INTRODUCTION TO ELECTRONICS: Bleeder Current in Multi-Tap Voltage-Divider Circuits

Adding a tap to a series circuit alters the untapped voltage ( V ) output across the resistor ( R ) downstream of it. We will now expand our studies to be inclusive of multi-tap voltage-divider circuits. The objective is to determine the magnitude of the bleeder current ( I₃ ) flowing through the lowermost tap. Please consider the following diagram:

The relationship between currents within each circuit branch is summed up as follows:

I_T = I_RL1 + I₂

I₂ = I_RL2 + I₃

Substituting I₂ into the first equation yields the magnitude of I_T:

I_T = I_RL1 + I_RL2 + I₃ 

Finally,

I_BLEEDER = I₃

I_BLEEDER = I_T – I_RL1 – I_RL2 

In order to determine each current value, the variables that comprise ohm’s law must be solved:

( V / R ) = I

Since we need resistor values to use the voltage-divider formula, we first determine the values of R_B, R_A, and R_T. Solving R_B and R_A requires usage of the double-reciprocal method of adding parallel circuits together. After resistors values have been determined, we then solve for V_A and V_B at points A and B:

R_B = [ R₃RL₁ / ( R₃ + R_L1 ) ]

R_B = [ ( 100kΩ )( 6.2 kΩ ) / 106.2 kΩ ]

R_B = [ ( 100kΩ )( 6.2 kΩ ) / 106.2 kΩ ]

R_B = 5.84 kΩ

We must next add R_L1 to the sum of R₂ and R_B ( R_2+B ) to obtain R_A:

R_2+B = 6.2 kΩ + 5.84 kΩ

R_2+B = 12.0 kΩ

R_A = [ R_L1R_2+B / ( R_L1 + R_2+B ) ]

R_A = [ ( 100 kΩ )( 12.0 kΩ ) / 112 kΩ ]

R_A = 10.7 kΩ

The circuit diagram has now been reduced to a series circuit:

Adding the remaining series resistors gives us the total resistance of the circuit:

R_T = R₁ + R_A

R_T = 12 kΩ  + 10.7 kΩ

R_T = 22.7 kΩ

We are now ready to use the voltage-divider formula to solve for R_A and R_B. Each of these resistances are a fraction of the source voltage. Since R_L1 and R_L2 are attached at points A and B, V_A and V_B are the R_L1 and R_L2 voltage values as well:

V_A = V_RL1 = ( R_A / R_T )( V_s )

V_RL1 = ( 10.7 kΩ / 22.7 kΩ  )( 24 V )

V_RL1 = ( 0.47 )( 24 V )

V_RL1 = 11.3 V

With the voltage and resistance values solved for R_L1, the magnitude of current traveling through it can be determined:

( V_RL1 / R_L1 ) = I_RL1

( 11.3 V / 100 kΩ ) = 113 µA

NOTE: Point B is energetically downstream of point A; thus, the voltage at point B will be a fraction of V_A. When using the voltage-divider formula here, we must use V_A, not V_s! Additionally, the net resistance from point A to ground is the sum of R₂ and R_B:

V_B = V_RL2 = ( R_B / R_2+B )( V_A )

V_RL2 = [ 5.84 kΩ / ( 5.84 kΩ + 6.2 kΩ ) ]( 11.3 V )

V_RL2 = ( 5.84 kΩ / 12.0 kΩ )( 11.3 V )

V_RL2 = ( 0.49 )( 11.3 V )

V_RL2 = 5.50 V

( V_RL2 / R_L2 ) = I_RL2 

I_RL2  = ( 5.50 V / 100 kΩ  )

I_RL2  = 55 µA

There are two methods of solving for the bleeder current ( I₃ ):

I_BLEEDER = I_T – I_RL1 – I_RL2 

I_BLEEDER = I_T – 113 µA – 55 µA

I_T  = ( V_s / R_T )

I_T = ( 24 V / 22.7 kΩ )

I_T = 1.06 mA

I_BLEEDER = 1.06 mA – 113 µA – 55 µA

I_BLEEDER = 892 µA

Additionally, if we divide the voltage drop from point B to ground by R_B, we obtain a similar result:

( V_B / R₃ ) = I₃ = I_BLEEDER 

V_B = V_RL2 = 5.50 V

I_BLEEDER = ( 5.50 V / 6.2 kΩ )

I_BLEEDER = 887 µA

The difference between the two values derived are due to rounding errors. Each value represents a very small number, and for our purposes, the differences between them are negligible.