# Use of DC source in Zener clamping circuit

A question in my engineering degree asked me to design a clamp circuit that gives a 10V 50Hz AC signal a +5V DC bias. I designed the following circuit as my answer:

I assumed 0.6V forward drop for the regular diode, but otherwise I assumed ideal operation. (The capacitance and resistance were chosen so that the capacitor didn't discharge too quickly between cycles). I tried it in a circuit simulator and it worked, however the solution included a DC source:

According to our textbook,

"If the desired clamp voltage requires the diode to be reverse biased, it is necessary to return the discharge resistor to a suitable dc supply voltage to ensure that the diode conducts and performs the clamping operation." ~ Electrical Engineering: Principles & Applications (5th Edition), Allan R. Hambley

This explanation doesn't really help me understand why the source is needed. Assuming the diode branch is ideal except for the 0.6V forward drop, it should have an I-V characteristic that conducts whenever the voltage across it dips below -5V. This should achieve the desired outcome, no? The capacitor would become charged to have 5V across it, as needed. Why do we need the DC source? And why was it 15V? (We were told we only had +/-15V available as a DC source.)

• Perhaps you should ask someone who can have access to the book. Like someone you can meet in person. – Dmitry Grigoryev Jun 18 '18 at 6:51
• The people running my university course don't know the answer themselves... Maybe I could send a letter to the publisher. – hddh Jun 19 '18 at 14:04
• I don't think the publisher is likely to help. Try re-reading the book and see if you can find a paragraph mentioning the DC source. Include it in your question if you want someone to explain it to you. – Dmitry Grigoryev Jun 20 '18 at 8:31
• I have scanned the book. There's no mention. – hddh Jun 20 '18 at 11:28