Help for a 5V to 15V level shifter

Question

I'm using FPGA DE2 board cyclone II from altera (for PWM generation) to control a DC chopper via ARCAL 2108 driver ;The problem is that the FPGA GPIO output gives a 3.3V but the driver is +15V input . I used a buffer IC (CD4009 ) to shift the voltage from 3.3V to 15V but when visualising the buffer's output with an oscilloscope I get a Vmax=15V(using measure) but the voltage diagramm has an offset!

Can any body help me !!!I thought to use an IC level shifter or even build an analog circuit (using transistors ,R,C...)

update:

I think I have found the right IC MC 34152 its high speed dual MOSFET. I've just tried it on scope but not with FPGA.

Your statement is not particularly understandable. With 3.3 volts on Vcc, 15 volts on Vdd, and Vss at ground, please answer the following questions. 1) With 0 volts input, what is the output? 2) With 3.3 volts input, what is the output? — WhatRoughBeast, Jul 09 '15 at 22:20
I believe the CD4009 is a CMOS to TTL level converter. That is, from 15V to 5V, not 5V to 15V. — EM Fields, Jul 09 '15 at 22:41
1-)With 0 volts input I get 0v 2-)with 3.3 volts input I get 10V with an offset of 5V (total of 15V but shifted) — Zak Mit, Jul 10 '15 at 00:11
The MC 34152 is not a MOSFET. It is a MOSFET driver. And there are lots of them out there. Try, for instance, the MAX4426/4427 series. — WhatRoughBeast, Jul 11 '15 at 20:47

score 1 · Answer 1 · edited Apr 13 '17 at 12:32

It sounds like you want some sort of interface glue between a FPGA DE2 board cyclone II from altera (generating PWM generation) and a ARCAL 2108 driver.

The 2108 requires a 15 VDC power supply connected to its VDD and GND pins. I'm guessing your FPGA is getting power from a 9V DC power adapter plugged into a mains outlet, right? You need a similar 15 VDC power adapter plugged into the same power source -- mains power -- to power the 2108.

The 2108 also has a "logic input voltage" Vim, and two data inputs InA and InB. The datasheet claims that those inputs support "CMOS or HCMOS input level ... All logic inputs are Schmitt trigger type.". So I'm guessing the 2108 board runs InA and InB into something like a CD4093, which runs fine from a "logic input voltage" of 5V or 15 V. (Could you look at the board and tell us the part number of the IC on the 2108 board connected to those inputs?)

The datasheet says that with Vim connected to 5 V, the input thresholds are 2.6 V and 1.3 V. It appears that you can directly connect the pins from the DE2 board -- "VCC5" pin (+5 VDC) and 2 digital output pins that PWM from 0 V to 3.3 V -- to the 2108 board -- with Vim set to +5 VDC -- without any interface glue.

(If you did need to do logic-level translation, Step up 3.3V to 5V for digital I/O and "3V Tips ‘n Tricks" have some ideas).

score 0 · Answer 2 · answered Jul 10 '15 at 01:16

You are trying to convert TTL-level (3.3 volts) to CMOS-level (15 volts) and this is not possible with the CD4009. The chip is described as a CMOS to TTL/DTL converter or CMOS buffer. The output level is determined by Vcc, while the input level is determined by Vdd. The first page requires that Vcc =< Vdd. So you can convert 0-15 volt swings to 0-5, but not the other way.

Also, if you put in 0-3.3 volts and get out 0-10 volts, you do not have a 15 volt swing with a 5 volt offset - you have a 10 volt swing. I have no idea what you mean when you say "visualising the buffer's output with an oscilloscope I get a Vmax=15V(using measure) but the voltage diagramm has an offset", but it does not mean what you think it does. I suspect that you are getting a very narrow 15 volt spike on the leading edge, and the scope is detecting it and telling you the total voltage swing is 15 volts. Alternatively, you are getting a negative spike on the falling edge to -5 volts, and the scope is reporting that as 15 volts (I suspect the latter, since I bet you're using a solderless breadboard and bad grounding practices).

Help for a 5V to 15V level shifter

2 Answers2