please what will the output waveform of a pwm signal look like after being switch by a mosfet or transistor. Especially in smps. Will the frequency of the pulse increase or decrease or will everything remain the same.

 

please what will the output waveform of a pwm signal look like after being switch by a mosfet or transistor. Especially in smps. Will the frequency of the pulse increase or decrease or will everything remain the same.

The frequency of the pulses will depend on the circuit driving the MOSFET. In a power supply it may vary with the demands of the load or the frequency may stay the same and the duty cycle may vary. The shape of the MOSFET output waveform will depend on what kind of load is being drivenT.
Regards,
Keith

 

The rate (or frequency) at which the power supply must switch can vary greatly depending on load and application. For example, switching has to be done several times a minute in an electric stove; 120Hz in a lamp dimmer; between a few kilohertz (kHz) and tens of kHz for a motor drive; and well into the tens or hundreds of kHz in audio amplifiers and computer power supplies. The main advantage of PWM is that power loss in the switching devices(Your MOSFET) is very low. When a switch is off there is practically no current, and when it is on and power is being transferred to the load, there is almost no voltage drop across the switch. Power loss, being the product of voltage and current, is thus in both cases close to zero.

 

The rate (or frequency) at which the power supply must switch can vary greatly depending on load and application. For example, switching has to be done several times a minute in an electric stove; 120Hz in a lamp dimmer; between a few kilohertz (kHz) and tens of kHz for a motor drive; and well into the tens or hundreds of kHz in audio amplifiers and computer power supplies. The main advantage of PWM is that power loss in the switching devices(Your MOSFET) is very low. When a switch is off there is practically no current, and when it is on and power is being transferred to the load, there is almost no voltage drop across the switch. Power loss, being the product of voltage and current, is thus in both cases close to zero.View attachment 206649

ok you mean that the load will automatically influence the switching frequency of the pwm ic, and also the output waveform of the switching mosfet . Is that it?

 

The frequency of the pulses will depend on the circuit driving the MOSFET. In a power supply it may vary with the demands of the load or the frequency may stay the same and the duty cycle may vary. The shape of the MOSFET output waveform will depend on what kind of load is being drivenT.
Regards,
Keith

Are you saying that the load will automatically influence the switching frequency of the pwm ic, and also the output waveform of the switching mosfet .
Ok please can you explain more about it. Or perhaps use any example to elaborate further please?

 

There are a number of different types of power supply that switch MOSFETs on and off to regulate the supply to a load. Each type has it's own unique way of handling feedback and controlling the FETs. Some use a fixed frequency and vary the duty cycle. Some use a fixed pulse length and vary the frequency. Some just switch on the FETs whenever the load demands it and switches off when the requirements are met. I hope this answers your question.
Regards,
Keith

 

The power transistor (either BJT or MOSFET) just acts as a switch in a SMPS, so it has no significant effect on the frequency or duty-cycle of the signal.

 

Are you saying that the load will automatically influence the switching frequency of the pwm ic,

"automatically influence" We have a power supply with a load. It has been working for a long time. The output voltage is what we want. (5.000 volts) The error amplifier see 5V and is happy. The load increases. This causes the voltage to start down. The error amplifier now sees 4.999 volts and it calls for the duty cycle to increase. The duty cycle will keep increasing until the output is back to 5.000 volts. Now the duty cycle will hold at this new duty cycle.

If the load is reduced the output will start up. 5.001V will cause an error voltage and push the duty cycle down until the voltage is back to 5.000 volts.

Automatically? There are parts that make this happen. (error amp and reference voltage)

 

There are a number of different types of power supply that switch MOSFETs on and off to regulate the supply to a load. Each type has it's own unique way of handling feedback and controlling the FETs. Some use a fixed frequency and vary the duty cycle. Some use a fixed pulse length and vary the frequency. Some just switch on the FETs whenever the load demands it and switches off when the requirements are met. I hope this answers your question.
Regards,
Keith

Ok thanks, the feedback tends to adjust the duty cycle of the pwm signal that was generated by the ic, before it is fed to the gate or base of the power transistor by the help of a reference voltage.

