LM3447 control law

For example:

My design needs the following feed-forward circuit parameters:
At Vin(rms)=250V Duty cycle should be 0.174 at peak of Vin sine
At Vin(rms)=190V Duty cycle should be 0.229 at peak of Vin sine

How to calculate Rff resistor value for preselected Rac resistor to achieve these parameters?

Hello,

The LM3447 regulates output power so there is an additional compensation to the duty-cycle from the input voltage.  How are you planning on using this?

Now it does not matter how I want to use this!
I need more detailed information about error amplifier to design own circuit to control duty cycle as I need. Tell me please an equation that determines how duty cycle depends on voltage on INV input. This is the only thing that I need.

I can determine this by series of experiments, but I prefer calculation before experiments.

Hello,

This is the information I received from the engineer involved with the LM3447:

In open-loop the duty cycle is controlled by the COMP voltage as it is compared to the internal ramp voltage by the PWM comparator.  To control the duty-cycle, the external sense voltage (via a resistor divider) can be connected to INV pin and the compensation can be connected between COMP and INV pin:

The relationship between the comp voltage and duty cycle (based on typical 2.22V ramp magnitude) is D = VCOMP/2.22;

Hope this answers the question.

Thank you very much! But I need more information about open-loop gain. Datasheet gives some information about gain of error amplifier  - Gm=100umho. But this value is unusable without additional information. Transconductance means an output current in relation to the input voltage. But COMP (output of amplifier) is voltage output, and COMP pin doesn't assumes resistive load to convert current to voltage. How I should treat this transconductance? Chip contains internal resistor to convert current from transconductance amplifier to voltage?

What is the overall gain: V(COMP)/(Vref-V(INV)) (or may be another equation? ) ?

Sorry, took a while to get this.

First, it will be average voltage not RMS which changes this slightly (I think you may have done average as the numbers ended up the same).  What is also added internal to the IC is duty cycle so it will try to regulate at 1.0 volts.  The Rff you have list is quite a bit higher than the EVMs and TI designs.  This means your duty cycle will be lower than these designs, which is fine if you prefer to do that.

I(Rac, average)  220*1.414*.636/824K = 240 uA

Iff is /10 = 24 uA

V(Rff, average) is 24 uA * 232K = 5.57 volts which means the duty cycle needs to be 1/5.57 or about 18%.  If Rff, was 143K then duty cycle would be about 29%.  If VAC drops duty cycle goes up if VAC goes up duty cycle drops.

Yes, you are right, i said "RMS" but meant "average". Sorry.

Why 5.57 volts from Rff, connected to INV input of transconductance amplifier with Gm=100umho results in 1/5.57 duty cycle?

Based on previous message:

Transconductance amplifier converts input voltage in to output current. ICOMP = gm*(V(+)-V(-)). The comp voltage is then based on the compensation network, Gc(s), such that VCOMP = ICOMP*Gc(s). The complete transfer function, Vcomp/(V(+)-V(-) = gm*Gc(s).

5.57 volts on INV input will be converted to Gm*(Vref-5.57) = 100u*(1-5.57) = -457 uA current from COMP - this current discharges capacitor up to near zero level and results in zero duty cycle. Or here is some undocumented on-chip feedback from gate driver (duty cycle) to FF output or INV input ?