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I need to run 9 leds (7 white and 2 red). I found this circuit

enter image description here

but there is no limiting resistor in the upper line of leds. Is it ok? If no, how I can modify this circuit?

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but there is no limiting resistor in the upper line of leds. Is it ok?

No, it's not okay. Your upper line of LEDs will most likely burn up.

If no, how I can modify this circuit?

Do as the other LED lines, add a resistor.

Assuming that a white LED has a forward voltage of 2.8 V, then 4 of them will give you a forward voltage of 11.2 V, I'll assume you want 133 mA through your upper LED line. Because that is what you got through your second line of LEDs.

Then the resistance of the resistor that you will add on your upper line of LEDs will be \$\frac{12-11.2}{0.133}\approx6\text{ Ω}\$. You can probably get away with a 5 Ω resistor, though I have absolutely no idea because I don't have a datasheet in front of me.

If I were you, then I would use a 10 Ω resistor because it's safer, if the LEDs has a forward voltage of 2.5 instead of 2.8, then with a 6 Ω resistor you will get 333 mA, more than twice. With a 10 Ω resistor you will get 200 mA, not as good but also not as bad.

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Your proposed circuit is a bad idea.

It seems you are trying to run over 100 mA thru the LEDs. At these power levels, it is better to drive LEDs with a current source than to use series resistors. You can use a NPN as a controlled current sink:

The emitter current is the voltage across R1 divided by its resistance. Since the B-E drop is reasonably constant, putting a fixed voltage on the base keeps a reasonably fixed voltage on R1. That in turn keeps the current thru R1 reasonably constant. Since the collector current is mostly the emitter current due to the gain of the transistor, the base voltage ultimately sets the current thru the LEDs to a good enough approximation for most purposes.

In this example, I targeted 1 V across R1, so the LED current is about 100 mA. The B-E drop is assumed to be 750 mV. Higher R1 voltage increases regulation and makes the current less sensitive to B-E drop changes. However, it also increases the power dissipation of R1. I'd try to keep the R1 voltage in the 500 mV to 1 V range.

In this example, R1 only dissipates 100 mW. I showed it as being a ½ W resistor so that you would keep its dissipation in mind. It's not clear what your intended LED current is, but it looks like it's more than 100 mA. You will need to adjust R1 and the base voltage to get the desired current.

The collector of Q1 looks like a fixed current sink to the rest of the circuit. You can therefore connect a string of LEDs to it without any additional current limiting. Use as many LEDs in series as possible, but such that their maximum voltage drop does not exceed the supply voltage minus the minimum drop across Q1 and R1. I'd try to keep at least 500 mV across Q1 C-E so that it can still provide some regulation.

With this circuit, a higher supply voltage is better because it allows for more LEDs to be run together. The remaining voltage drops are fixed, so they are a smaller fraction of a higher supply voltage.

Keep the dissipation of Q1 in mind. You can only use integer numbers of LEDs, assuming their maximum voltage drop. Then you have to go back and find their minimum voltage drop at the intended current. The supply voltage minus that minimum voltage drop minus the R1 drop will be across Q1. That times the current is the power that Q1 must be able to safely dissipate.

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It depends a lot from the circumstances and the characteristic of the LEDs.

Assume a single LED with 20mA@3.2V.

You can run it perfectly on a 3.0V power supply without resistor, since it will only draw may be about 10mA.

But be aware, that if the supply exceeds 3.2V, the current also exceeds 20mA, which will burn the LED. Also, in a hot environment, the LED will draw the 20mA at a lower voltage. If that's below 3.0V, the LED will draw more than 20mA, and will burn. Contrary to a filament bulb, heat leads to more current in LEDs, and so more heat...

The 27Ohm resistor in your circuit simulates an LED, thus the LEDs are driven at 3.0V and 111mA. (I neglected the transistor here) If this is below the nominal value of the LEDs, and you can ensure that the voltage will not exceed 12V and the LEDs will not run too hot, then the circuit might be OK.

Now, about 100mA at about 3V is pretty much for an LED, so it might heat up under normal conditions. In this case, yo should add a resistor in the first row, too, as proposed by the other answer.

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