Hello there,
I'm asking about how to make a cheap 15V floating supply from pre-existing +/- 80V supplies. The output current for the 15V supplies is not required to be much, maybe 100mA or so. I would need four of them.
Right now, the only solution I have is to use 15V DC wall adapters (wall warts), but that requires multiple outlet cords for one amp (I'm trying to use just one power cord for the whole amp, not five).
I've looked at charge pump IC's and the like, but I can't seem to find an IC/regulator/switching IC that provides a completely floating output voltage with minimal parts (I'm trying to avoid transformers and expensive magnetics).
==> Any ideas y'all could offer?
I'm asking about how to make a cheap 15V floating supply from pre-existing +/- 80V supplies. The output current for the 15V supplies is not required to be much, maybe 100mA or so. I would need four of them.
Right now, the only solution I have is to use 15V DC wall adapters (wall warts), but that requires multiple outlet cords for one amp (I'm trying to use just one power cord for the whole amp, not five).
I've looked at charge pump IC's and the like, but I can't seem to find an IC/regulator/switching IC that provides a completely floating output voltage with minimal parts (I'm trying to avoid transformers and expensive magnetics).
==> Any ideas y'all could offer?
Hi!
If you reduce a little the rail voltage from 80V to approx. 70v, by using pass transistors and zeners, you can use Texas Instruments' Power Modules PTB48510C or PT4711 depending on the needed power.
I'm using them within many of my projects / products with rails up to 55V with no problems.
The best of the PT4711 Module is the separate outputs. So I think 2 of these will satisfy your needs.
If you don't want to make a pre-regulation of your rails, you can use a 24V transformer and put it inside your case and use two of these PT4711 modules.
Also, you don´t need to use extra outlet cords, just get the power for this extra transformer at the input of the existing transformer (don't forget to use a fuse)
If you reduce a little the rail voltage from 80V to approx. 70v, by using pass transistors and zeners, you can use Texas Instruments' Power Modules PTB48510C or PT4711 depending on the needed power.
I'm using them within many of my projects / products with rails up to 55V with no problems.
The best of the PT4711 Module is the separate outputs. So I think 2 of these will satisfy your needs.
If you don't want to make a pre-regulation of your rails, you can use a 24V transformer and put it inside your case and use two of these PT4711 modules.
Also, you don´t need to use extra outlet cords, just get the power for this extra transformer at the input of the existing transformer (don't forget to use a fuse)
When ever I need an Isolated extra supply in a curcuit Idea I use these very small Transformers that I have.....
These transformers have a 120v/240v Primary and a 24v,0v,24v secondary (or 12v,0v,12v when useing the 240v Tap with 120v ac power) and are only about an Inch square and output about 150mA of current...I use these with a adjustable regulator and a couple small filter caps to get the desired Voltage for whatever curcuit I need them for and they are so small that they are very easy to fit into existing enclosures and give me the Flexability to get a Variety of Voltages from as low as +/-9v up to +50v or +/-50v if I use 2 transformers......
I got these particular transformers on e-bay for $1 each and they have found use in several of my projects were I needed an extra supply voltage and didn"t have much space.....
Just a thought.....
Cheers
These transformers have a 120v/240v Primary and a 24v,0v,24v secondary (or 12v,0v,12v when useing the 240v Tap with 120v ac power) and are only about an Inch square and output about 150mA of current...I use these with a adjustable regulator and a couple small filter caps to get the desired Voltage for whatever curcuit I need them for and they are so small that they are very easy to fit into existing enclosures and give me the Flexability to get a Variety of Voltages from as low as +/-9v up to +50v or +/-50v if I use 2 transformers......
I got these particular transformers on e-bay for $1 each and they have found use in several of my projects were I needed an extra supply voltage and didn"t have much space.....
Just a thought.....
Cheers
Hello fellow Buffalonian.
I have two 18v 1.6VA dual secondary transformers that would work out well. You would need to regulate the outputs, but regulators are cheap. I will also throw in four 1A bridge rectifiers, one for each secondary. Then all you will need is some prototyping board from radioshack or ebay and a few capacitors.
Here is a link to some cheap 15v regulators on ebay. Shipping is cheap also.
http://cgi.ebay.com/L7815CV-15V-1Am...yZ117000QQrdZ1QQssPageNameZWD1VQQcmdZViewItem
Here is a pic of your free gift.
I have two 18v 1.6VA dual secondary transformers that would work out well. You would need to regulate the outputs, but regulators are cheap. I will also throw in four 1A bridge rectifiers, one for each secondary. Then all you will need is some prototyping board from radioshack or ebay and a few capacitors.
Here is a link to some cheap 15v regulators on ebay. Shipping is cheap also.
http://cgi.ebay.com/L7815CV-15V-1Am...yZ117000QQrdZ1QQssPageNameZWD1VQQcmdZViewItem
Here is a pic of your free gift.
An externally hosted image should be here but it was not working when we last tested it.
Here's what I came up with for a high voltage DC gate drive supply:
http://batee.com/projectsanddesigns...ate_drive/fet_high_side_gate_drive_v_3.5.html
I use a $15 switching power supply to convert +165VDC to +12VDC required by the board shown.
I have built it and tested it from DC to 400KHz+. Works great.
Bryan A. Thompson
bryan@batee.com
http://batee.com/projectsanddesigns...ate_drive/fet_high_side_gate_drive_v_3.5.html
I use a $15 switching power supply to convert +165VDC to +12VDC required by the board shown.
