Rusty Senile - or both??

Been years since I had to do conversions calculations from AC to DC. So let me check my thinking and memory.

AC voltage is the RMS of DC.

The peak to peak in AC is 1.41 x RMS

So if I rectify with full wave bridge I should get 1.41 X VAC (RMS) NOT counting volt drop across rectifiers (1.7 V ??). Rectifier is a full wave bridge purchased from Miller welding equip.

Am building an AC and DC variable power supply for testing motors solenoid valves etc. A lot of this will be bench testing with no load on motor. I dont need precision mostly I just need go no go ie did the thing work? Any measurement will be -- is the current draw about right?

Variable voltage obtained with Variac fused to 10 amp on 120 line. WAS thinking of using shunt on variac output to measure both AC and DC amps. Then the peak voltage thinking has put a stop to this idea.

So here is my thinking.
Load is 1000 W RESISTIVE which will be the same on both AC and DC (less heat loss of course -- assume -- to be neglibible)

AC line volts are 100 and current therefore is 10.

RECTIFIED side
Line volt 141 (no filtering and disregarding the ripple) and current is 7.07.

The shunt would work (will be in the neutral) IF I keep in mind the 0.707 correction.

I think I will use the Fluke VOM for appropriate voltage measurements and current will be obtained with clamp on attachment or the shunt in case of DC.

What am I not thinking of or have forgotten?
Dan Bentler

Load is 1000 W RESISTIVE

Click to expand...

AC line volts are 100 and current therefore is 10.

Click to expand...

The 1000W load merely states the total watts the load is capable of absorbing. What is the actual resistance of the load? There is no THEREFORE involved. The figure you gave is only valid if the load is 10 ohms. Is it?

If it is 10 ohms then the following

RECTIFIED side
Line volt 141 (no filtering and disregarding the ripple) and current is 7.07.

Click to expand...

doesn't work. Yous state the peak rectified voltage. If the load is truly 10 ohms then the peak rectified current will be 14.1

Bernie

Let us start over with 10 ohm load.

I am tired and probably (oviously cant think or type strait)
FIRST if I rectify 100 VAC what is the Dc voltage?

Dan

Isn't it "usable" DC voltage from a full wave rect. is the RMS of AC.. So if AC is 100V effective DC is around 70 V ? (soory don't have a plastic brain handy.

We could get into the flaming match a few years back about DC versus merely unidirectional.

Rectified AC with no filtering will be pulses. With a full wave bridge and ignoring the diode drops (as you suggested) you will get peaks at 141 volts (approx) 120 times per second.

If you place this voltage across your 10 ohm load you will get pulsations of current. The heating effect of this current on your load should be the same (again ignoring the diode drops) as the AC voltage.

In one sense this is DC with some serious ripple. Depending on the type of meter used you could see a wide range of values.

Now, if your original AC supply is capable of lots of current and you use really big capacitors after the bridge you would begin to see some smoothing of the pulsating waveform and the measurement of average and RMS voltage should start moving toward that 141 volts.

So it's kind of a 'it depends' situation.

Bernie
Will be using Fluke 73 VOM for both DC and AC voltage and current measurements. AC current with a clamp on attachment and DC with shunt.
AC supply is a variac fused for 10 amp and output 0 to 140 VAC.

I intend to use this only for bench testing of motors and solenoids to mainly check to see if they run and get a rough idea of what the current is. I am tired of installing 90 VDC motors labeled good that do not run.

I did not include ripple in calcs and that was my mistake I believe.

For now I am going to live with the ripple on the DC.

I think I will go ahead and hook it all up and use known load ie 100 W lite bulb(s) and take a bunch of readings volts amps etc and report back.

At this stage I may install smoothing capacitors.
Thanks for help

Dan Bentler