Fast switching isolation

Thread Starter

Dritech

Joined Sep 21, 2011
901
Hi all,

I want to drive a motor using an MCU. The motor supply (both positive supply and ground) have to be totally isolated from the MCU supply. What component can I use to achieve this please? OPTOcouplers will not handle the 2A consumed by the motor. On the other hand, a relay will not be fast enough for switching with PWM. Are there any other simple components that I can consider?
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
I did consider that, but for that to work I have to use a common ground for the transistor (+motor) and the MCU.
 

Shagas

Joined May 13, 2013
804
Not really . You have to use common ground for the transistor and motor yes , but the mcu can be completely isolated if you are using an optocoupler. You are sending the information by light , you don't need a common ground
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
Another quick question.
I want the motor to operate with different voltage supply (range from 5V to 12V). If I use the attached circuit, will it work?

Circuit description:

When the OPTO coupler switches on, the supply (5V to 12V from J1) passes to the zener diode which is used to regulate the voltage to 3.9V. This causes the transistor to switch on, thus completing the path to ground.
Are the values good for this application? or is there a better way to implement this?

Thanks in advance.
 

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Last edited:

crutschow

Joined Mar 14, 2008
34,468
The transistor base current needs to be at least 1/10th of the 2A maximum motor current or 200mA to be properly saturated (fully ON). I doubt the Opto-coupler will supply that, so instead you can use a 2-transistor Darlington stage to reduce the base current or use a logic-level type N-MOSFET.

In either case you shouldn't need the added resistor and zener. For the MOSFET you just need a 10k resistor to ground. For the Darlington just select the series resistor to provide the desired base current at 5V. The higher base current at 12V won't be a problem.
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
Thanks for the reply. Is there the need to connect a resistor in series with the gate and the optocoupler? Will it work with both 5V and 12V if I connect it as shown in the attached schematic?
 

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bertus

Joined Apr 5, 2008
22,278
Hello,

You can use a resistor of 10 - 47 Ohms from the optocoupler-receiver to the gate of the mosfet.
If you use the IRF530 you can only use the circuit on 12 Volts (basicaly higher as 10 Volts, but not higher as 15 Volts (higher voltages can damage the mosfet)).
If you want to use the circuit on 5 Volts you must use a logic gate mosfet like the IRL530.

Bertus
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
Hi Bertus,

I updated the schematic. Why is the 47ohms resistor necessary? Also, Will the RFL530 work with 12V input?

Thanks in advance.
 

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crutschow

Joined Mar 14, 2008
34,468
..............
I updated the schematic. Why is the 47ohms resistor necessary? Also, Will the RFL530 work with 12V input?
The 47 ohms is to avoid any oscillations at the gate due to stray inductance. The ON resistance of the opto-coupler may be sufficient for that purpose as long as the connection lead length between them is short.

Yes, the RFL530 will work fine at 12V. You just need to keep the gate-source voltage below its maximum rated value as given in the data sheet.

What is the PWM frequency?
 

bertus

Joined Apr 5, 2008
22,278
Hello,

The gate of a mosfet is comparable with a capacitor.
The resistor will reduce the current peak when you are putting the voltage on this gate.

Bertus
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
Yes, the RFL530 will work fine at 12V. You just need to keep the gate-source voltage below its maximum rated value as given in the data sheet.
The datasheet states that the absolute maximum gate-source voltage of the IRL530 is 10V. 12V will exceed the limit.
 

crutschow

Joined Mar 14, 2008
34,468
The datasheet states that the absolute maximum gate-source voltage of the IRL530 is 10V. 12V will exceed the limit.
You are correct. I had not looked at the data sheet and expected a higher limit. :p So you could add a series resistor of 2.5kΩ from the coupler to the FET gate. That, in conjunction with the 10kΩ resistor to ground will reduce the 12v to 9.6V. It will also reduce the 5V to 4V but that's still sufficient to give a 0.22Ω max. turn-on resistance.

If you want the slighter lower on-resistance (0.16Ω) that 5V or more Vgs gives then you could use a smaller series resistor (say 1kΩ) and add a zener to ground (with value somewhere from 5.6V to 8.2V).

As I previously asked, what is your PWM frequency?
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
The transistor base current needs to be at least 1/10th of the 2A maximum motor current or 200mA to be properly saturated (fully ON).
Does this also apply to the 2N2222 transistor? I am going to use this transistor to switch on a relay with a coil current of 185mA. I will connect the the base to the microcontroller with a 1Kohms resistor in series.

5V/1000ohms = 5mA

5mA * 10 = 50mA
(i.e. not enough to for the relay coil)
 

crutschow

Joined Mar 14, 2008
34,468
Does this also apply to the 2N2222 transistor? I am going to use this transistor to switch on a relay with a coil current of 185mA. I will connect the the base to the microcontroller with a 1Kohms resistor in series.

5V/1000ohms = 5mA

5mA * 10 = 50mA
(i.e. not enough to for the relay coil)
The rule of thumb for a BJT transistor is that for full saturated turn-on, the base current should be at least 1/10 of the collector current. If you look at the transistor data sheets, they normally measure the saturation voltage with that base current value. That being said, if you don't need the lowest possible saturation voltage, and you are well below the transistor's current max., and the transistor has a relatively high β, then you could likely go to a somewhat smaller base current such as 1/15th or 1/20th of the collector current. Thus for 185mA collector current I would use no less than about 10ma for the base current.

Note that to calculate the base current you need to subtract the base-emitter voltage from the supply voltage to get the drop across the base resistor. Thus the resistor value for 10mA base current and a 5V supply would be (5V - 0.7V) / 10mA = 430Ω.

I asked about the PMW frequency because I was concerned about the response time for the circuit. 50Hz seems abnormally low for a PWM switching frequency. Is that perhaps the modulation frequency?
 
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