I have built it and tested it from DC to 400KHz+. Works great.
Bryan A. Thompson
bryan@batee.com
Thank you ALL for your ideas, VERY helpful!!!
batee: I was surprised by your schematic in that each DC/DC converter is only rated for 66mA continuous output (according to the datasheet), but your circuit is able to switch large FETs (which can require transient gate currents up to 1A to switch "hard"). NOT that I doubt your results at all, I was just surprised! Let me ask you, how did you know that 66mA would be "enough" for the switch supply?
I guess my question is: if the power supply is rated at a continuous or RMS current of X amps, how many amps can it supply in a fast transient, such as to switch a FET on/off? If I have a "little current" available all the time, how large of a current can I draw from this in a transient burst?
batee: I was surprised by your schematic in that each DC/DC converter is only rated for 66mA continuous output (according to the datasheet), but your circuit is able to switch large FETs (which can require transient gate currents up to 1A to switch "hard"). NOT that I doubt your results at all, I was just surprised! Let me ask you, how did you know that 66mA would be "enough" for the switch supply?
I guess my question is: if the power supply is rated at a continuous or RMS current of X amps, how many amps can it supply in a fast transient, such as to switch a FET on/off? If I have a "little current" available all the time, how large of a current can I draw from this in a transient burst?
Yes, the energy the DC-DCs generate is cached in C1-C4 (470uF) continuously.
Instantaneously the Optos are rated for 2.5A. Even with 10 Ohm gate resistors, they don't heat up at all.
You should be able to calculate the energy required to switch your FETs once (E=0.5*Cgs*Vgs*Vgs) J, multiply that by the number of switches per second, and get the total energy the PS needs to supply. Then calculate the power needed: Energy (in Joules) = Power (in Watts) * Time (in seconds), so Power Required = Energy/time). To get power in Watts, set t=1sec, then the units will be Joules/Sec = Watts. Next calculate current required = Power/Voltage. In my case, I think I calculated power required for 20KHz switch rate at about 5mW. Remember that Vgs isn't 15V, it's the output of the Opto driver, which is about 1-2V less than the input.
Ignore the energy dissipated in the gate resistors (again, they don't get hot = they're only 1/6W). Or calculate the average current for the decaying exponential that results from the capacitor charging action and use it to calculate energy lost to the resistor. Opto driver circuitry current draw is available from the datasheet. Add this to the value you calculated above.
From measurements, I see about 123.3mADC draw on the +12VDC power supply input when the top FET is left on and the bottom FET is switched at a rate of 22KHz. The fact that it works at 400KHz backs up the calculation - even then power consumption should be at about 1W/bottom FET while switching, or about 0.5W / bottom FET averaged over time. I don't think I saw much change in draw between no switching events and full speed operation.
Also remember that in a full bridge operation, you're only providing power to switch the bottom FETs for 50% of the time, and the top FETs for 1/2*(switchrate) time, so the DC-DC converters have time to recharge C1-C4.
If I had it to do over, I'd combine the power supplies for the bottom FETs into one, and I'd use the 1W modules instead of the 2W DC-DC converter modules, as they're only about 1/2 the cost. I'd also eliminate the inductors at the output of the DC-DC. Poor manufacturing tolerances result in radically different voltage outputs at each of the Caps C1-C4. I'd also place a high value resistor across the outputs of the DC=DCs to bring their voltage outputs down to the rated level.
Bryan
Instantaneously the Optos are rated for 2.5A. Even with 10 Ohm gate resistors, they don't heat up at all.
You should be able to calculate the energy required to switch your FETs once (E=0.5*Cgs*Vgs*Vgs) J, multiply that by the number of switches per second, and get the total energy the PS needs to supply. Then calculate the power needed: Energy (in Joules) = Power (in Watts) * Time (in seconds), so Power Required = Energy/time). To get power in Watts, set t=1sec, then the units will be Joules/Sec = Watts. Next calculate current required = Power/Voltage. In my case, I think I calculated power required for 20KHz switch rate at about 5mW. Remember that Vgs isn't 15V, it's the output of the Opto driver, which is about 1-2V less than the input.
Ignore the energy dissipated in the gate resistors (again, they don't get hot = they're only 1/6W). Or calculate the average current for the decaying exponential that results from the capacitor charging action and use it to calculate energy lost to the resistor. Opto driver circuitry current draw is available from the datasheet. Add this to the value you calculated above.
From measurements, I see about 123.3mADC draw on the +12VDC power supply input when the top FET is left on and the bottom FET is switched at a rate of 22KHz. The fact that it works at 400KHz backs up the calculation - even then power consumption should be at about 1W/bottom FET while switching, or about 0.5W / bottom FET averaged over time. I don't think I saw much change in draw between no switching events and full speed operation.
Also remember that in a full bridge operation, you're only providing power to switch the bottom FETs for 50% of the time, and the top FETs for 1/2*(switchrate) time, so the DC-DC converters have time to recharge C1-C4.
If I had it to do over, I'd combine the power supplies for the bottom FETs into one, and I'd use the 1W modules instead of the 2W DC-DC converter modules, as they're only about 1/2 the cost. I'd also eliminate the inductors at the output of the DC-DC. Poor manufacturing tolerances result in radically different voltage outputs at each of the Caps C1-C4. I'd also place a high value resistor across the outputs of the DC=DCs to bring their voltage outputs down to the rated level.
Bryan
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