All that matters or does the work is the pwm generator ic, not the switching component.
The mosfet or transistor is all dependent on it base signal.Am I getting it right?

 

Some web tutorials do tell me that the feedback loop adjust the duty cycle of the mosfet or transistor, as if they are the one switching their self.

My thoughts are that the pwm ic does all the work, and the switching component (mosfet, transistors, thyristors.)are all dependent on their base or gate signals.
And also the feedback loop adjust the duty cycle of the pwm generator ic and not the switching component.

 

The power transistor (either BJT or MOSFET) just acts as a switch in a SMPS, so it has no significant effect on the frequency or duty-cycle of the signal.

Ok it reaction is all dependent on the control ic or pwm generator ic?

 

"automatically influence" We have a power supply with a load. It has been working for a long time. The output voltage is what we want. (5.000 volts) The error amplifier see 5V and is happy. The load increases. This causes the voltage to start down. The error amplifier now sees 4.999 volts and it calls for the duty cycle to increase. The duty cycle will keep increasing until the output is back to 5.000 volts. Now the duty cycle will hold at this new duty cycle.

If the load is reduced the output will start up. 5.001V will cause an error voltage and push the duty cycle down until the voltage is back to 5.000 volts.

Automatically? There are parts that make this happen. (error amp and reference voltage)

Ok thanks, but when we talk of error amp, is it the optocoupler?

 

"automatically influence" We have a power supply with a load. It has been working for a long time. The output voltage is what we want. (5.000 volts) The error amplifier see 5V and is happy. The load increases. This causes the voltage to start down. The error amplifier now sees 4.999 volts and it calls for the duty cycle to increase. The duty cycle will keep increasing until the output is back to 5.000 volts. Now the duty cycle will hold at this new duty cycle.

If the load is reduced the output will start up. 5.001V will cause an error voltage and push the duty cycle down until the voltage is back to 5.000 volts.

Automatically? There are parts that make this happen. (error amp and reference voltage)

Ok thanks, but when we talk of error amp, is it the optocoupler?

 

"automatically influence" We have a power supply with a load. It has been working for a long time. The output voltage is what we want. (5.000 volts) The error amplifier see 5V and is happy. The load increases. This causes the voltage to start down. The error amplifier now sees 4.999 volts and it calls for the duty cycle to increase. The duty cycle will keep increasing until the output is back to 5.000 volts. Now the duty cycle will hold at this new duty cycle.

If the load is reduced the output will start up. 5.001V will cause an error voltage and push the duty cycle down until the voltage is back to 5.000 volts.

Automatically? There are parts that make this happen. (error amp and reference voltage)

Ok thanks, but when we talk of error amp, is it the optocoupler?

 

but when we talk of error amp, is it the optocoupler?

no
Optocouplers are used on isolated power supplies. (isolation not amplifications)
Somewhere in most power supplies there is a amplifier with one input looking at the output voltage and the other input looking at a reference voltage. (output voltage or some fraction of it) and (5v or 2.5V or 1.25V or some stable know voltage)
Example: you have a 10V output voltage. Two resistors divide 10V down to 5V. Reference is 5V. The amplifier compares the two 5Vs. The difference is amplifier and sent to the "PWM". If there is no difference the duty cycle remains constant.

 

Error amp with isolator:
R3 & R5 in this case divides VOUT down to 2.5V. The TL431 has a reference voltage of 2.5V, and has a op-amp.
The TL431 amplifier the difference of Vout and Vref. Current goes through R1 and the LED in the isolator. The isolator output sends information to the PWM. If VOUT is low then the "R" pin on the TL431 will be low. This causes the "K" pin to be high and little current will pass though U1. This tells the PWM to work harder. If Vout is too high, K is low passing large current in to U1. This tells the PWM to work less.

