Boosting pwm peak to peak voltage

[andesam]

Hi.

I am working on a circuit to control a dc motor using a pwm signal. The problem is that the pwm source (a LabJack UE9 DAQ) doesn't send a ttl signal, but a 0-3.5 volt signal (3.5 peak to peak). The power transistor needs about 5 volts to open completely between source and drain.

A simple op-amp does the job, but supposedly it will cause the transistor to overheat (my "mentor" told me so).

Can someone please shoot some ideas at me? I'd be really thankful.

- Andreas
(Please exuse my english)

 

[berkeman]

Hi.

I am working on a circuit to control a dc motor using a pwm signal. The problem is that the pwm source (a LabJack UE9 DAQ) doesn't send a ttl signal, but a 0-3.5 volt signal (3.5 peak to peak). The power transistor needs about 5 volts to open completely between source and drain.

A simple op-amp does the job, but supposedly it will cause the transistor to overheat (my "mentor" told me so).

Can someone please shoot some ideas at me? I'd be really thankful.

- Andreas
(Please exuse my english)

You should be able to use a logic gate that converts the TTL compatible input signal to a CMOS compatible output signal. The HTC series or VHCT or ACT series of logic gates would be good places to start looking. The "T" in those logic families' names are the tip that their inputs are TTL input voltage compatible.

 

[andesam]

Thanks a lot.

I tried with a NAND gate (4093), but i can't get the output signal to make any sense on the oscilloscope, it's very noisy. I connected the input (a and b) to the DAQ output and gnd, and powered the 4093 using an external 5 volt supply. i tried reading the 4093 output with respect to the DAQ gnd, but it's very noisy. I can resemble the input signal, but at the best it's just the same peak to peak voltage with a higher mean value.

Please let me know if this doesn't make any sense..

- Andreas

 

[vk6kro]

You could add a simple amplifier like this:

[PLAIN]http://dl.dropbox.com/u/4222062/switch%20cct.PNG

R1 may need adjusting. The bipolar transistor should switch at about 1.6 to 2 volts.

Note that this circuit is an inverter, so the FET will be conducting when there is no input voltage, and will switch off if there is an input voltage greater than about 2 volts.

(I have borrowed MatLabDude's drawing from another post. Thanks MLD.)

 

[MATLABdude]

w00t! PhysicsForums citation!

Your mentor may be worried about overheating ASSUMING your op-amp has +/-15V rails--if you use just 5V for the positive and 0V for the negative (obviously, you need to find an opamp which can accept this), you should be fine. Or better yet, use a comparator (since your Op-amp would just be used in comparator mode anyways).

Despite the fact that it is my circuit, it may not respond fast enough for your PWM (depending on what frequency you use--the MOSFET switches comparatively slow). You may wish to replace the FET with another NPN BJT (configured like the first) in order to get a non-inverted signal. That or compensate and generate the inverted PWM at the source and go with the single NPN.

If you happen to be XORing two phase-shifted PWM signals, did you know that the A XOR B is equivalent to ~A XOR ~B ? This may also come in handy...

 

[vk6kro]

:)

Yes, I borrowed the drawing, but changed the resistors.
I made R2 1000 ohms because of the high input capacitance of Mosfets. This gives a little more force to the drive, being able to rapidly charge the capacitance.

I was also worried that the input signal might not be going to zero on downward parts of the input wave form. In this case, it may not turn off the BJT. Hence the voltage divider at the input of the BJT.

I have used Mosfets like the MTP3055 at 3.5 MHz so the devices are capable of high frequency operation, but that high input capacitance makes them difficult to drive. Their legendary high input impedance becomes worthless if they have 100 ohms of capacitive reactance as an input.

 

[andesam]

Thanks a lot guys.

I think the concern regarding the op-amp was that it would slip some voltage out and make the transistor (a irf 3415 by the way) draw some current when den pwm signal is "down".

Since time is of the essense here, i think i will go for the op-amp / comparator option. I'll try out the complete circuit later today. You'll hear back from me :).

- Andreas

 

[berkeman]

Thanks a lot.

I tried with a NAND gate (4093), but i can't get the output signal to make any sense on the oscilloscope, it's very noisy. I connected the input (a and b) to the DAQ output and gnd, and powered the 4093 using an external 5 volt supply. i tried reading the 4093 output with respect to the DAQ gnd, but it's very noisy. I can resemble the input signal, but at the best it's just the same peak to peak voltage with a higher mean value.

Please let me know if this doesn't make any sense..

- Andreas

The CD4093 is a CMOS gate, with CMOS levels required at its input. You need to use a logic family with a "T" in it...

 

[vk6kro]

i think i will go for the op-amp / comparator option

Your PWM generator will be producing nice square waves and it is important to maintain this shape.

If a FET (or a BJT) is turned on, it has very little voltage across it, so it doesn't get very hot.
If it is turned off, it carries very little current, so it doesn't get very hot.
However, if it takes a long time to turn on or off, it will get hot because, at that time, it has voltage across it and a large current flowing in it.

What this means is that the rise time and fall time of the square wave must be very short.

An opamp or a comparator probably won't do this unless you choose a very fast chip. They will tend to put sloping edges on the square waves and this will cause the FET to heat up more than it would otherwise.

You only need a small amount of gain, so a simple transistor amplifier will give this to you without degrading your PWM signal.