Electronic – Voltage to PWM Circuit, need to understand frequency

pwm

This is an off shelf component that might fit my need. Unfortunately the documentation is terrible. It should convert a 0-5v or 0-10v analog to PWM (0-5v/0-10v is a jumper). The analog voltage signal controls the duty cycle. The potentiometer controls "Precision" per the description. I think they mean it can offset the duty cycle from the control signal. I don't care about that, I can program the control signal. This should be perfect for me if the PWM frequency is 21-28KHz.

I can't find/get any info about the frequency. It was cheap enough that I picked one up and started mapping out the traces. I've been trying to learn for a few hours now and I can't quite figure it all out. I think the 555 is controlling the frequency, but this doesn't look like any examples I have found. I want to know specifically which resistors/capacitor are controlling frequency and what it calculates to. I have the ability to rework the board and change them if needed. I don't have an oscilloscope right now, but I'm working on borrowing one. I'm hoping to figure this out and get any possibly needed components before I borrow the scope.

The picture should show connections, sorry if it is ugly. I will gladly clarify anything if confusing. FYI, I just used purple when it would cross over another red line.

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Best Answer

Your circuit will look something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Help understanding how to use a SG3525

And the PWM frequency will depend on \$C_2\$ and \$R_6\$ values.

\$Q_1\$ together with \$R_6 , D_2 , D_3\$ forms a constant current source.

And this current will be around \$I_S = \frac{0.66V}{R_6} = 140\mu\textrm{A} \$

And this current will charge the \$C_2\$ capacitor. And if the voltage across the capacitor reaches the NE555 threshold value (2/3 Vsup). The 555 will quickly discharge the capacitor to 1/3Vsup.

So, the equation for PWM frequency will look like this

\$\Large F \approx \frac{3 \cdot I_S}{C_2 \cdot V_+} \$

So for 5V supply, as you have in your circuit the equation becomes

\$\Large F \approx 0.6 \cdot\frac{I_S}{C_2} \$

I'm trying to simplify all the resistors and jumper leading to U2+ (R9,R10,R1,R3,R4). Suppose I don't need the option to jumper to a 0-5V input and only using 0-10V input. Also would like to eliminate adjustability with R1. From testing I know that for best function R1 is set at 8k Ohm between jumper above R3, and 94k Ohm to ground below R3.

In theory, this voltage divider at the input should do the job:

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