 

Error amp with isolator:
R3 & R5 in this case divides VOUT down to 2.5V. The TL431 has a reference voltage of 2.5V, and has a op-amp.
The TL431 amplifier the difference of Vout and Vref. Current goes through R1 and the LED in the isolator. The isolator output sends information to the PWM. If VOUT is low then the "R" pin on the TL431 will be low. This causes the "K" pin to be high and little current will pass though U1. This tells the PWM to work harder. If Vout is too high, K is low passing large current in to U1. This tells the PWM to work less.
View attachment 207187

ok thanks a lot, u have really put me at ease.
while i was still researching i came across a compensation circuit and a current sensor still in the feedback loop, i was confuse because i've never seen that before in any smps i've explored. So pls can you explain that a bit please?

 

i came across a compensation circuit and a current sensor still in the feedback loop

More information please. Schematic?

Compensation: A power supply of any kind can oscillate. It is important to have a very high gain error amplifier at DC or low frequencies, but high gain at 1khz will cause the supply to oscillate. (phase shift oscillator)
If you want 10.000V not 10.001V or 9.999V you need a amplifier that can see 0.001V and amplify it. But a gain of millions at 10khz will drive the power supply crazy. Look at the amp in post #16. There is a capacitor from input to out put. At low frequencies the capacitor is "open" and allows the amp to have high gain. At high frequencies the cap reduces the gain. This next graph shows the gain of an error amp verses frequencies. My guess is that this supply wants to oscillate at 10khz. Red line is gain. The amplifier has a max gain of 63db. The gain gets smaller at high frequencies. A little complicated and lots of math.

Here is a isolated power supply. If it was a dc to dc supply with no isolation then every thing in the blue box can be removed and the error amp in the green box would do the job.

 

More information please. Schematic?

Compensation: A power supply of any kind can oscillate. It is important to have a very high gain error amplifier at DC or low frequencies, but high gain at 1khz will cause the supply to oscillate. (phase shift oscillator)
If you want 10.000V not 10.001V or 9.999V you need a amplifier that can see 0.001V and amplify it. But a gain of millions at 10khz will drive the power supply crazy. Look at the amp in post #16. There is a capacitor from input to out put. At low frequencies the capacitor is "open" and allows the amp to have high gain. At high frequencies the cap reduces the gain. This next graph shows the gain of an error amp verses frequencies. My guess is that this supply wants to oscillate at 10khz. Red line is gain. The amplifier has a max gain of 63db. The gain gets smaller at high frequencies. A little complicated and lots of math.
View attachment 207289
Here is a isolated power supply. If it was a dc to dc supply with no isolation then every thing in the blue box can be removed and the error amp in the green box would do the job.
View attachment 207292

Ok thank you very much for bringing the circuit as well.
Hmm, I really have a long way to go !!!.
But should I understand all these things before I can successfully build a well standardized smps circuit?

 

More information please. Schematic?

Compensation: A power supply of any kind can oscillate. It is important to have a very high gain error amplifier at DC or low frequencies, but high gain at 1khz will cause the supply to oscillate. (phase shift oscillator)
If you want 10.000V not 10.001V or 9.999V you need a amplifier that can see 0.001V and amplify it. But a gain of millions at 10khz will drive the power supply crazy. Look at the amp in post #16. There is a capacitor from input to out put. At low frequencies the capacitor is "open" and allows the amp to have high gain. At high frequencies the cap reduces the gain. This next graph shows the gain of an error amp verses frequencies. My guess is that this supply wants to oscillate at 10khz. Red line is gain. The amplifier has a max gain of 63db. The gain gets smaller at high frequencies. A little complicated and lots of math.
View attachment 207289
Here is a isolated power supply. If it was a dc to dc supply with no isolation then every thing in the blue
box can be removed and the error amp in the green box would do the job.
View attachment 207292

Ok so the capacitors acts as the compensation circuit or ...