# Help with Amplifying the Output

Hi,

I have a very basic circuit with a resistor and an LED. I connected one of the PWM output pins from Arduino Uno to the resistor and LED. When I measure the current it is far less than what I want.

Let's say when I apply 5V PWM to the circuit, I measure current as 3mA, but I want to have 20mA.

I can't use a resistor with low resistance for now. I need to amplify the output voltage/current somehow.

What is the PWM duty cycle?
How are you measuring the current?

Actually, I'm changing the duty cycle via designed GUI. However, to obtain the measurements I set the duty cycle as %100.

I measured the current with a multimeter by placing the probes between LED and resistor.

Let me give you the details:

I designed a GUI to control pulse width, period and the voltage of the output.

I connected my Arduino Uno to a resistor and LED. I tested my design both with 100ohms and 1k resistors. I wanted to see pulses on LED.

By the way, when I use 100ohms resistor and apply 5V I measure 22mA. But I measure 4.3V instead of 5V.

I need to amplify the output voltage/current somehow.

You can't make the Arduino output more than 5V. You can't control, on the Arduino, how much current a device pulls. The Arduino does NOT push current; devices pull current (as much as they need, which can be more than the Arduino can (safely) supply).

But I measure 4.3V instead of 5V.

How are you powering the Arduino?

I powered up the Arduino with USB adapter.

passas:

An op-amp can be part of a circuit to amplify voltage. But, you don't get something for nothing. You'll get a corresponding reduction in current.

What are you really trying to do, and why are you concerned about the LED pulling enough current?

Your meter will measure RMS (AC sine waves) or DC. It will NOT measure PWM accurately... Some meters will give you the peak, some will give you the average and some (most?) will jump around randomly. And, there are some higher-end meters that will calculate true-RMS for "odd" waveforms like PWM.

By the way, when I use 100ohms resistor and apply 5V I measure 22mA.

That's about right... Assuming 2V across the LED and 2.3V across the resistor. Ohm's Law says 2.3V/100 Ohms = 23mA.

But I measure 4.3V instead of 5V.

You will get some voltage drop/loss through the Arduino... The more load (the lower the resistance and the more current) the more voltage drop you'll get through the Arduino. But, I don't know if 0.7V drop/loss is typical, and I don't know if you've got 5V applied to your Arduino.

No, an op-amp probably won't help. But what's the problem? You said you wanted to 20mA and you measured 23mA.

Thanks for all the replies.

Currently, I am at testing stages of a GUI controlled stimulator circuit. So, the voltage and current values are very important for me. For the stimulator design I am open to suggestions.

What "stimulator design"?

Kind of a medical device: Transcutaneous electrical nerve stimulation (TENS).

Still need help!!

passas:
Still need help!!

Still need to clarify what you need help with. You can’t amplify how much current something draws. Current is not pushed, so you can’t push more of it. You’ve been told what you need to do to amplify voltage.

passas:
Hi,

I have a very basic circuit with a resistor and an LED. I connected one of the PWM output pins from Arduino Uno to the resistor and LED. When I measure the current it is far less than what I want.

Let's say when I apply 5V PWM to the circuit, I measure current as 3mA, but I want to have 20mA.

I can't use a resistor with low resistance for now. I need to amplify the output voltage/current somehow.

How did you measure the current and with what?
You would be best to measure the voltage across the series resistor and using Ohms Law to calculate the current.
Placing a DMM in current mode in the circuit may cause false readings due to the DMM shunt resistor dropping 1V or more across it.
Tom....

Hi passas,

Consider using the arduino to power a transistor, which in turn powers your LED. You should be able to supply more voltage, and thus current: