# Negative buck converter?

#### PickyBiker

Joined Aug 18, 2015
105
I have a project that need both positive and negative regulated output from a higher voltage + and - unregulated supply.
Simple and very cheap positive buck converters are everywhere, but I haven't been able to find any negative buck converters.

Here what I have:

+25V at 2A unregulated
-25V at 2A unregulated

What I need for output is +15 and - 15
Here is the schematic I came up with for both positive and negative buck converters. I'm pretty sure the positive converter is right, my brain doesn't work too well in the negative polarity. I need someone to verify that I got the negative converter right.

#### ISB123

Joined May 21, 2014
1,236
You need inverting buck converter.Its achieved by switching the places of diode and inductor.

Here the is the circuit from the datasheet.But its only capable of outputting 700mA as inverting converter.

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#### Lestraveled

Joined May 19, 2014
1,946
Refer to page 29 of the TI data sheet for an inverting negative output supply.

#### PickyBiker

Joined Aug 18, 2015
105
I see the negative inverter in the data sheet, but those configuration only produce about 25% of the current the device is capable of. That's why I created the negative circuit the way did. The configuration I drew should produce the full 3A capability (if I have designed it correctly). I actually only need about 1.5 amps so half of the devices capability.

#### AnalogKid

Joined Aug 1, 2013
11,200
Hold on there, bucko. The circuit in post #1 will not work for several different reasons. More importantly, the circuits in posts #2 and #2 will not work either for you - because a) the part works only with a positive supply voltage; and b) it does not have enough current capability to invert +25 V to -15 V at 3 A.

No matter what a chip is called, it's operating mode is determined by the algebraic sum of the input and output voltages. In your case, this part will not operate with a negative input voltage so it must run on the +25 V source. You have a +25 V input and a -15 V output, for a total voltage difference of 40 V. This is the maximim allowable input voltage for this part, so it had better be well regulated. 25 Vin and 40 Vout means that the circuit is a boost converter, and like all boost converters with a fixed energy input the output current goes down as the output voltage goes up. 40/25=1.6 From the circuit's point of view, the output voltage is 1.6 times the input voltage, and the device has a switch current limit of 3 A, so the maximum possible output current is 3/1.6=1.875 A - not counting deratings due to device efficiency. Figure a max output current of 1.5 A, and make sure the chip has a heatsink.

If the -25V source is completely isolated from the +25V source, then things can be made to work. But if, for example, the +/-25V come from a power system with a center-tapped full-wave configuration, then we'll have to take a different approach.

ak

#### AnalogKid

Joined Aug 1, 2013
11,200
The configuration I drew should produce the full 3A capability (if I have designed it correctly). I actually only need about 1.5 amps so half of the devices capability.
Hold on there, bucko. The circuit in post #1 will not work for several different reasons. More importantly, the circuits in posts #2 and #2 will not work either for you - because a) the part works only with a positive supply voltage; and b) it does not have enough current capability to invert +25 V to -15 V at 3 A.

No matter what a chip is called, it's operating mode is determined by the algebraic sum of the input and output voltages. In your case, this part will not operate with a negative input voltage so it must run on the +25 V source. You have a +25 V input and a -15 V output, for a total voltage difference of 40 V. This is the maximim allowable input voltage for this part, so it had better be well regulated. 25 Vin and 40 Vout means that the circuit is a boost converter, and like all boost converters with a fixed energy input the output current goes down as the output voltage goes up. 40/25=1.6 From the circuit's point of view, the output voltage is 1.6 times the input voltage, and the device has a switch current limit of 3 A, so the maximum possible output current is 3/1.6=1.875 A - not counting deratings due to device efficiency. So you might get an output current of 1.5 A, but make sure the chip has a heatsink. For input voltage margin, change to a module based on the HV version of the chip. For output current margin, change to a module based on the 5 0r 6 A version of the chip, such as the LM2678.

If the -25V source is completely isolated from the +25V source, then things can be made to work with what you have. But if, for example, the +/-25V come from a power system with a center-tapped full-wave configuration, then we'll have to take a different approach.

ak

#### PickyBiker

Joined Aug 18, 2015
105
AnalogKid, I don't see the part working as you describe. If you look at the Lm2596 in the negative leg (the bottom half) you can see the GND of the chip is tied to the -25 supply so it is working (from its perspective) with a positive supply. As far as that chip is concerned it is operating in a normal way with the other active pins at its GND or more positive level.

