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An introduction to electronics
An introduction to electronics

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3.1 Voltage dividers

Voltage dividers are widely used in electronic circuits to create a reference voltage, or to reduce the amplitude of a signal. Figure 12 shows a voltage divider. The value of cap v sub out can be calculated from the values of cap v sub cap s, cap r sub one and cap r sub two.

Described image
Figure 12  A voltage divider circuit
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This shows a voltage divider circuit, consisting of a voltage source VS and two resistors in series, R1 and R2. Vout is measured between the two resistors.

Figure 12  A voltage divider circuit

In the first instance, let’s assume that cap v sub out is not connected to anything (for voltage dividers it is always assumed that negligible current flows through cap v sub out). This means that, according to Kirchhoff’s first law, the current flowing through cap r sub one is the same as the current flowing through cap r sub two. Ohm’s law allows you to calculate the current through cap r sub one. It is the potential difference across that resistor, divided by its resistance. Since the voltage cap v sub cap s is distributed over two resistors, the potential drop over cap r sub one is cap v sub cap r times one equals cap v sub cap s minus cap v sub out.

The current through cap r sub one (cap i sub cap r times one) is given by

cap i sub cap r times one equals left parenthesis cap v sub cap s minus cap v sub out right parenthesis divided by cap r sub one

Similarly, the current through cap r sub two is given by

cap i sub cap r times two equals cap v sub out divided by cap r sub two

Kirchoff’s first law tells you that cap i sub cap r times one equals cap i sub cap r times two, and therefore

cap v sub out divided by cap r sub two equals left parenthesis cap v sub cap s minus cap v sub out right parenthesis divided by cap r sub one

Multiplying both sides by cap r sub one and by cap r sub two gives

cap r sub one times cap v sub out equals cap r sub two postfix times left parenthesis cap v sub cap s minus cap v sub out right parenthesis

Then multiplying out the brackets on the right-hand side gives

cap r sub one times cap v sub out equals cap r sub two times cap v sub cap s minus cap r sub two times cap v sub out

This can be rearranged to

cap r sub one times cap v sub out plus cap r sub two times cap v sub out equals cap r sub two times cap v sub cap s

giving

left parenthesis cap r sub one plus cap r sub two right parenthesis times cap v sub out equals cap r sub two times cap v sub cap s

and therefore the fundamental result is obtained:

cap v sub out equals cap r sub two times cap v sub cap s divided by left parenthesis cap r sub one plus cap r sub two right parenthesis

SAQ 3

Suppose cap v sub cap s = 24 V and cap r sub two = 100 Ω. You want cap v sub out = 6 V. What value of cap r sub one do you need?

Answer

Rearranging the equation for cap v sub out gives

cap v sub out times left parenthesis cap r sub one plus cap r sub two right parenthesis equals cap r sub two times cap v sub cap s

and therefore

cap r sub one plus cap r sub two equals cap r sub two times cap v sub cap s divided by cap v sub out

which means the equation for cap r sub one is

cap r sub one equals cap r sub two times cap v sub cap s divided by cap v sub out minus cap r sub two

Substituting in the values given,

cap r sub one equals 100 postfix times normal cap omega multiplication 24 postfix times cap v divided by six postfix times cap v minus 100 postfix times normal cap omega equals 400 postfix times normal cap omega minus 100 postfix times normal cap omega equals 300 postfix times normal cap omega