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I have to build the circuit shown below as a homework, it is a power supply with frequent energy switching; but the schematic is incomplete... I should add another circuit which controls the voltage polarity converter (Four transistors). The circuit generates high frequency current pulses.

enter image description here

Anyone have an idea?

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  • \$\begingroup\$ @CamilStaps Actually I didn't know what to do! Where to start! I have no idea about the "voltage polarity converter" and its controlling circuit. Sorry about any inconvenience, but I will appreciate any though would explain the thing. \$\endgroup\$ – Siraj Muhammad May 16 '13 at 18:41
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    \$\begingroup\$ @SirajMuhammad What objectives do you have to reach in this homework? (This is more like a practicum assignment since you have to physically build the device.) \$\endgroup\$ – Nick Alexeev May 16 '13 at 18:48
  • \$\begingroup\$ @NickAlexeev Yes, it is practicum assignment. But the schematic is not complete, I need that controlling circuit. \$\endgroup\$ – Siraj Muhammad May 16 '13 at 18:55
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    \$\begingroup\$ Are you familiar with the H-bridge? \$\endgroup\$ – jippie May 16 '13 at 19:10
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    \$\begingroup\$ @pjc50 50 or 60Hz transformer would be much larger than a high-frequency transformer (say, 100kHz). The circuit in the question has more moving parts, but it's smaller and cheaper than having a low frequency transformer. \$\endgroup\$ – Nick Alexeev May 16 '13 at 19:24
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The starting point I would take with this circuit is to break the diagram down into functional blocks (or sub systems). Reading the circuit from INPUT -->--OUTPUT (Left to Right as drawn)

Block 1.

On the left hand side we have 4 DIODES arranged in a conventional (full wave) bridge rectifier configuration converting the AC input into (bumpy) DC. C1 is the smoothing capacitor that gets rid of the bumps (hopefully) and R is either a real or 'implied' circuit resistance to limit the initial charging current to the capacitor. The output across C1 is (or should be) smooth DC. ~ (charge time constant(Seconds) = R(ohms) x C(Farads)) - so far this is a very standard circuit and requires no further changes.

Block 2

The four NPN transistors (T1 ->- T4) and the PRIMARY COIL of the transformer.

The transistors are basically switched PAIRS.

When T1 and T4 are turned ON a current will flow through the PRIMARY COIL (N1) so that the TOP connection is POSITIVE and the BOTTOM connection is NEGATIVE - i.e. conventional current flow is TOP(+) to BOTTOM(-).

When T2 and T3 are switched ON the BOTTOM of the coil becomes POSITIVE and the TOP connection becomes NEGATIVE. i.e. current flow is BOTTOM(+) to TOP(-) thus producing and Alternating Current (AC) through the primary coil by switch the direction of the DC.

What is MISSING from this block is the electronic control system to switch each pair of transistors ON and OFF ensuring that AT NO TIME are ALL the transistors turned on (nasty things will happen and the magic blue smoke that makes all electronic devices work will escape and never work again).

Block 3

The split SECONDARY of the transformer with 2 DIODES, an INDUCTOR (L) and a smoothing CAPACITOR C2.

This is a standard circuit for FULLWAVE rectification and smoothing from a centre tapped transformer. It does not require any addition or modification. The inductor is very good for taking out spikes that will be generated by the switching action of the transistors.

Towards a solution.

Having analyzed the circuit in this way it is only BLOCK 2 that needs additional circuitry.

In the simplest case this would consist of a SQUAREWAVE oscillator (50% duty cycle) with COMPLIMENTARY outputs. e.g a simple 555 astable followed by a 'divide by 2' circuit such as a JK or D type flip flop. The 555 would run at TWICE the switching frequency. . The divider will set the duty cycle to 1:1 (ON to OFF)

I would also add FOUR snubber diodes across the transistors (C-E) (diodes connected the 'wrong' way to prevent transistor damage from back emf when the current though the primary inductor is turned OFF).

From the 'Q' output of the divider connect the bases of T1 and T4 through separate resistors. From the 'NOT Q' output connect the bases of T2 and T3 via separate resistors.