Once again, the CHIPS GND is on the -25 input and that makes the GND INPUT more positive. The bottom line is that both LM2596 chips are working with positive voltages.

This is the difficulty with working with a negative polarity mindset. I don't claim this circuit will absolutely work, but I think it will.

I'm going to try to figure out how to make LTSPICE work. I have never used it before but that will say for sure if I have this figured out correctly.

BTW: You are correct in thinking this is a center tapped transformer with a full wave bridge so the + and - 25 are not isolated.

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#### AnalogKid

Joined Aug 1, 2013
11,200
Think it through. Your 2596 circuit is a non-isolated regulator, and your two source supplies are not isolated from each other. You want a +/-15 V output, which means that the +15 V regulator and the -15V regulator must share the same output ground. The 2596 has only one GND pin, and it must be common to both the input and output circuits On the input side you have it tied to the -25, but on the output side it must be connected to the system GND. That is a problem.

However...

There is an old technique that popped up shortly after the introduction of the first 3-terminal regulator, called "regulating the ground". At the time there were no negative regulators to compliment the first positive regulators, and people really wanted to use the new-fangled easy parts for things like +/-12 and +/-15. The problem is that it adds a series impedance to that center-tapped ground, and analog circuits do not like that. Large unipolar ground currents modulate the "0" volts that ground should be, and can cause positive feedback into circuits.

ak

#### PickyBiker

Joined Aug 18, 2015
105
For my own sanity, I redrew the schematic with the negative half having the -25 input at the bottom. I relabeled GND as "Common" so it is easier to talk about chip GND vs the transformer center tap which is common. This arrangement makes it much easier for me to see which way is UP or Positive.

Now I will consider the circuit based on the last input from AK.

#### AnalogKid

Joined Aug 1, 2013
11,200
Your second schematic tells the tale. The output of a 2596 is regulated against its own GND pin, no matter what signal name that pin actually is connected to. So your IC2 is a +15V regulator, with its output +15V above its GND pin. Since its GND pin is connected to -25V, its output is in fact at -10V with respect to Common.

AND (not trying to pile on) your "-15" output cannot sink any current, certainly not amps of it. For proper operation of whatever this thing is powering, the -15 current path is from the -15 connector pin, through a regulator circuit that can move amperes of current from right to left in your drawing, to the -25 connector pin, and fro there to the negative bulk supply. Nothing in your negative regulator drawing supports a multi-amp current path from right to left.

The 2596 has an NPN bipolar switching transistor that connects pin 1 to pin 2. That is the only power path through the device. On your schematic, draw a big thick arrow on both 2596 symbols starting at pin 1, with the arrow head at pin 2. Then trace trace a 1.5 A balanced load current from J1 pin 1 toJ2 pin 1, through your load circuits, to J4 pin 2, to J3 pin 2.

ak

#### PickyBiker

Joined Aug 18, 2015
105
Yes, I see what you are talking about now. All I had to do was turn the thing right side up to see it. Thank you AK.

Now to find an alternate solution for the -leg.

#### AnalogKid

Joined Aug 1, 2013
11,200
As I said, there is an old trick for linear regulators that might work, but it has issues with unbalanced +/- load currents. I did some quick hunting for old app notes and datasheets, but can't find it online.

Starting with your second schematic, it goes like this:

1. Break the ground connection between U1 pins 3 and 5 (leave them connected to each other) and the vertical net from the bottom of C2 to the top of C4. This separates the positive regulator GND net from the rest of the supply.

2. Connect that net to the J4 pin 2 net, the output of the negative regulator.

This stacks the pos regulator on top of the neg regulator output, like two positive power supplies in series.

ak

#### PickyBiker

Joined Aug 18, 2015
105
Don't quite follow that. I attached the eagle schematic. Would it be possible for you to make the change in that so I can see it?

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#### AnalogKid

Joined Aug 1, 2013
11,200
Graphics file or pdf...