# is is possible to drop voltage without knowing a components amperage

I am new to circuits/electrical stuff

Correct me if i am wrong.

if i have an led (2.1V / 0.02A) and if i give it 0.02A it will only take 2.1V or if i give it 2.1V it will take 0.02A

Ohms law ($$\I=\frac{V}{R}\$$).

If start with 9V i can use ohms law to get the correct power for the led because i know how many amps it requires

My question is:

If i started with 9V. Could i drop the voltage down for a component that takes 5V without knowing how many amps it uses?

Sub-question:

Wouldn't a circuit like this drain a battery or cause the circuit to short out?

• This is the purpose of voltage regulators. – DKNguyen Jul 24 '20 at 14:20
• The voltage specification for an LED is usually the nominal measured voltage when the LED is driven with a specific current. If you drive 20mA you should see about 2.1V across the LED, with maybe 10% variation for different LEDs from the same batch. If you drive 2.1V across the LED you would expect an average current of 20mA across many different parts but the minimum and maximum extremes could be large. In practice, we try to drive an LED with a specified current, not a specified voltage. – Elliot Alderson Jul 24 '20 at 18:10

No need for a voltage dropper with two resistors. Just assume (a fairly accurate assumption I might add) that, the LED has 2.1 volts across it then, put a resistor in series with it to the 9 volt rail. The value of the resistor sets the current: -

$$I = \dfrac{\text{9 volts - 2.1 volts}}{R} = \dfrac{6.9}{R} = \text{20 mA}$$

Rearrange to find R (it equals 345 ohms).

Wouldn't a circuit like this drain a battery or cause the circuit to short out?

It's extremely sub-optimal.

If i started with 9V. Could i drop the voltage down for a component that takes 5V without knowing how many amps it uses?

You could (and waste a lot of power) or, just use a voltage regulator - that is what they are for.