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I am trying to design an isolated dc dc converter. Now, my input is approx 20V and I need an output of 7V at the final output. I understand that I need a PWM controller that is approx 100kHz or so. Now, how do I control the duty cycle of such an IC ? I need 7V from a 20V supply. So approx its 35% duty cycle. Also, I need a transformer for the isolation. How shall I select such a transformer ? I know this is a huge topic(selecting a transformer), but any pointers will be appreciated. Also, will I need a feedback or such ?

schematic

simulate this circuit – Schematic created using CircuitLab The above is a representative schematic of this circuit. Now, regarding the PWM generator, can an IC 555 suffice ? Will it give out sufficient gate drive ? My main concern is regarding the selection of the transformer.

The load is a max of 4Amps.

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closed as too broad by Adam Lawrence, PeterJ, Daniel Grillo, nidhin, Brian Drummond Feb 14 '16 at 22:13

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ "I need 7V from a 20V supply. So approx its 35% duty cycle." That would be valid for a forward converter with a 1:1 transformer. A flyback converter is different: you are storing energy in the transformer mutual inductance when the transistor is on, and releasing that energy to the load when the transistor switches off. Thus the voltage on the secondary side is related not only to the duty cycle but also the current draw of the load. You also need a diode on the output, or it's not a flyback. \$\endgroup\$ – jms Feb 8 '16 at 14:54
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    \$\begingroup\$ There are a number of good books on SMPS design.. I think this question involves far more than could be expected in an answer in this venue. In general you will require an inductor (XFMR1) with an air gap, some kind of feedback, and a lot of other details covered off. \$\endgroup\$ – Spehro Pefhany Feb 8 '16 at 14:56
  • \$\begingroup\$ I recommend you do more reading first, this is a huge topic. \$\endgroup\$ – MadHatter Feb 8 '16 at 14:59
  • \$\begingroup\$ @OlinLathrop. I need an output voltage of 5Amps and 10V. So a wattage of 50W approx. \$\endgroup\$ – Board-Man Feb 8 '16 at 15:19
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    \$\begingroup\$ Voting to close as 'too broad'. You need to gather some fundamental knowledge first, then ask specific questions about specific problems with your design process. Asking how PWM controllers work, how to choose a transformer, and whether or not you need feedback is waaaaay too broad for a single question. \$\endgroup\$ – Adam Lawrence Feb 8 '16 at 15:53
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In the question you say you want 4 A at 7 V (28 W), but in a comment 5 V at 10 A (50 W). That rules out cheap and available POE (power over Ethernet) transformers, which I often use for such things. Flyback transformers at the power levels you want are more scarce.

In any case, here are roughly the steps:

  1. Pick a transformer. Start with the power requirement, then something that has roughly the right input to output voltage ratio.

  2. Read transformer datasheet carefully.

  3. Look at the primary inductance and compute the switch on time to reach saturation with your input voltage.

  4. Look at the secondary inductance and compute the initial output current at the start of the output pulse, and how long that pulse will last given your output voltage.

  5. From the previous two points, you know how long a complete switching cycle should take, and how much energy will be delivered per cycle. Verify that you can still get the power you want with some margin. If not, go back to step 1.

  6. Pick a PWM period and duty cycle that should transfer a bit more power than you really need, but doesn't violate the saturation limits. You can set the on time right at the saturation limit, but it's good to leave a little more off time to make sure the energy in the core is really zero when the next pulse is started. Continuous mode is possible in a flyback converter, but it gets tricky. If you have to ask basics here, don't try to do continuous mode.

A trick I often use at this point is to set up a PWM output of a microcontroller to produce the switching pulses. If left open loop, this would make a little more voltage than you need under full load.

The trick is to use a shutdown input of the PWM module. Many micros have these. On the isolated side, drive a opto-coupler when the voltage hits the regulation threshold. On the primary side, this activates the shutdown input of the PWM. Make sure the PWM recovers when the shutdown signal is no longer asserted. Basically, that kills the oscillations when the voltage gets to where you want it, then resumes them when it drops below the threshold. This all happens in hardware with 0 CPU intervention after initialization.

This method results in a little more ripple than something that carefully and smoothly adjusts the PWM duty cycle in a control loop, but its a lot simpler and very robust.

I'm doing this in one of my current projects. In this case I need isolated 5 V for some communication interfaces. I use a PNP transistor around a 5 V LDO to sense when the LDO input is the B-E drop above its output. 700 mV or so is a nice comfortable headroom for this LDO. The transistor ultimately drives the feedback opto, which then shuts down pumping power into the isolated section. There is about 100 mVpp ripple on this input to the LDO, but the output is quite flat.

Here is a snippet of a schematic that implements what I described above:

Q5 is the switch, which is driven from the PWM output of the micro via a FET gate driver. The basic power supply on the isolated side is the secondaries of TR3, D6, and C23.

IC7 is the LDO that makes the clean and regulated 5 V power. Q6 is connected in such a way that it turns on when the input of the LDO gets to about 700 mV above its output. That in turn turns on opto IC8, which activates the shutdown input to the PWM, which stops dumping power to the isolated section. Eventually the voltage drops to where the LED in the opto is no longer on enough to keep the PWM off, and power is again transferred to the isolated section.

The input of the LDO has about 100 mVpp ripple on it, which is cleans up nicely to make its regulated output of 5 V.

Note that you can sometimes use the rougher 5.7 V supply directly. In this case I have a couple of LEDs, which actually take more power than most everything else. I have the LEDs connected to their current comes from the 5.7 V supply, thereby requiring less current out of the LDO. The extra 700 mV gets wasted as heat either way, but this way distributes it instead of concentrating it in the LDO.

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  • \$\begingroup\$ This is what's engineering is all about. \$\endgroup\$ – Bip Feb 8 '16 at 17:17
  • \$\begingroup\$ @Olin Lathrop. Regarding the transformer that you selected. How did you zero in on the part that you used. \$\endgroup\$ – Board-Man Feb 9 '16 at 12:23
  • \$\begingroup\$ @Boar: I started on the CoilCraft site, found what I needed, so didn't look any further. \$\endgroup\$ – Olin Lathrop Feb 9 '16 at 12:38

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