a charge pump because this would be a high side switching application for an N type device.
You can make a simpler circuit with several LM1084 (5 amp. in a TO-220 package) regulators together with .1 ohm resistors on the output legs of each device.
Whatever you use make sure to heatsink the package.

Electronics Question | 12 comments (12 topical)

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