Find attached analog signal conditioning circuit. It scales and offset signal from +-5 V voltage range to 0-3 V. Output signal is equal to Vout=(R155/R154)*Vin + Vref. By adjusting resistors R155=R153 and R154=R152 you can adjust signals from different voltage ranges. If jumper (J12 in the Fig.) is in position 2-3 it reference signal to Vref=0 V, and use it for unipolar signals adjusting, like for measuring inverter dc-link voltage. If jumper is in position 1-2 it reference signal to Vref (in your case generate Vref=1.5V), and use this configuration for conditioning bipolar signals, like inverter line currents. OP amp IC (LM348) is powered with bipolar supply +-15V. Protection diodes V23 and V24 assures that signal on controller analog inputs (P9:2 designation,here) does not over-range 0-3 volts.

I do not know much details about your project. But an amplifier is needed to change process output voltage level from 5 V to 15 V.

We can use amplifier. But for high frequencies the wave shapes are not good ( Not a pure square pulses).

But you indicated that frequency is 12 KHz (this is very low value)

Hi, what I can understand is you are using an IGBT based drive circuit for driving a high power load. The best way of driving the IGBT would be to use a high side or low side gate driver IC such as IR2110, IR2101, etc. You find this on Digikey if you search as IGBT drivers. The input to the driver IC will be from your processor which would be TTL or CMOS. The output shall be 10-20V with a totempole to drive the gate of IGBT or MOSFET.

I used a PWM @ 62.5kHz and voltage at min. 1V depending on the motor the signal is controlling.

The used µController has been an MSP430. Very easy to set values via CodeComposer.

http://img850.imageshack.us/img850/9806/ezyw.png

so far as i know, semikron provides IGBT driver for specific type of IGBT. Otherwise, you can make it by yourself. To develop the driver, you may consider two type of switching device positions: low side and high side. Low side driver is used for switching devices direct connected to ground, such as in boos chopper. The driver can be simply one like op-amp. By using op-amp you can rise the voltage from processor to 15V. If your switching devices is high side, you can use any type of opto-coupler to isolate between switching devices and processor. You can set the supply voltage to opto-coupler at 15V for the IGBT.

A transistor switch. The transistor has the higher power supply of course and there is an inversion. You could use a MOSFET for example. Even a Bipolar would suffice. If you are drawing a lot of current you will need a heat sink and a reverse diode for an inductive load.

Well, there are some problems with your circuit...

If you really want to use an N-ch MOSFET for M2, you'll need a high-side driver.

It'll be a lot easier if you just use a P-ch MOSFET.

Even then, you don't show an inductor on your schematic; that's pretty much a necessity. Perhaps that's off the right edge of your screen.

Anyway, have a look at the attached. The LTSpice .asc file is attached as well.

Everything to the left of M1 is just to convert the 5v PIC PWM output to the level and current required to charge/discharge the P-ch MOSFET gate in a reasonable amount of time. PWM frequency is 10kHz.

please start follow this website:

http://forum.allaboutcircuits.com/showthread.php?t=31790

Find attach schematic of one PWM channel conditioning. Inputs are 5 V (or it can be even 3.3V) from controller, and circuit outputs are 15 V for IGBT driver. Voltage shifter is done using transistor (BS170 in schematic). You have input buffer (74HCT244) powered by 5V and output buffer (HEF4049) powered by 15V.

I have a question about that circuit which was posted. Would that work on inductive loads? Do you need a reverse diode somewhere?

This circuit only converts 5V PWM to 15V PWM. It is only logic level shifter, and not intended to directly run IGBT's. As I understood the original question, author use Semikron inverter with integrated IGBT drivers which have necessary circuitry to directly run IGBT's. At its inputs drivers expect 15V PWM signals, but his processor is power with 5V and generates 5V PWM signals. So he need voltage translator, and the posted circuit can do this job. Of course, if drivers are transformer based, this circuit is not appropriate and additional driver must be used (maybe like before mentioned IR2110, TC4420 or similar). Originally I have used this circuit to run Semikron SKHI22AH4 integrated drivers, which expect at its inputs 15V (CMOS inputs). Processor was with 3.3V logic (TI F2812 DSP).

