Calculation behind 90V DC from bridge rectified 120vac

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synchro said:
Agree, except with the peak voltage instead of P-P.


The average of the rectified voltage and its 90% relationship to the RMS input voltage is relevant with choke input filters that have a sufficiently high inductance to keep the current flowing through the rectifier throughout the AC cycle. The choke input filter type has better inherent voltage regulation and less peak current than a capacitor input filter (and therefore less stress on the rectifier). Choke input filters were somewhat common with higher power vacuum tube equipment. But the weight and cost of such a choke usually makes it unattractive, especially compared to modern switching supplies.
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Some early tube radios and phonographs used a coil which also provided the stationary magnetic field of the speaker as a choke. You did not want the speaker to be very efficient at or below 60Hz to minimize AC hum from current in the "choke".
 
GoldDigger said:
Some early tube radios and phonographs used a coil which also provided the stationary magnetic field of the speaker as a choke. You did not want the speaker to be very efficient at or below 60Hz to minimize AC hum from current in the "choke".
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Ah, yes. At the time those were called "electrodynamic" speakers, although today that is a more general term. Those disappeared after Alnico permanent magnets became available.

 
210216-1009 EST

The title of the thread is 90 V not 90%. I suspect that 90 V is sufficiently low that for some 120 V 60 Hz sine wave sourced systems that one can get adequate torque from the motor with various mechanical loads and actual input voltages.

synchro:

I have one of those electrodynamic speakers sitting on top of one of my benches. Have another one or more in radios.

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I'd always assumed ... and we know what that means ... that 90V and 180V permanent magnet motors were designed to operate across the world where 100V, 110V, 115V, and 120V (and double those in some systems, we're not getting into split phase, balanced, and other things here) exist. I've only been around this stuff since the mid-60s but think those "90/180 standards" predate then. I'd assumed the voltage drop on selenium rectifiers came into play as well. We've REAL engineers, @synchro and @gar come to mind, not just little 'ol me with an EE degree but little real EE experience, and my degree is more electronics than power, or what it was in 1970.
 
210216-1653 EST

Following are some measurements on two different motors:

All measurements were with 123 V 60 Hz AC sine wave source. Either a half wave or full wave rectifier was the source. No filtering except the motor and its effect. Current measured with a Fluke Hall device in DC mode. The scope was 2 mS/cm synced from the AC line, and amplifiers set in DC mode. A Rigol scope was the measuring instrument. No external mechanical loading on the motors. Single shot waveforms were recorded. Motors were at steady state when recorded. Both motors were around a nominal 120 V rating.

One motor was a Bodine 24A4BEPM 3F. This is a 130 V DC 0.48 A 1/17 HP 10 to 1 250 RPM unit with an integral worm gear reducer.
Other motor was and Electrocraft 065200 004 servo motor with no gear box, but did include an integral DC tach, No major friction load.

When you look at the voltage and current at the motor terminals the following is seen:
The current waveform is a short sort of sinusoidal pulse close to in phase with the source AC voltage peak, slightly lagging.
The voltage waveform is the top of a sine wave, turns on at the counter-EMF voltage, and continues below the counter-EMF voltage before turn off because of stored energy in the motor armature inductance. When no current is flowing to the motor during the cycle the amature is working like the armature of DC generator with no load, and is a good measure of armature RPM. These characteristics are easily seen when viewing scope plots. There are many designs from a long time ago that used this counter-EMF voltage as a tachometer for speed control.

Results of my experiments:

Bodine motor:
Single phase half wave rectifier.
Current pulse width = 5.2 mS. Off time 11.2 mS. Sum 16.4 mS, close enough.
175 V peak. Counter-EMF = 125 V.
Ipeak = 0.6 A .
Inductive kick 60 V below counter-EMF.

Bodine motor:
Single phase full wave rectifier.
Current pulse width = 2*4.6 = 9.2 mS. Off time 2*3.7 = 7.4 mS. Sum 16.6 mS, close enough.
175 V peak. Counter-EMF = 140 V.
Ipeak = 0.4 A .
Inductive kick 50 V below counter-EMF.

Electrocraft motor:
Single phase full wave rectifier.
Current pulse width = 2*3.3 = 6.6 mS. Off time 2*5.0 = 10.0 mS. Sum 16.6 mS, close enough.
175 V peak. Counter-EMF = 160 V.
Ipeak = 1.0 A .
Inductive kick 30 V below counter-EMF.

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Hmmm!

After all this conundrum and obvious exemplary articulation--none of those (introspection) come close to OP's curious wont that asks:
“Calculation behind 90V DC from bridge rectified 120vac”
Where is the calculation?

