Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

values and number of capacitors at the input

Status
Not open for further replies.

Achiever

New Member
I'm designing a project that would need a 5 volt DC supply. The supply voltage will be coming from an external AC DC adaptor. The voltage will go to the voltage regulator producing a 3.3Vdc output necessary to the circuit. According to some readings I made the Adaptors available in the market varried on thier quallity as some have very large ripple voltage at the output. In order to support large variations in the ripple voltage bulk capacitors might be required.

My problem is I dont know how to calculate the required values of the capacitors that would be necessary to the design assuming I wanted the device to work with the very low quality adaptor. Also in most designs I have noticed placing different values of the capacitors at the input. Is there a way to determine the correct values and number of capcitors at the input part of the circuit.
 

MikeMl

Well-Known Member
Most Helpful Member
Things I would need to know in order to do the calculation:

Current (i) drawn by the 3.3V load?

Drop-Out Voltage spec (Vdo) for your 3.3V regulator.

Open Circuit voltage (Voc) out of your DC adapter if you hang a big capacitor (~1000uF) across its output?

Does your adapter have full-wave or half-wave rectification?

Line frequency, 50 or 60Hz?

After you determine the answers, we will apply this formula:

C*ΔV = q = i * Δt, rearrange to:

C = i * Δt/ΔV

where C is the required input filter capacitor ahead of the regulator.

i is the load current in A.

Δt is either 8.33ms or 10ms, if 60Hz or 50Hz, full wave rectification. Double these for half-wave.

ΔV = Voc-Vdo
 

Achiever

New Member
Thanks MikeMI you helped me understand it better.

Now it make sense to me. However was it always necessary to place a bulk capacitor across the output of AC to DC adaptor in addition to the capacitor that will be place ahead of the voltage regulator? If for example the calculated value of the capacitor is approx. 100nF would it make a difference if I use capacitors of different values in parallel equal to 100nF.

My chosen AC to DC adaptor is the switchmode type and I'm not sure if its based on full wave or half wave rectification. if I want to support either rectification would it be okay if I always assume that the adaptor is based on halfwave rectification?
 

MikeMl

Well-Known Member
Most Helpful Member
Thanks MikeMI you helped me understand it better.

Now it make sense to me. However was it always necessary to place a bulk capacitor across the output of AC to DC adaptor in addition to the capacitor that will be place ahead of the voltage regulator? If for example the calculated value of the capacitor is approx. 100nF would it make a difference if I use capacitors of different values in parallel equal to 100nF.
Everything I posted assumed that you had an unregulated plug-in wall transformer ("wall wart"). The ones marked DC usually have rectification built in, some have a filter capacitor, some do not. The calculation above would provide filtering and energy storage assuming that there is the classic rectification ripple at the output of the wall wart.

My chosen AC to DC adaptor is the switchmode type and I'm not sure if its based on full wave or half wave rectification. if I want to support either rectification would it be okay if I always assume that the adaptor is based on halfwave rectification?

Since your adaptor is already produces regulated 5V DC, you do not need any additional capacitance between its output and the input to an add-on 3.3V linear regulator. All of the calculations above assumed an "unregulated" adapter.

There is a possibility that you could modify your existing adaptor to output 3.3V instead of 5V. I have "opened up" some adapters, and just by changing one resistor, was able to "retrain" them to put out a different voltage.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top