Power management circuit design(help)

Hello, Everyone.
My name is zhen.
I am designing a board as a voltage regulator for a ultrasound imaging device, LM96550 Ultrasound Transmit Pulser- Texas Instruments

The desired function of this board is to receive a 110AC input from power source and regulate it in to 11 different DC voltages for 11 different input.
-5V
5V
1.8V
3.3V
-55V
50V
-10V
10V
50V
40V
-40V
Here is my way of doing it.

I first managed to use an ATX to convert the 110AC into 12V/5V dc, and I am planing to use LM5001 High Voltage Switch Mode Regulator (TI chip) to boost the 12V to 50V. Then i will use an linear regulator to drop the 50V voltage to 40V. (the same way to obtain the 10V from 12V and 1.8 V from 5V.
By now, I have got 6 desired voltages.
Except all the negative ones,

The challenging ones for me is the -55 and -40, I could not find any TI chip regulators to obtain a negative voltage as low as -55/-40.

In the mean while, i accidentally found this AN-H59DB1 demo-board which is to convert a 12VDC input to 6 different voltages. Even though those 6 voltages are close to the desired voltages and even includes a negative voltages as low as -75, but those voltages do not match perfectly to the desired voltages.
Otherwise, i will just order it to save the trouble

I do not want/could to use any external circuit other than IC chips to do the job.
Is there a anyway to obtain the -55 or -40 VDC?(I was suggested to use transistor to build a regulator my self, but I have trouble doing it since i am a first year EE student)


Thank you for reading this and sorry for the wall of texts.

Any advises would be really appreciated.

Zhen

Hi Zhen,

According to the AN-H59's application note (**broken link removed**), all of the output voltages are easily adjustable using a potentiometer.

ZekeR said:

Hi Zhen,

According to the AN-H59's application note (**broken link removed**), all of the output voltages are easily adjustable using a potentiometer.

Click to expand...

Thank you ZekeR, I guess buying the board is the easy way to do this.
The problem is that the AN-H59'board is ganna take 3 weeks for me to get it.

Let us say if i use this board. One of the negative output is -65V(I can adjust it with potentiometer to -55V) , however, I need 3 negatives voltages ,-55V,-50V,-40V, is there any way to regulate the -55V to -50V/-40V?

Do you know any other ways of obtaining the -55V?

The easiest thing to do is to buy multiple AN-H59's. Another approach is to use an LM337, as long as you're careful to protect it in case of a shorted output. Another approach is to build your own LDO. Another approach is to buy a switching regulator and configure it as a Cúk converter. And yet another approach is to use a boost converter with its "ground" attached to the –55V rail, which allows it to operate as a negative buck. And there are other ways. Your choice.

Thank you ZekeR, it is really helpful.
Since I am a first year engineering student, a lot of them are really difficult for to implement it.
Except the LM337 one , Could you elaborate on that? I checked the datasheet, it only goes down to -37V and takes input range at -4.2 to -40. Since i do not have a input range from -4.2 to -40. Does it mean that i have to use an other chip to pull the 12V down to -4.2? and how could i go down to -55V or -50V while the maximum is -37V

Sorry for the keep asking you, this problem has been bothering me for last 2 days.

The LM337 will break if you apply too much voltage across it—that is why it is limited to 37V. We should be able to take advantage of the fact that the AN-H59 is using a current-limited chip to generate its negative rail. Without doing any hard math, I estimate the 55V output it shouldn't be able to deliver more than 200mA. So, you can protect the LM337 by placing a 7.2V, 2W zener diode across it (attached between its Input and Output). When the LM337's output is short-circuited, there will be no way to have more than 7.2V across it; the zener will deliver the current and drag the AN-H59's negative output down to -7.2V. The power dissipated will be ~200mA*7.2V=1.5 watts. Get a 2W zener diode to make sure it doesn't burn up.

I am a little bit confused right now, The -37V is the lowest output of LM337 right? What you do you mean by"will break if you apply too much voltage across it"? why would we apply voltage onto the output?

Maybe i was not clear about my question, let us say that I will get a -55V from AN-H59 by adjusting the potential meter. and then i want to convert this -55V to -50/-40 V,

you suggested to use the LM337 to do it by applying this -55V on the input of LM337. And then, I checked the data sheet, it seems that the LM337 can only take input range from -4.2 to -40(higher than -55), and product a out put as low as -37V which is higher than the desired -50/40.

Did i understand the wrong idea? or you are suggesting otherwise?

The reason why the LM337's datasheet says it can't take more than –40V at its input is only because, under short-circuit conditions (or otherwise when its output capacitor is discharged) it will have 40V across the chip (–40 at the input, 0 at the output). Under normal operation, it will only have about 5V or 10V across it.

Because the LM337 doesn't actually have a pin that connects to ground, however, we can supply it with more than –40V at its input—as long as we're careful to ensure it never has more than 40V across it. Placing a zener diode across it from its input to its output will ensure it never has more voltage than the zener will allow. For a 7.2V zener, this will obviously be 7.2V. The concern, though, is that the zener must be able to take a fair amount of power—because it may be called on to take the –55V power supply's full current, and you don't want it to break. So, get a 2W zener.

For the –50V to –40V conversion, you can place two 7.2V zeners in series for a clamping voltage of 14.4V.