Transformer isolation drive circuit question

大小家伙好

Transformer isolation drive circuit question [Copy link]

 

The circuit is shown in the figure below, the driving frequency is 28kHz

During the test, it was found that at the primary side of the transformer (T1B), when the duty cycle was not close to 50%, the upper and lower amplitudes of the primary waveform measured with an oscilloscope were asymmetrical;

When the duty cycle is 48%

When the duty cycle is 15%

When the duty cycle is 8%

Measure the waveforms of pin 6 and pin 10 of transformer T1 to ground respectively.

Under different duty cycles, the waveform of pin 10 to ground is a square wave with an amplitude of 23V and a minimum value of 0V;

At different duty cycles, the amplitude of pin 6 is 23V, and the minimum value is a square wave of 23V~0V. The waveform of pin 6 of the transformer to ground is shown in the figure below.

Duty cycle 0%

When the duty cycle is 8%

When the duty cycle is 20%

When the duty cycle is 35%

Try replacing T1B with a 100k resistor, the waveform is normal, as shown below

When the duty cycle is 45%

When the duty cycle is 15%

Tried changing C17 to 10uF, but the waveform did not change

Remove diodes D9 and D11, and the waveform of transformer pin 10 to ground remains unchanged, but the waveform of pin 6 to ground changes (the peak-to-peak value is constant at about 20V, and the waveform gradually shifts downward as the duty cycle increases. When the duty cycle is 45%, the amplitude is 20V and the minimum value is 0V)

Question:

1. Why is the waveform of pin 6 to ground different from that of pin 10 to ground?

2. How should this circuit be modified so that it can work properly?

PowerAnts

The transformer cannot transmit DC components and needs to perform "DC component recovery" on the secondary side.

maychang

1. Why is the waveform of pin 6 to ground different from that of pin 10 to ground?

The waveforms of the two feet to ground should be different. The difference between the waveforms of the two feet to ground is the voltage across T1B.

maychang

2. How should this circuit be modified so that it can work properly?

This circuit works perfectly fine and conforms to circuit theory.

I don't know what kind of waveform you want to achieve at both ends of the transformer to be considered "normal".

maychang

Note that in the first three waveforms, although the amplitudes are different, the waveform areas on the upper and lower axes are the same. This is exactly the manifestation of the inductor "not being able to transmit DC components". For example, the second waveform (see below).

maychang

The red and green parts of the second waveform are the same area.

The same is true for the first and third waveforms.

大小家伙好

PowerAnts posted on 2019-8-29 16:32 The transformer cannot transmit the DC component and needs to perform "DC component recovery" on the secondary side

Why can't capacitor C17 remove the DC component?

大小家伙好

maychang posted on 2019-8-29 16:42 2. How can this circuit be changed to work properly? This circuit works completely normally and conforms to circuit theory. I don't know if you want to change the two ends of the transformer...

I thought that according to this circuit, the waveforms at both ends of the transformer should be symmetrical in upper and lower amplitudes.

Why does capacitor C17 fail to isolate the DC component?

The waveform areas on the upper and lower axes are the same, the volt-seconds are balanced, and when the duty cycle is not 50%, the charging time of the inductor is inconsistent with the discharging time, and the amplitude is different. How can it be shown that the inductor "cannot transmit DC components"?

大小家伙好

Posted by big and small guy on 2019-8-29 17:10 maychang posted on 2019-8-29 16:42 2. How to change this circuit to work properly? This circuit works completely normally and meets the circuit...

After thinking about it, I realized that the upper and lower amplitudes are different because the charging and discharging times are different, but how can this reflect that the inductor "cannot transmit DC components"? I didn't understand.

maychang

Posted by big and small guy on 2019-8-29 17:10 maychang posted on 2019-8-29 16:42 2. How to change this circuit to work properly? This circuit works completely normally and meets the circuit...

"Why doesn't capacitor C17 play the role of isolating the DC component?"

Capacitor C17 has played the role of "isolating DC components".

maychang

Posted by big and small guy on 2019-8-29 17:10 maychang posted on 2019-8-29 16:42 2. How to change this circuit to work properly? This circuit works completely normally and meets the circuit...

