MOSFET for higher voltages (30-40 V) - Page 1
Forum section: electronics / beginners

author: twam
author_rank: Unknown
timestamp: December 29, 2019, 03:57:22 pm
content: Hi everybody,For switching my flipdots I finally decided to use a capacitive-discharge approach. It currently works fine when using with the attached schematic and a voltage of 24 V. For switching the voltages I useDMG6602-SVT.Now I want to increase the supply voltage to something between 30 V and 40 V. According to the data sheet the DMG6602-SVT has a breakdown voltage of 30 V, so I'm looking for alternatives. I was thinking of something like theDMP4025LSDfor the P-Channel andBSS123for the N-Channel MOSFET. DMP4025LSD has a breakdown voltage of 40 V which should be fine, however the maximum gate-source voltage is +- 20 V. Also almost all other P-Channel I found with higher breakdown voltage have a similar maximum gate-source voltage. How can I use them in a similar setup?

author: Yansi
author_rank: Super Contributor
timestamp: December 29, 2019, 04:34:42 pm
content: How is that circuit supposed to work? You are not only using unsafe supply voltage (V is littel too much for a 30V fet), not mentioning the Vgs maximum voltage rating, that you are already exceeding.What is this circuit supposed to do?

author: magic
author_rank: Super Contributor
timestamp: December 29, 2019, 06:09:35 pm
content: Big distributors like mouser or digikey have comprehensive databases and parametric search which will allows you to find thousands of FETs with just about any parameter you may desire, as long as such parts exist.That being said, MOSFET gates are quite delicate so 30V rating is somewhat uncommon and more than that is very rare. Add 10k resistors at the drains of the N-FETs and then the P-FETs will see only half the supply voltage at their gates.

author: twam
author_rank: Unknown
timestamp: December 29, 2019, 06:55:53 pm
content: Quote from: Yansi on December 29, 2019, 04:34:42 pmCharge and Pulse are controlled by the µC to charge and discharge the capacitor. The output on the right is VCC for an UDN2981/TBD62783 transistor array which is then connected to the flipdots.

author: Yansi
author_rank: Super Contributor
timestamp: December 29, 2019, 07:05:00 pm
content: I still don't understand the purpose of this circuit. What is this circuit good for, in a flipdot driver?

author: Wimberleytech
author_rank: Super Contributor
timestamp: December 29, 2019, 07:36:46 pm
content: I have no clue about flipdots. That being said, why not just use a voltage divider as shown here (first stage only).

author: twam
author_rank: Unknown
timestamp: December 29, 2019, 07:41:13 pm
content: Quote from: Yansi on December 29, 2019, 07:05:00 pmThe capacitor is charged and then decoupled from supply. Then it is decharged via the flip dot which is connected to GND. The flip dot is basically just a coil.

author: Yansi
author_rank: Super Contributor
timestamp: December 29, 2019, 08:26:18 pm
content: Professionally made flipdot drivers do not contain any such circuits.I am still failing to understand, what your intention here is. Probably, sort of "charge metereing device". But what for? Unlikely to work this way anyway. You will crap the charging mosfet out on first turn on.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: December 29, 2019, 08:28:00 pm
content: But the current pulse from charging the capacitor directly is ... probably even bigger than the load you're trying to isolate?

author: twam
author_rank: Unknown
timestamp: December 29, 2019, 09:22:23 pm
content: This method was suggested by in thispostand alsothis projectuses the same approach.The displayed schematics works fine so far with several days of runtime switching every 150 µs.

author: Yansi
author_rank: Super Contributor
timestamp: December 29, 2019, 09:48:40 pm
content: With the exception, that the charge power supply is a constant current source there.Nowhere one can see it in your drawings.

author: Zero999
author_rank: Super Contributor
timestamp: December 29, 2019, 11:19:45 pm
content: I'd replace the MOSFET with a BJT and add a suitable emitter resistor to give the required gate voltage.

author: jonroger
author_rank: Regular Contributor
timestamp: December 30, 2019, 01:04:12 am
content: +1 on adding a resistors to form a voltage divider. Without it, Vgs on QxB will exceed maximums.