Clamper circuits - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: February 17, 2014, 05:06:54 pm
content: I'm having trouble understanding how clamper circuits work. While I understand what they do just fine, I don't understand how they do it. To keep things simple, I'm assuming a proper cap value has been selected for the circuit frequency and ignoring any effects a load might have on the circuit. The 5V is a peak measurement.On the positive half of the cycle the current passes through the diode and a voltage is built up across the capacitor equal to 5V - 0.7V = 4.3V. The right side of the capacitor should have a positive charge built up. My voltage across the diode (Vo) should be 0.7V.On the negative half of the cycle the diode does not allow current flow. The voltage on the capacitor (4.3V) plus the source voltage (5V) should add and a 9.3V drop should exist across the diode. So I get a 10V peak to peak sine wave that alternates between -0.7V and 9.3V.Is that a correct summation? Just typing this up has helped, but I'd appreciate if someone could verify if this is correct or I have something wrong.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 17, 2014, 05:52:50 pm
content: Yup!

author: Legion
author_rank: Frequent Contributor
timestamp: February 17, 2014, 07:08:31 pm
content: When I try to apply the same logic for a negative clamped circuit it doesn't seem to work.All that's changed is the direction of the diode.For the first 90 degrees the voltage across the diode goes from 0V to -5V. The next 90 degrees is -5V to 0V. Somehow we've lost half of our voltage.On the negative half of the cycle the capacitor charges, but does not allow current to flow through the diode. I'm not clear how this affects the output voltage.Something is wrong with how I'm approaching the negative clamper, but I can't see it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 17, 2014, 09:24:00 pm
content: The arrows on the diagram get reversed, of course.Suppose you get a positive half-cycle followed by a negative half-cycle... at the diode, the positive half-cycle is shunted to ground, so the cap charges to +4.3V (positive towards the voltage source). On the negative half-cycle, the cap remains charged to 4.3V, so the negative peak becomes -9.3V.Tim

author: Legion
author_rank: Frequent Contributor
timestamp: February 18, 2014, 02:20:09 am
content: Quote from: T3sl4co1l on February 17, 2014, 09:24:00 pmCan you explain what you mean by shunted to ground? During the positive half cycle the diode acts like an open switch. You say the cap charges to +4.3V on the voltage source side during the positive half cycle, but shouldn't the +4.3V be on the right side of the cap (right side of the diagram)? The current flows counter clockwise during the positive, so wouldn't the capacitor build up a positive charge on the right side?For such a simple circuit I find it very confusing. Especially the changing polarities and how they relate.

author: Nerull
author_rank: Frequent Contributor
timestamp: February 18, 2014, 03:08:00 am
content: I think you've got them backwards. In your first circuit - a positive clamper - the diode conducts and the capacitor charges, with the positive charge on the right side, during the negative half-cycle. During the positive cycle the diode stops conducting and the voltages on Vin and the right side of the capacitor add.The second is a negative clamper. During the positive half-cycle, the diode conducts and the capacitor charges to Vin on supply side - meaning the charge on the right side of the capacitor opposes the voltage on the input. During the negative half-cycle, the diode blocks and the negative voltage and charge on the right side of the capacitor add.

author: Legion
author_rank: Frequent Contributor
timestamp: February 18, 2014, 03:13:08 pm
content: Quote from: Nerull on February 18, 2014, 03:08:00 amBut how can the diode conduct during the negative half cycle? It's reverse biased.