This is a very theoretical and simplistic circuit solution - No consideration is given to properly interfacing control signals , signal timing, type, feedback control etc. The switching input is just a simple square wave.

EDIT 1 (additional information)

Protecting the driver transistors from back e.m.f. when switching an inductive load.

Switching the load ON is not a problem as the inductor will cause the current to rise slowly (L/R time constant). The problem happens when the transistor is turned OFF - the inductor's magnetic field collapses and induces a very high (E proportional to dB/dT) voltage in the reverse direction (back e.m.f). The result is that this voltage appears across the transistor in the wrong direction and promptly destroys it. A diode connected across the transistor in the reverse direction to the normal flow of current will act as a short circuit and limit the reverse voltage across the transistor to about 0.7V.

enter image description here

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  • \$\begingroup\$ Big thanks to you! Great explanation, actually it took me the whole day to understand the diagrams, but you made it easy. One last thing, the square wave oscillator which will control the transistors, should it be some special kind of oscillators as we are dealing with power BJTs? Thank you a lot, would you mind to have a look on the final circuit after I build it? \$\endgroup\$ – Siraj Muhammad May 17 '13 at 5:52
  • \$\begingroup\$ There is a lot of information missing on the diagram such as the voltages/currents/ frequency used/required. There is no mention of the BJT type e.g. 2N3055, Tip42A etc. In order to design and build the circuit for real you'll need to know a lot more. The oscillator circuit is nothing special. You could do the same thing with a two transistor multivibrator and have a variable resistor to give you a 1:1 mark/space ratio. Interfacing with the BJTs is the tricky part. Usually these devices have very low current gain (10-20). Also the top transistors are acting as SOURCEs and the bottom as SINKs \$\endgroup\$ – JIm Dearden May 17 '13 at 12:45
  • \$\begingroup\$ Actually I have made the required calculations... and chose components values. I will post the final circuit in a minutes. \$\endgroup\$ – Siraj Muhammad May 17 '13 at 13:03
  • \$\begingroup\$ Here's what I figured out... however, I didn't get nice number on the multimeter.. it shows values of nano volts, keeps increasing to millivolts. What's wrong with that? s2.postimg.org/6yucocgc9/image.jpg \$\endgroup\$ – Siraj Muhammad May 17 '13 at 13:28
  • \$\begingroup\$ the basic circuit FUNCTION is correct but the DRIVE (Voltage/Current) is totally inadequate. The voltage across the capacitor C1 is about 300V (1.4 x 220). The Tip50 is rated at 400V (OK) and 1 Amp (OK) current gain about 10. You've chosen a 7400 series chip(5V) driven from the 12V output of a 555.Wrong! The output from the 7474 is 0 - 5V, its current capability is a few mA. Neither of which is sufficient to drive the Tip50's. First step is just concentrate on the oscillator and divider 12V section using a 555 astable running at say 2KHz followed by a CMOS divider (all at 12V). \$\endgroup\$ – JIm Dearden May 17 '13 at 20:11
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What you have is a full bridge converter in power supply parlance.

T1 and T4 are driven ON together, followed by T2 and T3, exciting the primary winding with opposing polarity. The secondary is a simple center-tap rectifier which along with the LC filter gives you DC output.

In continuous mode, the output is a function of the input voltage, the transformer turns ratio and the duty cycle at which you drive the bridge:

\$V_{out} = \dfrac{N_s}{N_p} \cdot V_{in} \cdot D \$

where \$N_s\$ is the number of secondary turns per winding section (half of the total turns), \$N_p\$ is the number of primary turns and \$D\$ is the duty cycle.

For example, at 400V input, 14:1:1 turns ratio and duty cycle of 45%,

\$V_{out} = \dfrac{1}{14} \cdot 400V \cdot 0.45 = 12.857V\$ (assuming ideal diodes, etc.)

This topology is controlled by pulse width modulation - an error voltage in the secondary is compared with a reference voltage, which yields a change in duty cycle that tries to zero out the error voltage. There are many off-the-shelf control ICs that implement a complete full bridge converter control scheme.

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