What’s the voltage required to light one string?

What current is drawn by one string when the above voltage is supplied?

~ 3.5V 150mA. Supply voltage is 5V

 

I just ran a sim on the circuit in #19 it works fine.

But @ 5 volts the voltages across the LED and resistor is 1.2 to 2.7.

A 5k feedback resistor from pin 3 to CV brings that to .9 to 3.3.

 

@AnalogKid, ElectricSpidey, and all the contributors a big thank you. I will build the #19 and let you know

 

I built a circuit based on the what I posted in post #2. I used a triangle wave from a signal generator to modulate the pulse width and added three transistors to drive the LEDs.

 

I built a circuit based on the what I posted in post #2. I used a triangle wave from a signal generator to modulate the pulse width and added three transistors to drive the LEDs.

And that raises a point. While it might be that the reason the fade part is so short - it looks like the LED comes close to full brightness quickly, then slowly tops off - is camera sensor saturation, another driteria is the shape of the driving waveform. R-C circuits ramp up quickly and then approach the max value more slowly. Maybe a better driving waveform is a triangle waveform.

But wait - that's what dl used. OK, then maybe a better driving shape is an inverse exponential, something that as a voltage waveform starts increasing slowly, then increases more rapidly as time goes on. OR, a linear ramp as dl used, but a linear *current* ramp rather than a voltage ramp. A Howland current pump is voltage-to-current converter. Drive that with a triangle wave, and you might get the breathing visual effect you want.

ak

 

I forgot one thing: in circuit #19 how can I change the ton and toff time? what's the component that determines that?

 

Your question indicates that you have not read the datasheet. Read it.

ak

 

In the datasheet I've found the classic formula:

t1 = 0.693 (RA + RB) C
t2 = 0.693 (RB) C

but in your schematic the 555 is wired completely different (for example pin7 and pin8 are not connected) so the formula is not applicable. It's that C2 that determines the t? Or perhaps is C1? R1 cannot be because it's only the Rbase of the transistor.

 

I would think you would start with the t2 formula, with RB=base resistor (R1) but then it gets tricky because there is also some current flowing in that resistor from the follower circuits.

But, this should be minimum so the above is a good starting point...adjust from there.

I really don't think you are going to be happy...did you notice the voltage range in post #22?

You can move the entire range upward by using the CV pin, but I will leave that to after you try the circuit.

 

To get near-50/50 duty cycle out of the standard 555 astable circuit, one resistor must be much larger than the other (you can figure out which one). In this case, one resistor can be dropped out of the math and the two equations combined.

- or -

The part in the schematic is the CMOS version of the 555. In that datasheet there is an alternate astable circuit that works for all versions. Note the similarity of the two circuits' equations.

ak

 

R1 cannot be because it's only the Rbase of the transistor.

"only" - ? No.

R1 is the circuit timing resistor. That is why it is connected to the Trigger and Threshold inputs. Along with C1 it creates a pseudo-triangle wave at the R1-C1 node. All 555 astable circuits have this waveform across the timing capacitor. A common use is to filter it to make a low-to-medium quality sine wave.

Back to the triangle wave ... some of the current through R1 does not go in/out of C1; it goes through the bases of the output transistors. This will increase the oscillator frequency. To minimize this, the transistors can be replaced with Darlington types to reduce the base current. Or, for any particular frequency, go with a smaller resistor and larger capacitor. This will make the transistors' base currents a smaller percentage of the total resistor current.

ak

 

The schematic at #19 doesn't work at all

 

And that raises a point. While it might be that the reason the fade part is so short - it looks like the LED comes close to full brightness quickly, then slowly tops off

That was just me not taking the time to match the triangle wave used to adjust duty cycle to the triangle wave output from the opamp. Here's an updated clip; ignore the two LEDs in the corner. That was an experiment with a 555 timer.

 

The schematic at #19 doesn't work at all

Do you have a meter to check the voltages across the LED strips while the circuit is running?

If so do it and post those voltages.

 

So the circuit worked with the values in AK's circuit. so I just ran it again with the LED current raised to 150mA and nothing....flatline.

So, I started reducing the base resistor and the circuit started to work again.

 

The two transistors are emitter followers, so the base current is proportional to the emitter (and LED) current. And, as above, the more base current there is, the less current there is to charge the timing cap. Also, the more base current there is, the higher the voltage drop across R1. Round numbers:

150 mA into the LEDs, transistor hfe of 150 >> 1 mA base current
1 mA, 47,000 ohms >> 47 volts across R1.

So no, it won't work with higher emitter loads. C1 never charges up to 0.67 Vcc. A solution is to use darlington transistors, but this would require a higher operating voltage.

NOTE: post #19 has the work concept highlighted.

ak

 

Well I got the circuit working in the sim @ 10k with .35mA base current using 222s and 907s.

But the voltage range on the cap only allows the output current to reach 8mA 1.6 to 3.3 volts

When I boost the input voltage to 12 it flatlines again until the base resistor is about 5k. But the output current still only reaches 80mA.

These problems are why I use 12 volts for my LED projects and adjust the control voltage as well as buffering the timing network.