Author Topic: Switching negative voltage rail  (Read 2696 times)

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Offline Rachie5272Topic starter

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Switching negative voltage rail
« on: April 24, 2018, 06:12:10 am »
I'm trying to figure out something very simple: how to control a negative voltage with a transistor.  I have a voltage rail which is -200V with respect to mains earth.  The control circuitry is +3.3V with respect to mains earth.  Is there a simple way to switch a negative rail without isolation?
 

Offline NiHaoMike

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Re: Switching negative voltage rail
« Reply #1 on: April 24, 2018, 06:20:34 am »
How much current? One option could be a DC SSR.
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Offline Rachie5272Topic starter

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Re: Switching negative voltage rail
« Reply #2 on: April 25, 2018, 05:12:06 pm »
I estimate around 100 mA, and it needs to switch in the 100 Hz range.  I need about 100 of them, so I'd like to keep the unit price under $1.

Is there a way to do this with a simple transistor?
 

Offline Dajgoro

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Re: Switching negative voltage rail
« Reply #3 on: April 25, 2018, 07:02:12 pm »
What about this?
« Last Edit: April 25, 2018, 07:09:10 pm by Dajgoro »
 

Offline kony

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Re: Switching negative voltage rail
« Reply #4 on: April 25, 2018, 07:24:44 pm »
If you had one low voltage negative rail generated as well, you can make it much less roasty (R3) by using highside PNP instead of Q3 with biasing resistor against the low voltage negative rail (1V8). At assumed 100 channels, ~30W of power dissipation saved is well worth consideration. Also it removes need of the current mirror needing to be rated for a full voltage difference of the digital positive and power negative rail, so normal duals (BC856BS) are available.

EDIT: On second tought - is that current mirror even needed?  >:D
« Last Edit: April 25, 2018, 07:45:17 pm by kony »
 

Offline duak

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Re: Switching negative voltage rail
« Reply #5 on: April 25, 2018, 08:14:39 pm »
Does the load have much in the way of bypass capacitance?  The power that the switching element  has to dissipate can be as high as 1/2 FCV^2 where F is the frequency (100Hz), C is the total load capacitance  and V is the load voltage (200 V).  The dissipation in the switching element can be reduced by adding by adding a series resistor or, even better, a series inductor.   Note also that the switching element has to be rated for the peak current that may be drawn while switching.

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Offline Zero999

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Re: Switching negative voltage rail
« Reply #6 on: April 26, 2018, 08:09:07 am »
The current mirror is not needed, nor is the separate -1.8V supply, which could be a resistor.

Here's another solution, which switches the -200V rail, rather than 0V.

The 3.3V control signal needs to be able to supply the full base drive to Q1, so Q1 might need to be a MOSFET or a Darlington pair.
« Last Edit: April 26, 2018, 08:12:23 am by Hero999 »
 

Offline kony

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Re: Switching negative voltage rail
« Reply #7 on: April 26, 2018, 08:44:41 am »
I already explained why the extra rail makes sense due to total power dissipation of the switches array - it's kind of absurd to be burning substantially more power on control signals than what are the losses on transistors (would be fine for few channels, but not 100). Your design concept have same problem as one Dajgoro presented - R2 will be roasty as hell (over 3W Pd as shown!). Darlington would help here - that does hold true even for mine suggestion.
 

Offline Zero999

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Re: Switching negative voltage rail
« Reply #8 on: April 26, 2018, 10:09:35 am »
I already explained why the extra rail makes sense due to total power dissipation of the switches array - it's kind of absurd to be burning substantially more power on control signals than what are the losses on transistors (would be fine for few channels, but not 100). Your design concept have same problem as one Dajgoro presented - R2 will be roasty as hell (over 3W Pd as shown!). Darlington would help here - that does hold true even for mine suggestion.
I agree, but now there's the problem of an additional 1.8V supply.

I think Darlington pairs or MOSFETs are better options. If Q1 were a Darlington, such as the TIP150, then R1 could be 1k8 and R2 120k. R3 would no longer be needed, since the TIP150 has built-in base-emitter resistors.
www.nteinc.com/specs/original/TIP150_2.pdf

Switching speed is not a big consideration, since the frequency is only 100Hz, a MOSFET could be used, even with high value resistors and sluggish gate drive, although bypass capacitors could be added for more speed.
 


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