Want to use a small transformer that produces 8 / 16 / 24 VAC from 120VAC line voltage and rectify the 8V AC output to DC to use with a cheap pressure transducer that wants 6.5 VDC input power. Bridge rectifiers are readily available on Digikey.com as small integrated circuits, very inexpensive. There is an online calculator for sizing a smoothing capacitor at <redacted by moderator> but there is no discussion of how smooth the ripple voltage will be made or distinction between the output capacity of the transformer vs. the DC load requirement (in amps), so I don't know if I need a capacitor of a few micro-farads or hundreds of micro-farads. The transformer can produce more than 1 amp DC , but the DC current needed by the pressure transducer is probably only 20 mA. Also not sure how accurate the 6.5 VDC power to the pressure needs to be or what ripple in the DC voltage is tolerable. Would appreciate if someone with experience in these matters can offer some comments. Of course, I can't and don't expect anyone to provide precise answers when they aren't given by the manufacturer of the devices (like the pressure transducer, for example), but I am inexperienced and hope that someone familiar with these topics can offer some comments.

 

Hi WFD,
Welcome to AAC.
This circuit will give +6.5V up to 50mA from a 8Vac Transformer.
This will easily cover the 20mA you require.
Use the TO92 version of the LM317 regulator.
The 4 off, 1N4001 represent your full bridge rectifier
E

 

Use is formula:

C = I / (2 x f x Vripple)

Hence, if I = 1 A, f = 60 Hz, Vripple = 1 V

C = 1 / ( 2 x 60 x 1 ) = 8000 μF

 

Eric –
Thanks for your reply. Seems strange to me to have 2 capacitors in parallel as C3 and C1, likewise C2 and C4. Isn’t C3, C1 equivalent to 1 capacitor of 470.1 uF and C2,C4 equivalent to 1 capacitor of 100.1 uF ? Similarly R1 and R2 equivalent to 1 resistor of 870 ohms ? In the graph Vout is a dead straight line at 6.5 V, but in my application I would think a ripple of, say, 0.2 V which is about 3% would be acceptable. Then capacitor(s) might be the size of my thumb, not the size of my fist ? I anticipate that maybe the downside of having a 3% ripple in Vdd to the pressure transducer might be a ripple of similar percentage in its output signal ? In my application, that much uncertainty could be acceptable and worthwhile if the capacitors would be much smaller. A lot less fun, but simple solution to all this is just to buy a DC wall wart power supply ?

 

Eric –
Thanks for your reply. Seems strange to me to have 2 capacitors in parallel as C3 and C1, likewise C2 and C4. Isn’t C3, C1 equivalent to 1 capacitor of 470.1 uF and C2,C4 equivalent to 1 capacitor of 100.1 uF ? Similarly R1 and R2 equivalent to 1 resistor of 870 ohms ? In the graph Vout is a dead straight line at 6.5 V, but in my application I would think a ripple of, say, 0.2 V which is about 3% would be acceptable. Then capacitor(s) might be the size of my thumb, not the size of my fist ? I anticipate that maybe the downside of having a 3% ripple in Vdd to the pressure transducer might be a ripple of similar percentage in its output signal ? In my application, that much uncertainty could be acceptable and worthwhile if the capacitors would be much smaller. A lot less fun, but simple solution to all this is just to buy a DC wall wart power supply ?

Simple beginner mistake.
Yes, two capacitors in parallel do add. But note that the tolerance of a 100 μF or 470 μF capacitor is about ±20%.
Hence adding 0.1 μF is not noticeable.

What is different is the high frequency response of the smaller capacitor. Physically large capacitors have very large ESR (effective series resistance and inductance). Hence the small capacitor is more effective at suppressing high frequencies.

 

Simple beginner mistake.
Yes, two capacitors in parallel do add. But note that the tolerance of a 100 μF or 470 μF capacitor is about ±20%.
Hence adding 0.1 μF is not noticeable.

What is different is the high frequency response of the smaller capacitor. Physically large capacitors have very large ESR (effective series resistance and inductance). Hence the small capacitor is more effective at suppressing high frequencies.

I realized that just getting 0.1 uF of additional capacity would have been a non-sensical reason and that I must be missing the real reason due to being a newbie. Thanks for adding an increment to my understanding. Maybe similarly (?) - is there a reason I shouldn't just use a 6V DC wall wart ? I do know that wall warts are typically switching power supplies which means that the output voltage is a non-uniform mess compared to the perfectly flat 6.5 V line shown by Eric, but the question is if it will actually matter in my application. I don't know if the cheap (from China via an eBay seller) pressure transducer is sensitive to Vdd being far from perfect and also I can tolerate a few percent error in the pressure voltage signal. I also know from testing that wall wart output voltages are often different than their nominal voltage. Maybe the only way to find out is to do some testing ?

 

I realized that just getting 0.1 uF of additional capacity would have been a non-sensical reason and that I must be missing the real reason due to being a newbie. Thanks for adding an increment to my understanding. Maybe similarly (?) - is there a reason I shouldn't just use a 6V DC wall wart ? I do know that wall warts are typically switching power supplies which means that the output voltage is a non-uniform mess compared to the perfectly flat 6.5 V line shown by Eric, but the question is if it will actually matter in my application. I don't know if the cheap (from China via an eBay seller) pressure transducer is sensitive to Vdd being far from perfect and also I can tolerate a few percent error in the pressure voltage signal. I also know from testing that wall wart output voltages are often different than their nominal voltage. Maybe the only way to find out is to do some testing ?

Most older "wall warts"used to all have transformers inside, and so they did weigh a bit more. The downside is that while their power had much less noise, they were not even close to regulated, while most switcher supplies ARE fairly well regulated, but rather noisy. An older wall wart and a three-terminal regulator such as the LM7806 can provide a regulated six volts which should work with the sensor.

 

The purpose of the 0.1 μF capacitor is not so much about high frequency filtering. It is to prevent the 3-terminal voltage regulator from going into oscillation.