high voltage electronic load = resistors + NPN transistor - Page 1
Forum section: electronics / beginners

author: 001
author_rank: Super Contributor
timestamp: November 05, 2018, 08:23:34 pm
content: HiI need for simple load to test supply 400V 0-0.4ACan I use common 30V transistor load and power resistor in series?

author: ArthurDent
author_rank: Super Contributor
timestamp: November 05, 2018, 09:00:49 pm
content: You will still be connecting a 400 VDC power supply to your 30 VDC electronic load. The resistor will only limit current, expect problems

author: 001
author_rank: Super Contributor
timestamp: November 05, 2018, 09:09:21 pm
content: Quote from: ArthurDent on November 05, 2018, 09:00:49 pmYou misunderstood meFor examplecan I set load for 0.5A at 30V and add 370/0.5=740 Ohm resistor in series?

author: ArthurDent
author_rank: Super Contributor
timestamp: November 05, 2018, 09:30:30 pm
content: I understood. Putting any resistor in series with your 400 VDC supply will not prevent anything you connect to the supply from seeing 400 VDC, even for a split second.If you're going to add a resistor anyway why not just use the proper value resistor (800 ohms) and remove the electronic load from possible destruction.

author: 001
author_rank: Super Contributor
timestamp: November 05, 2018, 09:37:48 pm
content: Quote from: ArthurDent on November 05, 2018, 09:30:30 pmTo trim load from 0 to 100%

author: aheid
author_rank: Regular Contributor
timestamp: November 05, 2018, 09:40:20 pm
content: To expand a bit.The 400V will be across the electronic load, regardless of how much current the load is drawing, within the specs of the power supply of course.While you can certainly use a (400V - 30V) / 0.4A = 925 Ohm resistor to make sure the the transistor does not see more than 30V at max rating, what happens if the transistor for some reason conducts a bit less (due to noise or temperature change or whatever)? Let's say it drops down to conduct only 0.38A. Then you got 0.38A * 925 Ohm = 351.5V across the resistor, which means the remaining 400V - 351.5V = 48.5V is across the transistor... woops!

author: ArthurDent
author_rank: Super Contributor
timestamp: November 05, 2018, 09:44:21 pm
content: You will not be trimming from 0-100% but about 93-100% and once the electronic load is adjusted for low current it will see 400 VDC.

author: 001
author_rank: Super Contributor
timestamp: November 05, 2018, 09:47:22 pm
content: Quote from: ArthurDent on November 05, 2018, 09:44:21 pmNoI can100% current is for 0 volts at transistor0% is for open circuit

author: ArthurDent
author_rank: Super Contributor
timestamp: November 05, 2018, 09:51:28 pm
content: ...and what about 50% ??

author: aheid
author_rank: Regular Contributor
timestamp: November 05, 2018, 09:58:07 pm
content: Quote from: aheid on November 05, 2018, 09:40:20 pmDoh, late at nightStill the point stands. At max rating of 0.4A, let the transistor conduct fully, so it has effectively 0V across it. Then the resistor must be 400V / 0.4A = 1k Ohm. As you reduce the current through the load, the voltage across the transistor will increase. The transistor can handle max 30V, at which point 370V will be across the resistor. The current at that point is 370V / 1k Ohm = 0.37A. If you reduce the current through the load further, you'll blow your transistor...

author: Zero999
author_rank: Super Contributor
timestamp: November 05, 2018, 11:00:20 pm
content: The transistor and resistor form a potential divider. When the transistor is fully on, then it has very little voltage across it and all of the 400V is across the resistor, but when the transistor is off, it will have 400V across it.You won't be able to find a BJT rated to take 400V and 0.4A simultaneously, the safe operating area of any BJT will be exceeded, even 200V at 0.2A will be a challenge, as that will be the point where there's maximum power dissipation in the transistor. Several transistors could be connected in cascode configuration to split the voltage across the devices.

author: ArthurDent
author_rank: Super Contributor
timestamp: November 05, 2018, 11:48:32 pm
content: This is the electronic load I now have and its maximum voltage rating is 360 VDC / 30 A / 300 watt. Trying to find some load that works at a higher voltage might be a problem at a reasonable cost. This load uses MOSFETs not BJTs.

author: 001
author_rank: Super Contributor
timestamp: November 06, 2018, 07:05:49 am
content: Quote from: ArthurDent on November 05, 2018, 11:48:32 pmThank Youthe idea to go with resistors came to me becouse Mosfet SOA is small for linear modeCan You post schematic of power stage of Yours load?

author: Wolfgang
author_rank: Super Contributor
timestamp: November 06, 2018, 10:59:25 pm
content: Quote from: 001 on November 05, 2018, 08:23:34 pmWhat you need is a classic case for a MOSFET design using LINEAR devices from IXYS. The normal switching types are not suitable because the can easily killed by thermal runaway in the linear region. Dont forget, you are talking about 160W of dissipation (I would use 4 transisors on a fan cooled heatsink).The idea of using series resistors to reduce MOSFET voltage and power is possible, but unelegant aqnd cumbersome.There are a lot of electronic load circuits, even here in this forum. Most are based on an Op-Amp regulator. Pick one, get the right linear MOSFETs, and it should work.