The waveform areas on the upper and lower axes are the same, the volt-seconds are balanced, and when the duty cycle is not 50%, the charging time of the inductor is inconsistent with the discharging time, and the amplitude is different. How can it be shown that the inductor "cannot transmit DC components"?

Precisely because the amplitude is large when the time is short and small when the time is long, the product of duration and amplitude, the so-called volt-second product (note that this product is the DC component), remains unchanged when the duty cycle changes, and the positive and negative DC components just cancel out to zero, which shows that the inductor "cannot transmit DC components".

PowerAnts

Posted by Big and Small Guy on 2019-8-29 17:05 Why can't capacitor C17 remove the DC component?

What do you do to remove the DC component? What you should do is to restore the DC component on the secondary side of the drive transformer, and then provide the MOSFET or IGBT with an isolated, unipolar drive pulse whose amplitude does not vary with the duty cycle.

大小家伙好

Thank you both for clarifying my doubts!

maychang

Posted by Big and Small Guy on 2019-8-29 17:31 Thank you both for clarifying my doubts!

I don't know what kind of load is connected to the secondary of your transformer. If it is driving a single-ended MOS tube or IGBT, the duty cycle cannot change too much, because the amplitude of the positive and negative pulses also changes when the duty cycle changes. If the amplitude of the positive pulse is too small, the MOS tube or IGBT cannot be turned on. If the amplitude of the positive or negative pulse is too large, the gate of the MOS tube or IGBT will be broken down.

But the transformer drive push-pull, half-bridge, full-bridge MOS tube or IGBT tube is OK, but the first post circuit PWM_R and PWM_L must have dead zone.

PowerAnts

maychang posted on 2019-9-3 11:43 I don't know what kind of load is connected to the secondary of your transformer. If it is driving a single-ended MOS tube or IGBT, then the duty cycle cannot change too much, because the duty cycle...

The DC component recovery circuit of the PWM isolated drive has a measured duty cycle of 5-95% and a small amplitude change.

zhuyebb

15. Thank you for sharing

maychang

PowerAnts published on 2019-9-3 12:53 The DC component recovery circuit of PWM isolation drive, the measured duty cycle is 5-95%, and the amplitude changes very little

If there is a DC component recovery circuit, the 14th floor "when driving a single-ended MOS tube or IGBT, the duty cycle change cannot be too large" is naturally another matter.

大小家伙好

maychang posted on 2019-9-3 11:43 I don't know what kind of load is connected to the secondary of your transformer. If it is driving a single-ended MOS tube or IGBT, then the duty cycle cannot change too much, because the duty cycle...

The secondary is connected to IGBT, half bridge

The actual half-bridge bridge drive is shown in the figure below. The original intention was to use the method shown in the figure below to obtain a complementary square wave drive half-bridge.

During the test, the driving duty cycle was 48%, several diodes were connected in series with +20V, and the measured driving positive and negative amplitudes were ±16V.

The circuit can work normally. The next step is to consider using a transformer to isolate the upper and lower bridge arms respectively, because the duty cycle needs to be adjustable.

In the picture on the 15th floor, if there is a series capacitor on the primary side of the transformer, can the capacitor in the picture on the 15th floor be removed?

maychang

Posted by Big and Small Guy on 2019-9-3 14:22 The secondary is connected to IGBT, and the actual half-bridge bridge drive is shown in the figure below. It was originally intended to use the method shown in the figure below to obtain a complementary square wave drive half-bridge. ...

"The next step is to consider using a transformer to isolate and drive the upper and lower bridge arms respectively, because the duty cycle needs to be adjustable."

If a transformer is used for each upper and lower bridge arm, then both upper and lower tubes need to add the DC component recovery circuit mentioned by PowerAnts on the 15th floor.

maychang

Posted by Big and Small Guy on 2019-9-3 14:22 The secondary is connected to IGBT, and the actual half-bridge bridge drive is shown in the figure below. It was originally intended to use the method shown in the figure below to obtain a complementary square wave drive half-bridge. ...

In order to make the duty cycle adjustable and the duty cycle variation range quite large, for the half-bridge, it is just right for the transformer to have two secondaries.

I will tell you the specific waveform after I draw it.