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enter image description here

This question originates from a difficulty to understand fly-back diode current in DC motors.

In the above figure which logic is ok: 1) 9V-2V = 7V, so diode is reverse biased and current will not flow. 2) diode is applied 2 volts forward bias in a loop so current will flow from the source V2

Which logic is valid?

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  • \$\begingroup\$ Can this question please be fixed to match the answers again? \$\endgroup\$ – Scott Seidman Oct 25 '14 at 20:43
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    \$\begingroup\$ instead of posting a simplified but maybe inaccurate translation just post the original setup and let people make their own translation. \$\endgroup\$ – Requist Oct 25 '14 at 21:01
  • \$\begingroup\$ yeah i know i messed it up. that was not in my mind. \$\endgroup\$ – user16307 Oct 25 '14 at 21:03
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The diode will only see the 2 V forward, so the current will flow. The diode is directly connected to the 2 V generator so everything outside doesn't matter (excepted short-circuit and so on of course), if you can measure 2 V at the generator then there is also 2 V at the diode.

But you'll not have 2 V at this point, you'll have the diode drop (0.6 to 0.7 V for silicon and 0.3 for schottky diodes for example) or a fried diode if the current is too high...

EDIT (Since the OP has edited his question my answer just above is no longer valid):

The diode will only see 2 V in reverse (or -2 V in the conventional reading direction of a schematic), so the current will not flow. The diode is directly connected to the 2 V generator so everything outside doesn't matter, if you can measure 2 V at the generator then there is also 2 V at the diode.

So you'll have 9 V accross the whole thing, -2 V accross the diode and +9 + -2 = 7 V accross R1.

EDIT² (answer to the secondary question: Where will the current flow when V1 = 9 V and then V1 = 0 V?)

Since the diode isn't conducting there is only one current path and it will be from V1, through V2, then trough R1, then to GND when V1 = 9 V.

When V1 = 0 V (connected to GND in other words) the path will be the same but the current will flow in the reverse direction (but the diode will still be reverse biased and not conducting, don't forget the diode is still directly connected to the generator V2 so everything outside doesn't matter).

From a different point of view: if you think that V2 is like a battery, when V1 = 9 V you recharge it through R1, and when V1 = 0 V then you discharge it directly through R1.

EDIT³ (new schematic and new secondary question: Where will the current flow when Q1 is closed and then opened?)

With this new schematic when Q1 is closed (let's assume it's a perfect transistor) the path and current direction will be the same as in my edit² when V1 = 9 V.

When Q1 is opened there is no path where the current can flow so there is simply no current flowing.

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  • \$\begingroup\$ if V1 is supply and V2 is back emf of a DC motor, it is said that back emf opposes the supply voltage. but if current flows through the diode(shorts) how can it oppose V1? will current from back emf flow through the diode in that case? \$\endgroup\$ – user16307 Oct 25 '14 at 19:25
  • \$\begingroup\$ It's because the back EMF isn't in this direction, it's opposed to V1 (in your schematic V2 is effectively in series in the same direction as V1), see this article and this video for more information. \$\endgroup\$ – Biduleohm Oct 25 '14 at 19:28
  • \$\begingroup\$ oh ok i updated the figure. here my confusion is this: lets say there is pwm and when 9V exits in ON mode or doesnt exist in OFF mode, where will the current from V2 flow through? thnx for video but they dont show current flows. my problem is icturing where the current flow from V2 (back emf current) \$\endgroup\$ – user16307 Oct 25 '14 at 19:34
  • \$\begingroup\$ The diode is reverse biased. 9V at the cathode, and 7V at the anode. \$\endgroup\$ – krb686 Oct 25 '14 at 19:44
  • \$\begingroup\$ @user16307 Edited my answer, it should be clear now ;) \$\endgroup\$ – Biduleohm Oct 25 '14 at 19:50
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Considering the edited original question, and considering a motor load to be an inductance in series with its winding resistance, as in 1), below:

2) shows the motor connected to a power source with charge flowing through the motor because of the closed switch. Note that with current through the motor, its supply side will be more positive than the switched (ground) side and that the current through the motor will vary depending on the supply voltage, the back EMF generated by the windings dynamically cutting/being cut by a magnetic field, the DC resistance of the windings, and the mechanical load on the motor.

In 3), the switch is abruptly opened causing the charge flowing through the inductance to stop, which will cause the magnetic field to change direction by starting to collapse. When that happens, the energy in the magnetic field will have no place to go, and as the field collapses, the voltage across the inductance will reverse because the field is cutting the inductor's turns backwards, and the voltage across the inductance will rise dramatically in a more or less vain effort to maintain a current through the inductance until an arc is produced across the open switch contacts, "discharging" the inductor.

In 4) we have the same situation as in 2), except that a diode is placed across the inductance so that it's reverse biased by the supply, making it essentially an open circuit with only its leakage current being supplied by the 9 volt source.

In 5, we have the same situation as in 3), except that when the switch opens and the voltage across the inductance changes polarity and starts to rise, the diode will now become forward biased, will start to conduct, and the energy in the inductor's collapsing magnetic field will be dissipated in the diode's forward resistance and the motor's resistance. Notice that since the diode is conducting and its cathode is connected to the 9 volt source, its anode cannot rise higher than one diode drop above the supply, limiting the voltage across the switch to about 10 volts peak during turnoff.

enter image description here

THE PUDDING:

Here are two identical circuits except for the flyback diode:

enter image description here

and here is the LTspice circuit list so you can run a simulation and see what the flyback diode really does.

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  • \$\begingroup\$ I hate to be the bearer of bad news, but I think you're slightly confused. goo.gl/6arjuU \$\endgroup\$ – krb686 Oct 25 '14 at 19:42
  • \$\begingroup\$ @krb686: Oops... I was working on the original question before it got edited. \$\endgroup\$ – EM Fields Oct 25 '14 at 19:53
  • \$\begingroup\$ my confusion is about current loops. but am asking this question maybe 10 times including other posts but no answer. and mentioning where will the "current flow". i write it again: here my confusion is this: lets say there is pwm and when 9V exits in ON mode or doesnt exist in OFF mode, where will the current from V2 flow through? thnx for video but they dont show current flows. my problem is picturing where the current flow from V2 (back emf current).where will the current flow through from V2(back emf) can you tell me the loop? \$\endgroup\$ – user16307 Oct 25 '14 at 20:21
  • \$\begingroup\$ @EMFields sorry the real figure would be here:i.imgur.com/1wRtR6F.png and the quesiton is when Q1 is off where will the current form back emf flow? will back emf same polarizaiton during Q1 is ON and OFF? \$\endgroup\$ – user16307 Oct 25 '14 at 22:14
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    \$\begingroup\$ @user16307 short answer: no current flows when Q1 is off; polarisation is the same (why it would change?), long answer: see my edit ;) \$\endgroup\$ – Biduleohm Oct 25 '14 at 22:42

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