Can this be expressed in quadratic equation outside Alnico magnets and Bodine motors that will result in 90 volts output after rectification from outdated selenium rectifiers , MOVs and modern solid state semi-conductors?
Hence, OP is curious on how we could end up with higher output compared to the input voltage. I've done (or been ) playing with this stuff too with simple test equipment and using capacitors and low voltage transformers. I always come up with the same result. . . higher output.

Pure DC is higher than AC in terms of usage benefit in 90 volts DC-- as opposed to 120 volts AC.

Forget the peak-to-peak unrectified voltage . . .just show in simple terms with a higher DC output through the combined roles of the rectifier and capacitance.

Show the mathematical expression. . . .not what you get from the bench toys.

That's what OP is asking.
 
210216-2123 EST

myspark:

There is no calculation based on circuit theory that gives you 90 VDC from a 120 V AC sine wave for a simple circuit. Can I make some circuit to do it? Yes.

If you have a full wave rectified 120 V sine wave and do something to get its average value, then the result is 120*0.636/0.707 = 107.9 V .

If I start with some sort of chopped +/- DC voltage that has an RMS AC voltage of 120 V, and appropriately adjust the pulse widths and heights, and full wave rectify this. then I can produce a an average output voltage of 90 V DC. That does not normally exist in nature.

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210216-2144 EST

Starting from post #1.

"90V DC motors are common and are intended to be thyristor controlled off bridge rectified 120vac mains. For simplicity sake, ignoring the foward voltiage drop, what is the math behind 90v DC output? Is that 90v DC average or RMS?"

The DC motor is going to roughly work based on the average DC voltage, not RMS. But you need to look at the input voltage, including shape, and its relationship to the motor counter-EMF. If it is a pulsating voltage, then it is a function of impedances and the amount of energy pushed into the motor relative its total mechanical load and efficiency.

"I would think that output (ignoring the 2 x VF drop of 2-3v) would still be 120V RMS, or in other words, 120v 600W heating element would still absorb 600W of real power when fed through a rectifier, but if fed from a 90v battery, it would only operate at 56% power. (90/120)^2."

For a full wave rectifier correct, and 56% is correct.

A resistive load fed from an ideal half wave rectifier will dissipate 1/2 the power of the same load with ideal full wave rectification or direct, no rectifier, drive from the AC source. This full power is Vrms^2/R . 90 V DC is equal to 90 V RMS into a resistive load.

"Does a 90v DC rated motor fed on a 90v battery pump the same cfm of air against a constant torque load (same psi diff) as when its powered from 120 vac through a bridge?"

No. It pumps more from the 120 V bridge.

.
 
210217-2234 EST

How many of you that responded to this thread understand the basic theory of how a permanent magnet field, commutated armature, DC motor works with various input voltage waveforms, including AC and DC? If you change the motor to a wound field unit with separate DC excitation to the field, then how does this change the equations?

.
 
210218-1225 EST

What is an equation for the shaft speed of a permanent magnet DC motor supplied from a variable voltage DC source, and for various mechanical loads on the motor? Change to essentially the same motor, but with a wound field that you can adjust the current to the field independent of the applied voltage to the armature, and now what is a useful equation?

This is basic knowledge that any electrical engineer should know, and might be useful for any electrician that works with motors.

What happens if you loose field excitation to a DC motor while the motor is running, and this motor has armature voltage still applied at the time of loss of the field?

.
 
gar said:
What happens if you loose field excitation to a DC motor while the motor is running, and this motor has armature voltage still applied at the time of loss of the field?
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The field starts putting out voltage as a generator?
 
S'mise said:
We use the term thyristor here Stateside also. Thyristor is just a broader category. It can also mean triac, diac, or several other types of components.
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That may be so for you. For us, it would not be a triac or diac or igbt or the like. Just tje thyristor. As it happens I have a book on my shelf by J.M.D. Murphy on the subject of thyristors, Dated 1973.

I digress.
 
210218-2120 EST

paulengr:

In post #5 you said:

"RMS sort of can be used but it has no meaning in DC."
RMS does have meaning in both AC and DC, and it is the same in both.

"In DC motors the field is almost immaterial."
The field is very important in a DC motor, and must be there. And its field intensity (magnitude) is extremely important.

"The armature power minus counter EMF which is usually under 10% so it’s a small “loss” is the power output."
No. But power output (shaft mechanical power) is slightly less than power input to the armature (input V * I to motor armature) minus armature I^2*R.
There is also windage and other losses that subtract from input power.

"So since the motor is almost just a resistor armature voltage is approximately equal to speed in strong field (field weakening destroys this relationship)"
Internal armature voltage is K*field intensity*RPM. This must equal applied voltage to the armature minus the armature I*R voltage drop.
Reduction of field intensity increases shaft speed for constant armature voltage and mechanical load.

"armature current is approximately proportional to torque"
Correct.
 
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