Trying to understand this biasing circuit for the voltages

This is a boost regulator 3.3 to 10 V where the switched output pulses (SW) feed above two linear regulators using pulse rectified doublers to emitter followers after Zeners.

The left side uses D5 as an AC coupled +ve clamp and doubler to get -20 V to drive the 7.5 V zener low to get -7 V out for a low current load.

The right side is similar but uses D4 to create a negative clamp and doubler to create +20 V to drive the 18 V Zener to get 17.5V out. The load is just a constant bias for the LCD to define black / white thresholds or similar effect.

Design details are in the datasheet.

D.A.(Tony)Stewart said:

The left side uses D5 as an AC coupled +ve clamp and doubler to get -20 V to drive the 7.5 V zener low to get -7 V out for a low current load.

Click to expand...

The left charge pump doesn't double the voltage to -20 V. The left charge pump inverts the input voltage to -10V. That -10 V (across C2) is applied to a series regulator formed by a zener diode and PNP pass transistor.

When you take into account the zener voltage and Vbe of the transistor you get around -7 V.

BlackHelicopter said:

The left charge pump doesn't double the voltage to -20 V. The left charge pump inverts the input voltage to -10V. That -10 V (across C2) is applied to a series regulator formed by a zener diode and PNP pass transistor.

When you take into account the zener voltage and Vbe of the transistor you get around -7 V.

Click to expand...

I missed that the right side rectifier clamp D4 was referenced to the +10V rectified pulses output added 10V to the output to get 20 another way.
yes left side is not a doubler.

FreshmanNewbie said:

In this link, can someone explain how the circuit in Figure 1 on page 3 works?



source: Texas Instruments TIDU683, "LCD Bias Power Reference Design with TPS61085"

Click to expand...

Which part do you not understand?
The Boost Reg?, The pulse Clamp and rectifiers? or the Zener regulators?