Dear Evgenije Adzic Sir

Thank you for sending the circuit to convert 5V/ 3.3 V to 15 PWM. Sir, i am using LA-25P Current sensor and LV-25-1000 voltage sensor. I need to convert the output to 3v to give the ADC input. Since the processor i am going to use is fixed point (TMS320F2812), I need the offset for ac sine wave.

Do you have any circuit for the reference

Thank you

With thanks and Regards

Rajin M. Linus

Find attached analog signal conditioning circuit. It scales and offset signal from +-5 V voltage range to 0-3 V. Output signal is equal to Vout=(R155/R154)*Vin + Vref. By adjusting resistors R155=R153 and R154=R152 you can adjust signals from different voltage ranges. If jumper (J12 in the Fig.) is in position 2-3 it reference signal to Vref=0 V, and use it for unipolar signals adjusting, like for measuring inverter dc-link voltage. If jumper is in position 1-2 it reference signal to Vref (in your case generate Vref=1.5V), and use this configuration for conditioning bipolar signals, like inverter line currents. OP amp IC (LM348) is powered with bipolar supply +-15V. Protection diodes V23 and V24 assures that signal on controller analog inputs (P9:2 designation,here) does not over-range 0-3 volts.

If your input signal is in +-5 V range maybe it is better to put resistors R155=R153=3k3 and R154=R152=11k to extend output signal voltage range to exactly 0-3 V, to get better resolution (only if your input signals are limited to +-5V, or use R154=R152=12k).

Why not just use a TTL-CMOS converter IC if it is required. It is not clear what you mean as like Evgenije has pointed out most Semikron driver modules require 15V for power but they do accept 5v direct as the PWM input. They are designed to take output from micro or DSP which are normally 5V. This has little to do whether the processor is fixed or floating point. All A/D converters are integer (0-2^n-1) so you have to also offset and scale this to correspond with the actual sensor voltages and the actual quantity they are measuring.

All the offset and scaling for the LEM modules can be done with a single OP-AMP.

Surely you must have such experience before trying to do this? I see you are in IT design and manufacture so this appears a little left field.

You will require 0-5v for the DSP, +-/15v for the LEM modules or even just +15 as the LEM modules just need a differential supply voltage and will always have its "zero" output midway of the supply voltage. ie 0v for +-15V or 7.5V for +15.

You need to do some reading and system design or else BIG BANG!

As simple as it sounds this is not a good idea as when all inputs are zero the outputs will all be high. Thus during power up or power down you can create a dead-short and damage the inverter. The SEMIKRON driver modules normally have protection for this but still it is not good practice.

Also you cannot correct in s/w as the DSP/Micro PWM generates the signals in h/w and outputs complementary signals for each phase.

I suggest to use LM319. I had a same problem the DSP output PWM signal was 0-3.3V while the driver required 0-15V. If you are using standard driver circuit like Skyper pro then you don't need to worry about protection issues. All you need is to boost your 0-3.3 or 0-5V signal to 0-15V using LM319.

Also look at the TI CD4504B-BP. It can do TTL-CMOS level all 6 in a single DIL package.

Agreed it is actually a very simple issue.

Some questions in RG are too basic and it appears like too many undergraduate students looking for quick answers / solutions for their design projects. I wonder whether any acknowledge the input from RG in their thesis / dissertations?

It makes one why their supervisors cannot advise them or are the students just lazy and looking for quick answers. Many of the questions asked on RG suggest the latter.

They can easily do what I did and Google TTL to CMOS converter.

You can use optocoupler with +15V VCC, the +5V signal is the input to the optocoupler while the output will be +15V.

@Evgenije Milan Adzic ,can we use your circuit mentioned above to change logic level from 12V to 5V PWM signals? but I don't want to mess-up its dutycycle