Why is my 555-timer PWM duty cycle range so small?

[ArdWar]

It should work as a summing amplifier.

I don't think that particular circuit values are correct however. I think they're just copying AP5724 circuit (which is correct within reason), correctly scaling the sense resistor for 20mA full scale load current, but forgot to scale control voltage injection resistor (R3, and/or R2) to 0-2V control voltage.

[Cory Parsnipson]

It should work as a summing amplifier.

I don't think that particular circuit values are correct however. I think they're just copying AP5724 circuit (which is correct within reason), correctly scaling the sense resistor for 20mA full scale load current, but forgot to scale control voltage injection resistor (R3, and/or R2) to 0-2V control voltage.

Thanks! This helped me figure out how to analyze the circuit. Now if I can only remember my fundamentals...

The feedback (FB) pin internal difference amplifier's reference voltage is 0.31V (nom.).   That means the current regulation loop acts to keep FB close to that voltage swinging above and below it slightly with every cycle of the internal switch.  The datasheet tells up the formula for the average LED current (no dimming) is I_f=0.31V/Rset.

By injecting current at the FB pin, you can fool the control loop that the LED current is higher than it actually is, thus reducing the LED current.  However its at an inconveniently low impedance, requiring a control current equal to the max. LED current, so R2 is added so the control current is reduced by a factor of 1+ R2/Rset.   Thus to fully extinguish the LEDs, Ictl_max=0.31V/(R2+Rset), and R3 scales that to the desired upper limit of the control voltage input range.

However I think there may be an error in the datasheet's maths (and thus the R3 resistor value I proposed earlier, which I scaled from theirs)!  It states that "For V_DC range from 0V to 2V, the selection of resistors in Figure 3 gives dimming control of LED current from 0mA to 20mA." which is patently absurd as Max. V_DC gives min. I_f and visa versa.  Also the potential divider R3:(R2+Rset) is wrong - we know it must develop 0.31V at FB when I_f=0.  Neglecting Rset as its three orders of magnitude smaller than the potential divider resistance, the fig 3 resistors give V_FB=V_DC/21 which is a max. control voltage of ~6.1V.

Sorry, I think I'm doing something really stupid. So you said that to fully extinguish the LEDs:



So if I plug in 5k for R₂ and 15 ohms for R_set, then the max control current is 6.18 * 10^-5 Amps? Is this supposed to be the same as the max LED current (20 mAmps)? I don't think I'm doing it right.

I'm also confused about how exactly R₃ scales the current...

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In the meantime, I tried to find a relationship between V_DC and I_SET by solving the op amp and surrounding circuit. I'm getting lost in the middle and not sure if this is the right direction...

Labeling the diagram like so:



I can try and do the node voltage calculation for V_N, like this guys does: https://electronics.stackexchange.com/questions/156719/opamp-inverting-amplifier-the-case-of-non-inverting-input-not-being-connected-t



I assume V_N is equal to V_REF. I get the same answer by solving for V_SET:



This is where I'm stuck. I think I need to figure out the relationship between V_DC and I_SET, and I have this hunch that I_SET is I_LED + I_CTL... And that I'm missing an equation or two to figure out the relationship between I_LED and I_CTL.

And then I guess plug in numbers for V_DC and I_CTL to decide which values to use for R2/R3?

[Ian.M]

No, the control current is not the LED current.  Its the current through R2 and R3 (neglecting the tiny FB input current), and thus R3 scales it into a voltage, as the control loop acts to keep FB at the reference voltage as long as the LED current is greater than zero.

[Cory Parsnipson]

No, the control current is not the LED current.  Its the current through R2 and R3 (neglecting the tiny FB input current), and thus R3 scales it into a voltage, as the control loop acts to keep FB at the reference voltage as long as the LED current is greater than zero.

Right, the LED current is different from the control current, but you mentioned before that to fully turn off the LEDs, the control current needs to be equal to the max LED current?

However its at an inconveniently low impedance, requiring a control current equal to the max. LED current, so R2 is added so the control current is reduced by a factor of 1+ R2/Rset.   Thus to fully extinguish the LEDs, Ictl_max=0.31V/(R2+Rset), and R3 scales that to the desired upper limit of the control voltage input range.

So, then:

I_{CTL_MAX} = 0.02 = 0.31 / (R₂ + 15) => R₂ = 0...

I'm still confused...

[b_force]

Just one question: why? You must be using a microcontroller, so skip the 555 and just generate the PWM directly, from that.

Or just use a NAND or schmitt inverter instead.

When doing this right you have control over frequency which is (relatively) more constant as well as duty cycle for the same package as a 555.