I am trying to design a DC-DC switching mode power supply to step down 10V to 6V and drive a 2A current through a motor (145mH load with 100 ohm series resistance). The attached file shows the schematic in Cadence ORCAD Capture 16.6. The present setup is giving me 200mA of current and I need to get more current. Presently I am using PWM with a period of 600usec and I think increasing the period will give me more charging time on the inductor and thus get a 2A current. However, the node voltage drops when I do this and so I boost the inductance values to around 2H each. Would you think that this is an advisable design or could you suggest any changes? Any help will be appreciated.
Here is the first file on simulation results showing the output node voltage.
Here is the second file on simulation results showing the load current.
Here's is the final design that gives me the desired output. The motor load is 5.6mH with series resistance of 1.65 ohms.
Simulations on my final design for the voltage converter. I am getting a 61% efficiency on this.
I was able to improve the efficiency of the converter to close to 90% after I added more N and P channel devices in parallel and increased the current drive. In the above design, the MOSFETs are operating above their maximum safe limits and so their resistance is high and the efficiency is low.
Moreover, I have found that these devices have been discontinued and so I need to use some other MOSFETs with similar characteristics now. I will logon to element14.com and find what's available.
Nirav, the inductor you are using in the Buck converter is incredibly big (150 mH). Also, the frequency you are using is very low (1 KHz). May I ask why you are using a hand-made converter and not use an industrial one like LM2596 ? It is achieved by using 3 parts. These industrial ones work at about 100, 200 KHz and use much smaller inductors like 33 through 100 uH (MICRO-HENRY, instead of mH like yours). LM2596 has a 3A drive capability.
Your Buck converter has a design that resembles a push pull ? and eliminates the CATCH DIODE ?
Have you read this on Wikipedia:
http://en.wikipedia.org/wiki/Buck_converter. This more traditional design works very well.
Also, one more question: Are you really sure that the internal resistance of your motor is 100 Ohms. To pass 2A from a 100 ohm resistance, you would need 200 V ??? And, your motor is only 6V ??? some things are not adding up. I would go back to the drawing board and confirm these numbers again. Also, the voltage levels I see on Orcad is in the microvolt range, which is a direct consequence of the numbers not adding up ...
Dear Sir,
Firstly I was not able to find any ICs with high current drive capabilities so I designed this power supply. I will look at the ICs that you have mentioned.
The frequency is a little low and that is an error in the simulation. I was planning to use the PWM output of Arduino Uno to generate the pulse waveform and this works at 16 MHz and thus the clock frequency will be 0.0625 micro seconds = 62.5 ms.
The internal resistance of the motor is 0.015 ohms and not 100 ohms and this is just the resistance of the motor winding. I don't think I mentioned 100 ohms anywhere.
The inductor is 150 mH to match the inductance on the motor. So the inductor on the supply charges to a 2A current and then in the off cycle this inductor provides a constant current. For the inductor to provide a near constant current to 150mH load for a 62.5 ms period, such a high value will be necessary.
The capacitor just achieves low pass filtering of the output signal to generate a constant DC voltage. Moreover, the supply is meant to drive a stepper motor, where each of the 4 windings will be powered one after the other. The windings will have to charge up to 2A current in a very short time for the motor to function smoothly. For this it is necessary that peak currents at charging are significantly higher. So the components need to be designed for a current rating of 10-20 A.
I hope this answered your questions satisfactorily.
As far as the diode is concerned, there is an equivalent circuit of the Buck Converter on the Wikipedia page you mentioned, where in there are 2 switches: one connects the inductor to the power supply and the other to ground. So my circuit is closer to that version of the Buck converter the first version you mentioned.
Nirav, your circuit diagram is showing 100 ohms, this is where I got 100 ohms.
Also, you said : 0.0625 micro seconds = 62.5 ms. ... you mean 62.5 ns (nano).
using transistors instead of diodes is totally fine, but, you have to have excellent control of the MOSFETs, otherwise, they will actually be less efficient. Also, it makes timing more challenging. The CATCH DIODE allows you to ignore that part of the circuit
Dear Sir,
My initial circuit was wrong and my simulations were not working and the 100 ohm resistor were one of the problems. I have shared my updated circuit in one my comments towards the end. This one works fine and I am not having any trouble with the simulations now.
Thank you for taking the time out to read my questions.
Regards--
Nirav Desai
Here is the updated circuit for your reference: This seems to be working fine as you can see in the simulations posted in the subsequent comment.
https://www.researchgate.net/file.PostFileLoader.html?id=520d789ad039b1a91051eaf7&key=72e7e520d789a2a936
Dear Sir,
Also the first simulation had convergence difficulties and I that is why I had to put in the high resistance values.
I later increased the error margins on the simulation and so removed the large resistance.
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