0
\$\begingroup\$

I tried to make a PMDC motor rotate through a step down transformer. Obviously I needed DC. So, I added a 1N4007 general purpose diode in two ways to have a half wave rectified input for the motor.

Case-1:-The motor rotates.

Case-2:-The motor pulsates (vibrates in to and fro motion) as if there is AC input across the motor.

Case-2 does not make sense to me as in both cases the diode rectifies the power supply. enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ Because in Case (2) there IS no DC component, \$\endgroup\$ – Brian Drummond Aug 26 '16 at 17:47
  • \$\begingroup\$ Case:-1 does not have it either. Both outputs have alternation. \$\endgroup\$ – Dwiparna Datta Aug 26 '16 at 17:50
  • 3
    \$\begingroup\$ Case 1 most certainly does have a DC component, unlike Case 2. \$\endgroup\$ – Brian Drummond Aug 26 '16 at 17:54
3
\$\begingroup\$

You can feed whatever you want into a transformer but you'll never get dc out without an output rectifier.

Transformers work because the alternating voltage on the input creates a magnetic field (\$\Phi\$) that is also alternating in the core. The induced emf in the secondary is N\$\dfrac{d\Phi}{dt}\$ where N is number of turns.

All the input rectifier does is force an average current through the primary which will saturate the core and likely cause the 1N4007 to fail.

The half wave primary voltage does create a rate of change of flux (ignoring saturation issues) that will be seen on the secondary as an AC waveform with no DC content.

\$\endgroup\$
  • \$\begingroup\$ Yes, I can't get DC output from a transformer if I feed DC. Actually no o/p at all. But, I can get a half wave rectified output if I feed the same to it. All the transformer is interested in is the alternation in input voltage. Half wave rectified signal has alternation and it is not pure DC. \$\endgroup\$ – Dwiparna Datta Aug 26 '16 at 17:30
  • 7
    \$\begingroup\$ The DC content of the input waveform is not passed to the secondary winding - only the AC content and this means AC out with ZERO (repeat zero) DC content. You are also likely to destroy your transformer with a rectified input because the core will saturate and get extremely hot. \$\endgroup\$ – Andy aka Aug 26 '16 at 17:34
  • \$\begingroup\$ @Andy aka - Why would the core saturate & get extremely hot? It's the same AC in whether it sees half or full phases, right? Or, would the ripple DC cause that effect in the core? If so, how? \$\endgroup\$ – zeffur Aug 26 '16 at 18:28
  • 1
    \$\begingroup\$ You are only passing current in one direction into the core - look up "walk into core saturation". \$\endgroup\$ – Andy aka Aug 26 '16 at 18:53
  • 1
    \$\begingroup\$ @DwiparnaDatta Half-wave rectified AC does not have alternation. It is pulsating DC, not AC, and feeding it into a transformer will not work. In fact, it may destroy the transformer. \$\endgroup\$ – marcelm Aug 26 '16 at 20:57
4
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's rectified mains input to transformer.

But, I can get a half wave rectified output if I feed the same to it. [From your comment to Andy aka.]

No you can't. You might get the same waveform (probably with distortion) but since the output can't have DC then the average voltage must be zero.

enter image description here

Figure 2. Top waveform is DC feed into transformer. Bottom waveform is transformer output.

Since there is no DC component in the output your DC motor can not run.


enter image description here

Figure 3. Note that even if you feed in a sinewave with a DC bias that the output would be a sinewave (provided you hadn't saturated the transformer).

\$\endgroup\$
  • \$\begingroup\$ Will you send me a screenshot of the result you got from the CircuitLAb simulation? I believe I am failing to do it properly. \$\endgroup\$ – Dwiparna Datta Aug 26 '16 at 18:29
  • 1
    \$\begingroup\$ I didn't run a simulation. \$\endgroup\$ – Transistor Aug 26 '16 at 18:37
0
\$\begingroup\$

Both cases are wrong anyway. Your rectifier, at least should be like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ This two-diode connection keeps the motor's rotor from demagnetizing during the 'inert' half of the AC cycle, so (at some motor rotation speeds) may be essential for smooth motion. \$\endgroup\$ – Whit3rd Aug 27 '16 at 7:08
  • \$\begingroup\$ @Whit3rd Not just this. The motot winding has its inductivity, therfore at the begining of the second (blank) half cycle, the current is the same as at the end of first half cycle, so the current makes a loop trough a transformer. A parallel diode gives a free path to the motor decaying current. \$\endgroup\$ – Marko Buršič Aug 27 '16 at 7:16
-2
\$\begingroup\$

Case 1: The transformer transfers both halves (phases) of the induced AC into the secondary winding. The rectifier blocks 1/2 of the AC out of the secondary winding & the motor works on the available ripple DC, but it isn't very powerfully driven because you are only providing it with half of the available power. Replace your rectifier with a full wave bridge rectifier to increase the power output to your motor.

Case 2: The rectifier blocks half of the AC at the input. This means only half of the available energy will be transferred into the secondary winding when the magnetic field collapses with each AC transition. it also means there will be less than half of the energy concentrating effect in the transformer core that would be present if a full AC wave was transferred though it--which also results in less power available to your motor.

\$\endgroup\$
  • \$\begingroup\$ Your last sentence reads as though some DC will be present on the output. Is this what you intended? \$\endgroup\$ – Transistor Aug 26 '16 at 18:43
  • \$\begingroup\$ I haven't tested it, but there should be a ripple DC on the secondary winding. I imagine it would be weaker than case 1. Do you disagree -- if so, why so? :) \$\endgroup\$ – zeffur Aug 26 '16 at 18:51
  • \$\begingroup\$ See my answer where I explain why. \$\endgroup\$ – Transistor Aug 26 '16 at 18:57
  • 3
    \$\begingroup\$ Your answer (case 2) is fundamentally wrong. -1 \$\endgroup\$ – Andy aka Aug 26 '16 at 18:58
  • 1
    \$\begingroup\$ @zeffur: "How is that mostly positive ripple on the output secondary different from half-wave ripple DC ..." That's the point. I can't be DC out. It is AC with an average voltage of zero. Half-wave ripple DC actually has DC and, usually, is all above the 0 V line. I positioned the zero line slightly below half peak-peak voltage to show that the part above the line will balance the part below the line. \$\endgroup\$ – Transistor Aug 26 '16 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.