# Current limiter for VOD3120AB

I want to supply VOD3120AB with a voltage of 24v/2A from Power Supply. What is the value of the Resistor that I must add to the VCC of the VOD3120AB so that the VOD3120AB works properly?

Information from the datasheet like this:

1. Wide operating VCC range: 15V to 30V
2. ICC = 3.5 mA maximum supply current
3. Output power dissipation Pdiss 250mW

I tried to calculate like this:

R = V/I = 24/0.0035 = 6857 ohm

P = I^2 * R = (0.0035^2) * 6857 = 0.084 watt

I must select R with resistance 6857 ohm and power dissipation = 0.084 watt. Maybe Resistor 6k8/0.125 watt, Its correct?  What is the value of the Resistor that I must add to the VCC of the VOD3120AB so that the VOD3120AB works properly?

You are thinking about this incorrectly. Just supply the device with 24 volts and that is sufficient. It'll take a quiescent current of less than 3.5 mA and it's job done. Test circuit from data sheet: - The maximum power dissipation is the losses of the device when driving a fully rated load such as a MOSFET's gate-source capacitance. Make sure you don't exceed this value by choosing your MOSFET and operating frequency carefully.

• Thanks for point out Jun 21 at 20:20

You can apply 15 V to 30 V to the Vcc pin, and the device will draw whatever current it requires - no need for any current limiting resistor there.

However, the input seems to be an LED, so you will need to add a current-limiting resistor in series with it to limit the LED current to 7 - 16 mA, as stated in the "Recommended Operating Condition" table.

• Thanks for point out. Jun 21 at 20:22

If you're referring to powering the LED from 24V, first pick the desired current, say 20mA and use Ohm's Law.

$$E = I \cdot R$$

$$24V = 0.02A \cdot R$$

Then realize that a forward-biased LED is going to drop some voltage, say 1.8V when powered at 20mA. So subtract this from the supply voltage.

$$24V - 1.8V = 0.02A \cdot R$$

$$22.2V = 0.02A \cdot R$$

Then rearrange to solve for R by dividing both sides by current (the opposite of multiplying current, so R is left by itself on the right side.)

$$\frac{22.2V}{0.02A} = \frac{\require{cancel}\bcancel{0.02A} \cdot R}{\bcancel{0.02A}}$$

$$\frac{22.2V}{0.02A} = R$$

$$1110 \Omega = R$$

• I have to keep the input (left) and output (right) with different ground (different power supply), because the input side will be connected to the microcontroller with a voltage of 3.3v and a current of 10mA, while the output side will be connected to a voltage of 24v and a current of 2A . Intentionally isolated so that the large electric current from the MOSFET does not damage the microcontroller. Thank you. Jun 21 at 20:27