Simple Solar Tracking circuit

Discussion in 'Renewable Energy' started by helpmonkey, Apr 27, 2010.

1. ColinMember

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Where did you get this rubbish from ?????

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The rubbish is coming from you.

Since you seem to think that a base resistor is needed in this circuit, please explain the plot shown below:

Why is the base current of Q1 Ib(Q1) equal to only ~380uA without a base resistor when the transistor is delivering 92mA to the load. Note that the emitter current is determined by R1, and that maxes out when V(R1) is 10V - Vbe.

Last edited: Nov 4, 2015
3. ColinMember

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This is the circuit I am talking about and there is no current limiting on the output of U2.

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5. dougy83Well-Known Member

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How does U2 (an LM339) drive an NPN transistor without a pull-up resistor?

6. Les JonesWell-Known Member

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As the emitter of Q1 is at -12 volts then R3 connected to zero volts will be a pull up resistor as the base of Q3 does not need to be driven more positive than about - 10.5 volts.

Les.

7. dougy83Well-Known Member

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Oh, thanks. I didn't pay any attention to the negative power supply. There is a 2k7 pull-up resistor.

8. ColinMember

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How does U2 (an LM339) drive an NPN transistor without a pull-up resistor?

The 2k7 is not needed. U2 is fighting against the 2k7. The circuit is just badly designed.
This is the second fault with the circuit.

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So you are welcome to redesign it, Colin... But be aware, your design will be critiqued...

10. dougy83Well-Known Member

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U2 has an open-collector output. It needs a pullup to drive the base of the NPN transistor; that is the function of the 2k7 resistor.

11. ColinMember

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Connecting the collector of the first transistor in the super-alpha arrangement to the collector of the output transistor means the output transistor cannot turn ON fully.
If we take Q3, it cannot behave as an amplifier if the collector is not at least 0.3v above the emitter.
If the collector voltage is less than 0.3v, the transistor simply acts to pass the base current to the emitter via the natural voltage drop of this junction.
With the collector 0.3v above the emitter, the minimum voltage on the collector will be 0.3v plus 0.7v across the base-emitter of the TIP3055. This means the collector of the TIP cannot go lower than 1v.

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You have accurately described the well-known limitations of the Darlington connection, where Vce(on) =~1V.

So What! The power dissipated in Q3/Q1 is less than 1W, it switches >1A to a 12V motor with about 4mA into the base of Q3 and if the neg power supply is -12.0V, the motor still gets more than 11V across it.

If you really, really wanted a more efficient switch then use an NFET instead of the Darlington. The original goal of the this design was easy availability of the parts, so I tried to use what in 2010 might have been available in a RadioShack, which has since closed most of its stores...

Last edited: Nov 5, 2015
13. Tony StewartWell-Known MemberMost Helpful Member

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The specs for the TIP3055 include'
Vce(sat) = 1.1V (Maximum) at Ic = 4.0A, Ib = 400mA. which is typical Ic/Ib ratios of 10:1 when saturated for a wide range of current.
Pd = 90W with suitable heatsink.

2N3904 is similar with 10:1 Ic:Ib but with Ic max= 200mA Vce(sat)=0.13V @100mA

So saturated base current gain at output drop = 1.1V is 100x
Meanwhile Ib depends on Supply voltages, LM358 2V saturation , 4k7 base resistor e.g. 12V motor Ib max= 2mA , then Motor can draw up to 200 mA except on surge , output devices come out of saturation and Vce rises as they get warmer.

So if using 12V motor and 3V motor will give different current limits determined by many factors including Rb and especially Rload.

But heatsinks are highly recommended.

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Tony, Q1 ('3055) is not saturated, so your analysis doesn't work. The base current of Q3 doesn't depend on the comparitor at all! R3=2.7K, not 4.7K. I have no idea where you pulled a LM358 from?

Look at the predicted voltages/currents in this simplified circuit with the motor switch on.
Note V(c)-V(e)=~0.75V.
Note the current in R3 = ~4mA
Note that Q1 would be dissipating only about 3/4W.

This sim was done with the nominal β for both Q3(300) and Q1(360). I reduced β to 50 for both Q3 and Q1 and the voltages/currents changed hardly at all.

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15. Tony StewartWell-Known MemberMost Helpful Member

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Most BJT's when Vce < 2 at high currents, it has begun the saturated state where hFE drops rapidly towards 10 ot 10% of linear hFE(typical)
Vce=0.75 @ 1A is certainly saturated so gain should be reduced to 10. Note Vce(sat) spec can be converted to an ESR value or Rce value, so Vce can be predicted.

Another change I would suggest is R load .. A 1 amp rated motor will have a series resistance about 1/5th to 1/8th of V/I (rated). THis will affect your results for stall current. Thus for stall current, R becomes 12/8 =1.5 Ohms now recalc and see how Vce rises based on Rb and hFE reduction to perhaps 20.

Initially the comparator drives full voltage thru Rb to the 2 stages. Since 2N3904 is limited to 100mA abs max. it is wise to drive much less than 10mA llimited by Rb. Thus Ic current is limited by Rb into a 1.5 Ohm Rs coil of motor. The actual operating point of TIP3055 changes as motor begins to move with back EMF.

My initial concern was arrangement of optical sensors and lost power from hunting , such that more effort to prevent this and make tracking as efficient as possible to gain only ~20% from a stationary south position which must be accumulated to break even after the power is drawn. Power required to hold a massive solar wing may be proportional to power capacity of the "wing" from aerodynamic effects unless gear reduction improves holding torque but that also extends motor duration.

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A 3055 with 0.7V Vce @ 1A is nowhere close to being saturated, and its β is still whatever β was. To get a single, non-Darlington connected NPN power transistor to saturate (so that Vce is as low as possible, typically about 0.4V, you overdrive the base with a base current of Ic/10, not because the β=10, but as a safety factor to minimize Vce and power dissipation in the transistor, especially at elevated temperatures.

An advantage of the Darlington connection is that you do not have to (nor can you) overdrive the output transistor or get it into deep saturation. None of the base drive = Ic/10 stuff applies to driving a Darlington. Go look at an integrated Darlington data sheet.

For example, look at Fig 2 of the TIP120 data sheet. It shows that to get a Vce of 0.6V at Ic=1A, the required base current is 4mA.

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17. Tony StewartWell-Known MemberMost Helpful Member

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The integrated Darlington here uses a ratio of 15 per stage instead of 10 so they get 250 for 2 stages. I used 10 per stage or 100 per pair.
The saturation begins when the comparator drives with 12V and with no load the output of 12V-Vce is available to the motor. But a 1A motor @12V always have start current 5~8x rated current meaning ESR or Rs of motor coil is 1/5 to 1/8th of 12 Ohms so with 1.5 Ohms , you might expect 11V/1.5 Ohms or 7.3Amps , then one expects it comes out of saturation from base starved current but then linear operation prevails so hFE is back to normal.

At 4A the hFE ranges from 20 to 70 so if the motor demands 7,3A what will the actual Vce and current be? Using the better TIP121 it would suggest the base current must be 7.2A/250=29 mA.
But it is only 4mA so even the TIP131 comes out saturation on start stop and Many watts are pulsed into the driver until the motor reaches full speed and current drops to 1A.

Make sense yet?