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How to get a constant voltage output from a dc to dc converter?
- Thread starter danish
- Start date
Hello danish, welcome to our forum.
Please show us a schematic of your circuit. Otherwise we cannot help you as there are so many ways to design a DC/DC converter. Also give us some details as for example:
- required output voltage and allowed tolerance
- required output current
- allowed range of input voltage
- whatever parameters are available
Please show us a schematic of your circuit. Otherwise we cannot help you as there are so many ways to design a DC/DC converter. Also give us some details as for example:
- required output voltage and allowed tolerance
- required output current
- allowed range of input voltage
- whatever parameters are available
Danish, your question could mean two things.
If you mean that the output voltage varies under load then the answer is that you need feedback to tell the dc to dc converter to stop when the voltage reaches the desired value and start up again when it drops.
If you mean that the output has too much ripple then you need additional filtering
If you mean that the output voltage varies under load then the answer is that you need feedback to tell the dc to dc converter to stop when the voltage reaches the desired value and start up again when it drops.
If you mean that the output has too much ripple then you need additional filtering
here is the schematic for the circuit that i have designed in proteus.
Required output voltage is 40v and output current should be 100mA.
Attachments
hi Steve
I have designed a circuit that i have uploaded now and this circuit has output voltage upto 30v. But this output voltage has ripples and it is not constant as it goes down to 3V after some time
KrisBlueNZ
Sadly passed away in 2015
As Steve suggested in post #3, you need to add circuitry to monitor the output voltage and adjust the duty cycle of the control signal feeding the transistor so that the output voltage remains constant.
There are various ICs available that will do this - for example the UC3842/3/4/5 series, and others that have the switching transistor built in.
What is the purpose of this whole project? Do you want to learn more about it or do you have a specific application in mind?
There are various ICs available that will do this - for example the UC3842/3/4/5 series, and others that have the switching transistor built in.
What is the purpose of this whole project? Do you want to learn more about it or do you have a specific application in mind?
Firstly I am looking for the schematic of this circuit. Although I have tried to desing by my own but this circuit (that I have uploaded) needs some changes in order to get a constant 40V output.
KrisBlueNZ
Sadly passed away in 2015
So what do you want to do?
I want to know what is the problem in my circuit. what changes do I need in this circuit. Is there another way to design the circuit in order to get the desired output.
KrisBlueNZ
Sadly passed away in 2015
The problem is that the amount of energy you inject into the inductor on each switching cycle is constant. It is determined by the input voltage and the ON time of the transistor. Each time the transistor turns OFF, this energy is dumped into the output circuit. This causes a constant amount of power to be transferred to the output.
Power is equal to voltage × current. If your load doesn't draw much current, the voltage will be high. (Low current × high voltage = fixed power.) If your load draws a lot of current, the voltage will be low. (High current × low voltage = fixed power.)
You need to add components that monitor the output voltage, and adjust the duty cycle of the signal that is driving the base of the transistor, so that the right amount of energy is injected into the inductor to give the output voltage you want.
As the load current varies, the amount of energy you need will vary as well. If the load is not drawing much current, you need to reduce the amount of energy injected into the inductor on each switching cycle, otherwise the output voltage will rise. This is done by decreasing the ON time of the transistor, by decreasing the duty cycle of the driving waveform.
If the load is drawing a lot of current, you need to increase the amount of current injected into the inductor, otherwise the output voltage will drop; you need to decrease the duty cycle of the drive waveform.
There are various ways to do this. A good way is implemented in the UC3843 current-mode controller IC which is available cheaply from Digi-Key and other distributors. It is made by several manufacturers. The data sheets are:
http://www.ti.com/lit/ds/symlink/uc2845a.pdf - Texas Instruments / Unitrode (the original designer of these devices)
http://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/UC3842B.pdf - ON Semiconductor
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000966.pdf - STMicroelectronics
http://media.digikey.com/pdf/Data Sheets/Fairchild PDFs/UC3842A,UC3843A.pdf - Fairchild
The UC3843 is designed to drive a MOSFET, not a bipolar transistor. It is also normally used in isolated flyback converters, but can be used in this circuit.
Power is equal to voltage × current. If your load doesn't draw much current, the voltage will be high. (Low current × high voltage = fixed power.) If your load draws a lot of current, the voltage will be low. (High current × low voltage = fixed power.)
You need to add components that monitor the output voltage, and adjust the duty cycle of the signal that is driving the base of the transistor, so that the right amount of energy is injected into the inductor to give the output voltage you want.
As the load current varies, the amount of energy you need will vary as well. If the load is not drawing much current, you need to reduce the amount of energy injected into the inductor on each switching cycle, otherwise the output voltage will rise. This is done by decreasing the ON time of the transistor, by decreasing the duty cycle of the driving waveform.
If the load is drawing a lot of current, you need to increase the amount of current injected into the inductor, otherwise the output voltage will drop; you need to decrease the duty cycle of the drive waveform.
There are various ways to do this. A good way is implemented in the UC3843 current-mode controller IC which is available cheaply from Digi-Key and other distributors. It is made by several manufacturers. The data sheets are:
http://www.ti.com/lit/ds/symlink/uc2845a.pdf - Texas Instruments / Unitrode (the original designer of these devices)
http://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/UC3842B.pdf - ON Semiconductor
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000966.pdf - STMicroelectronics
http://media.digikey.com/pdf/Data Sheets/Fairchild PDFs/UC3842A,UC3843A.pdf - Fairchild
The UC3843 is designed to drive a MOSFET, not a bipolar transistor. It is also normally used in isolated flyback converters, but can be used in this circuit.
Arouse1973
Adam
As both Kris and Steve have said you need feedback to monitor the output voltage and adjust the duty cycle either way depending on whether the voltage is higher or lower than what you need. Vout=Vin/(1-D) where D is your duty cycle.
Adam
Adam
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