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I have a big missunderstanding about how pwm (ground) works. Does input and output of a PWM (including step-down dc2dc) share the same ground ? If so how can i have an input in 12V 50% duty cycle (6V output) that use like 1Amp . Then 6V * 1A = 6W but the 12V input will use 0.5 Amp to have same power .

How can there be common ground if on output i have 1A and input 0.5 Amp. I tried to search google to better understand ground in PWM but everywhere i look it talks about duty cycle and stuff like that.

Can someone make me understand GROUND use in PWM ? (+including step-down regulators that use PWM i think).

Thank you very much.

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  • \$\begingroup\$ It only gets 0.5 A when you either average it out over time in your measurement or smooth it out with a capacitor. \$\endgroup\$ – winny Aug 1 '16 at 8:43
  • \$\begingroup\$ i dont understand , cannot visualize mentaly this . :( but why is there common ground then ? how can i have 1A one side and 0.5A other side if they use common ground ? the amount of electrons dat enter must come out ... if they use common ground then it should be same amps both sides. \$\endgroup\$ – 123onetwothree Aug 1 '16 at 8:45
  • \$\begingroup\$ Instantaneously they don't like you say. It's either 0 or 1 A in your example on both sides. Now, if you have a (large) capacitor on the input side, those 1 A pulses will be averaged out to 0.5 A to the outside. If you measure on the ground trace/wire inside your PWM circuit, you will still see those 1 A pulses. \$\endgroup\$ – winny Aug 1 '16 at 8:50
  • \$\begingroup\$ so i have 1A on INPUT and 1A on output but capacitor somehow smooth 1A on INPUT to 0.5 ? it's so depressing cause I DONT get it . Where can i learn this , what book, site, video, audio i have to listen to FULLY understand internat workings of PWN ground use ? \$\endgroup\$ – 123onetwothree Aug 1 '16 at 8:53
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    \$\begingroup\$ Current flows in loops, remember? The loop with the capacitor in is not the same as the loop with the 12V input. It might help you if you found a simulation where you could monitor the current at arbitary points. \$\endgroup\$ – pjc50 Aug 1 '16 at 9:30
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I see your dilemma. The buck output stage will look something like this: -

enter image description here

For simplicity I'm using the example of a synchronous buck regulator - this is equivalent to both MOSFETs being a kind of high frequency change over switch feeding a low pass filter formed by the L and the C.

With no load connected, do you see that the average current from the incoming supply HAS to be zero amps? This is important to grasp so take some time to ponder this. To help you understand, the inductor is recharged with current when the switch is connected to 12V and that current is discharged to a negative value when the switch conects to ground: -

enter image description here

Blue is the switching voltage fed to the left of the inductor and green is the current waveform through the inductor. It has to be this way because there are no power dissipating components connected to the switching node.

When the switch reconnects to the incoming 12 V, the current flow from the 12 V is negative i.e. energy is being returned to the 12 V source.

That current reduces in magnitude rising through the zero ampere point and reaches a positive maximum. After this point, the supply is disconnected and zero current is taken from that supply. Current (in red below) restarts at a negative value and the process repeats: -

enter image description here

Basically, the incoming supply current (red) has an average value of zero amps. As I said earlier, it MUST be this way because there are no power dissipating components that could force an average DC current from the incoming 12 V.

It's worth noting at this point that the output voltage (for a 50:50 duty cycle) will be 6V with a slight amount of ripple voltage. The ripple is due to the L and the C forming an imperfect low pass filter. A low value could be typically 20 mVp-p superimposed on the DC output of 6V.

In effect, the 50% switching process, the inductor and the capacitor behave like an ideal step-down transformer; 12 V is converted to 6 V and, with no load connected, zero average current is taken from the incoming supply.

When you do connect a 6 ohm load, 1 A DC flows into that load (ignore the slight bit of ripple, minimized to trivial amounts through sensible component selection) so, what happens to the inductor current....

Clearly, 1 amp DC has to flow through that inductor to feed the 6 ohm load and that will mean an inductor current that looks like this: -

enter image description here

The lower half of the picture shows the current taken from the 12 V supply - it has exactly the same current as the inductor when the switch connects it to the inductor but, for the other 50% of the time, that current is zero.

Hence, the average current taken is 0.5 amps.

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  • \$\begingroup\$ wow ... what a nice explanation . thanks a lot man ... i will read this like 20 times now for understanding completly . thanks again \$\endgroup\$ – 123onetwothree Aug 1 '16 at 11:24
  • \$\begingroup\$ can u give an example of an inductor and capacitor value ? to test in a simulator ? \$\endgroup\$ – 123onetwothree Aug 1 '16 at 12:13
  • \$\begingroup\$ Try switching at 100 kHz with a 10 uH and 100 uF. \$\endgroup\$ – Andy aka Aug 1 '16 at 14:08
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Ground use in PWM is the same as in any other circuit. Ground is just a reference point. PWM = Pulse Width Modulation

It is not DC.

If you have 12V in the input, it is switching on and off. Pulses. When it is on, it draws 1A, when it is off, it draws 0A. So it is pulsating, 1A, 0A, 1A, 0A, ... So there is not 1A at 12V DC all time.

When you smooth it by capacitor, you will get 6V output and 1A DC.

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  • \$\begingroup\$ and 0.5 A on input ? this makes me so weird ... i dont understand how does it do that. if u put an AMMETER to input (at the battery) what will i see ? \$\endgroup\$ – 123onetwothree Aug 1 '16 at 9:31
  • \$\begingroup\$ It is 1A, 0A, 1A, 0A... average it, and you'll get 0.5A \$\endgroup\$ – Chupacabras Aug 1 '16 at 9:32
  • \$\begingroup\$ at the output it's averaged . but how about input ... cause i dont get pulses on input , only output . \$\endgroup\$ – 123onetwothree Aug 1 '16 at 9:33
  • \$\begingroup\$ As I said, in the input it is 1A, 0A, 1A, 0A, 1A, 0A.... \$\endgroup\$ – Chupacabras Aug 1 '16 at 9:34
  • \$\begingroup\$ and on the output ? it's not the same like 1A , 0A, 1A, 0A because it pulses on the output \$\endgroup\$ – 123onetwothree Aug 1 '16 at 9:35

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