Switching negative voltage rail - Page 1
Forum section: electronics / projects

author: Rachie5272
author_rank: Regular Contributor
timestamp: April 24, 2018, 06:12:10 am
content: 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?

author: NiHaoMike
author_rank: Super Contributor
timestamp: April 24, 2018, 06:20:34 am
content: How much current? One option could be a DC SSR.

author: Rachie5272
author_rank: Regular Contributor
timestamp: April 25, 2018, 05:12:06 pm
content: 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?

author: Dajgoro
author_rank: Frequent Contributor
timestamp: April 25, 2018, 07:02:12 pm
content: What about this?

author: kony
author_rank: Regular Contributor
timestamp: April 25, 2018, 07:24:44 pm
content: 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?

author: duak
author_rank: Super Contributor
timestamp: April 25, 2018, 08:14:39 pm
content: 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.Cheers,

author: Zero999
author_rank: Super Contributor
timestamp: April 26, 2018, 08:09:07 am
content: 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.

author: kony
author_rank: Regular Contributor
timestamp: April 26, 2018, 08:44:41 am
content: 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.

author: Zero999
author_rank: Super Contributor
timestamp: April 26, 2018, 10:09:35 am
content: Quote from: kony on April 26, 2018, 08:44:41 amI 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.pdfSwitching 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.