Resistor values

[adrianvon]

Hi,

I am going to use an H-bridge to drive a DC motor. The manufacturer of the H-bridge suggest to use a 10Kohms resistor between the logic pins and the microcontroller. Now, that I am going to include an opto isolator between the microcontroller and the H-bridge, do I keep the same resistor values?

If I change the 10Kohms to 1Kohms and make the opto isolator resistor to 10Kohms, will that be better?

 

[ericgibbs]

hi Adrian,
The opto input should be about 15mA, so allow a 1.5V drop in the opto emitter diode.
Working from 5V thats [5V -1.5V]/0.015mA = 233R , so a 240R or 270R series resistor from the PIC pin to the opto diode.

E

 

[MikeMl]

What is the rise and fall time and duration of the signals you are sending to the driver? I worry that the 10K pullup on the collector of the opto will be too slow.

 

[adrianvon]

Thanks for the replies.

The opto input should be about 15mA, so allow a 1.5V drop in the opto emitter diode.
Working from 5V thats [5V -1.5V]/0.015mA = 233R , so a 240R or 270R series resistor from the PIC pin to the opto diode.

That is for the opto isolator input right?

What is the rise and fall time and duration of the signals you are sending to the driver? I worry that the 10K pullup on the collector of the opto will be too slow.

I will be using a PWM to control the motor, but I do not know the frequency yet. Is there a calculation to determine the resistor values with respect to the PWM signals?
Also, when the opto isolator is active, will the 10Kohms and the 1Kohms act in series (i.e. 11Kohms)?

 

[MikeMl]

...
I will be using a PWM to control the motor, but I do not know the frequency yet. Is there a calculation to determine the resistor values with respect to the PWM signals?
Also, when the opto isolator is active, will the 10Kohms and the 1Kohms act in series (i.e. 11Kohms)?

The power dissipation in the FETs is largely determined by how rapidly they turn on/off, and then by how frequently they turn on/off. All things being equal, you want to switch them as fast as possible (short rise and fall times), and as infrequently as possible (consistant with the motor not buzzing).

To make the input pin of the driver go high, you have to wait for the NPN inside the opto to turn off, then for the 10K in series with the 1K to charge the pin input capacitance of the driver. (slow?)

To make the input pin of the driver go low, you have to wait for the NPN inside the opto to turn on, then for the 1K to discharge the pin capacitance of the driver. The 10K just goes for the ride.

 

[adrianvon]

Thanks for the reply.

If I use the TLP621 opto isolator with a 1.9Kohms resistor, the turn-on time will be 2us and the turn-off time will be 25us (datasheet page 6). That is OK for a PWM right?

Also, can I eliminate the resistor at the input of the driver and just leave the 1.9Kohms of the opto isolator?

http://pdf.datasheetcatalog.com/datasheet/toshiba/2236.pdf

 

[MikeMl]

No, that is horribly slow.

I dont have model for TLP621, but look at this sim with a 4n25. There are two resistors you need to worry about. The one on the input of the opto, and the one used for the pull-up resistor. The one from the collector of the opto to the input of driver can be deleted.

 

[adrianvon]

I did some research and found the 6N137 which looks quite fast.

 

[adrianvon]

The one from the collector of the opto to the input of driver can be deleted.

If some current leak or noise occur in the ground line when the motor is turned on, will a small value resistor prevent from interfering with the IN pin? or I can totally eliminate the resistor without any problems.?

 

[adrianvon]

Another quick question. The datasheet of the BTN7970 states:

In order to optimize electromagnetic emission, the switching speed of the MOSFETs is adjustable by an external
resistor. The slew rate pin SR allows the user to optimize the balance between emission and power dissipation
within his own application by connecting an external resistor RSR to GND.

So if I increase the resistance at the SR pin, the power dissipation will increase and the emissions will decrease ; and vice versa ?

Since I am going to use PWM, shell I ignore the emissions and reduce the slew rate?

 

[MikeMl]

...
Since I am going to use PWM, shell I ignore the emissions and reduce the slew rate?

Depends on how hot your FETs get

 

[MikeMl]

If some current leak or noise occur in the ground line when the motor is turned on, will a small value resistor prevent from interfering with the IN pin? or I can totally eliminate the resistor without any problems.?

You have already taken a huge step in eliminating potential noise problems by using opto-isolation on the signals between your MCU and the H-bridge. You need to keep the motor supply separate from the MCU supply, however.

 

[adrianvon]

Yes I will be using two different supplies, but the enable logic pin of the H-bridge will be using the same supply off the motor or else it will not work (no common ground reference). So the problem is not about the MCU logic pins but on the H-bridge enable logic pins.

 

[MikeMl]

Why is the enable any different than the others?

 

[adrianvon]

Sorry I was referring to the IN and INH pins. Since the driver IC connects the logic ground and the power ground together (internally), I cannot separate the grounds. In fact this was the main reason why I used the optos, so at least I separate the MCU and other ICs ground line from the ground of the H-bridge. But unfortunately I cannot do the same for the driver logic ground.

 

[MikeMl]

If you are just tying it permanently high or low, that wont matter...

 

[adrianvon]

Just one more quick question; I am considering using the HCPL2631 since it can switch at 25ns with a 350ohms resistor. Now the manufacturer if the BTN7970 suggests to use a resistor value in the range of 10Kohms at the logic input as protection.

Quote from datasheet:

The digital inputs need to be protected from excess currents (e.g. caused by induced voltage spikes) by series
resistors in the range of 10 kΩ.

If I use the 350ohms at the opto isolator only, will that be enough for protection?

 

[remon7]

According, Ohm's law we know that “The current is directly proportional to the applied EMF <voltage> and inversely proportional to the resistance in the ckt." if resistor exists> resistance >And Decreases according to ohm’s law, but current are unswervingly proportional to voltage and current is inversely proportional to resistance it means as current increases> voltage increases. And resistance increases> current decreases same as voltage if there is no resistor, there should be no resistance excluding inner resistance of voltmeter & ammeter.