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I would like to understand how the oscillator of this PWM controller works, and what would happen if I do not apply an RC network but directly a PWM signal generated from a microcontroller for example.

enter image description here

It is indicated into the datasheet (https://www.ti.com/lit/ds/symlink/uc3845.pdf?ts=1590914073928) that the capacitor is charged from the reference voltage Vref until the voltage across the capacitor reach an upper threshold and then the capacitor is discharged via an internal current source. Nevertheless the oscillator as showned above generates pulses.

enter image description here

When the capacitor is discharging, we can think that a pulse is generated. It could be the inverse... I would like to know exactly how it works... ie when the pulse is generated ? Is it when the capacitor is discharging or when it is when the capacitor is charging.

Now, if I directly inject a PWM signal into the RT/CT pin, like this :

schematic

simulate this circuit – Schematic created using CircuitLab

When the PWM signal which rise (very rapidly as it is a rectangular waveform) to the upper threshold, the controller will draw current (for discharging the capacitor), it will be provided by the microcontroller, then the PWM will be low and the PWM controller will stop to draw current and will wait the the PWM controller rise again to draw again current. So it seems necessary to not insert a resistance as it could insert a voltage drop when the pwm controller draws current and the PWM controller could not able to see the different voltage level of the PWM.

Can anyone know how it is built internally and if what I want to do is possible ?

Thank you very much and have a nice day.

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2 Answers 2

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If you are trying to control the oscillator externally use a sync pulse as shown in the data sheet: -

enter image description here

This is also described here: -

enter image description here

There are several ways that the oscillator can work and this is probably the basis: -

enter image description here

Picture from here. If it has an offset duty cycle i.e. it produces more of a pulse than a square wave it may have an internal diode and resistor to forcibly shorten the time taken to charge or discharge the capacitor like this: -

enter image description here

Picture from here. \$R_d\$ and its series diode may be inside the chip. There isn't a requirement for \$R_c\$'s diode if \$R_d\$ is >> \$R_d\$.

There are a few other ways this can be done too.

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  • \$\begingroup\$ I saw this solution too. Nevertheless I would have appreciated to know how the oscillator works internally. I should go straight on ! I have seen also circuit which directly connect a microcontroleur pin with the RTCT pin \$\endgroup\$
    – Jess
    May 31, 2020 at 12:31
  • \$\begingroup\$ Added some bits @Jess \$\endgroup\$
    – Andy aka
    May 31, 2020 at 13:14
  • \$\begingroup\$ I will do what the datasheet recommand to do ;) Thank you for your edited post :D \$\endgroup\$
    – Jess
    May 31, 2020 at 17:01
  • \$\begingroup\$ Well I do not like what the datasheet recommand to do as the resitor at the synchronizing node is low (24 Ohm). It draws a lot of current that I cannot provide without to do an "adapative" circuit. I will ask some helps from TI \$\endgroup\$
    – Jess
    Jun 2, 2020 at 15:10
  • \$\begingroup\$ @Jess It also says that the peak voltage is 0.5 volts so, if you have a 5 volt clock signal, you can use a 220 ohm resistor to feed the 24 ohm resistor and get a 10:1 attenuation. Then you have to ask yourself if your 5 volt signal can suffer being loaded by 244 ohms. \$\endgroup\$
    – Andy aka
    Jun 2, 2020 at 15:19
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Probably something alone these lines:

schematic

simulate this circuit – Schematic created using CircuitLab

Where the voltage at the pin is something like this:

enter image description here

And the pulses are likely generated during the discharge of the timing capacitor.

If you want to force the output pulses to follow you would have to overpower the internal current sink to keep the pin voltage above a few hundred mV (estimated but you could measure or simulate it). If my estimated vale of 20mA is in the right ballpark, you might need a buffer or discrete transistor to do it reliably.

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  • \$\begingroup\$ Thank you for your answer ! \$\endgroup\$
    – Jess
    May 31, 2020 at 16:59

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