What is the dissipation of R1

[richsf]

I'm new to using the kicad/ngspice simulator.  I was using it to calculate the power dissipation on R1.

Not sure if I'm doing it right.  Of course this type circuit has large current spikes in the middle of the AC waveform.

How many watts will R1 dissipate ?

[ledtester]

How many watts will R1 dissipate ?

For a resistor, P = I^2*R.

In this case P = (110mA^2 * 1K) = 12.1 W.

Not sure if I'm doing it right.  Of course this type circuit has large current spikes in the middle of the AC waveform.

The current through R1 oscillates between 101 and 112mA, power between 10W and 12.5W.

[wasedadoc]

How many watts will R1 dissipate ?

For a resistor, P = I^2*R.

In this case P = (110mA^2 * 1K) = 12.1 W.

Not sure if I'm doing it right.  Of course this type circuit has large current spikes in the middle of the AC waveform.

The current through R1 oscillates between 101 and 112mA, power between 10W and 12.5W.

R1 is 75 Ohms, not 1k.

[SiliconWizard]

You are using a simulator. Why don't you use it? Just measure the current through R1, as you did for R3, and use P = R1*I(R1)^2 as ledtester showed you.

Note that the average current through R1 should be close to that of through R3, assuming that the capacitor is ideal. (With real components, the capacitor would dissipate more power.) So, that should be about 0.9W.

I'll add some correction though: to estimate the power dissipation, the power is not the average current squared times the resistance. This is true for DC, but you're in an AC situation here.
For R3, since the voltage across it is rectified and filtered, the approximation would be, okish.

But for R1, you need to use RMS values.

[richsf]

Thank you both.

For R1 I got 9.6 Watts RMS from my simulation, using a measurement function.  Although you gave the power dissipation for R3 you also gave an answer for power of R1 that is reasonably close to what I got.  So, my simulator passes a sanity test.

I added the waveforms image. 
Orange vC1
Blue is powerR1
White is iR1

This is a common circuit in old antique radios, and I realized I^2R would not work on R1, but that simple approach works on R3 of course because it is DC.

If I used my trueRMS meter to measure RMS amps through R1 and RMS voltage across it, would those values give my a true answer using I^2*R ?

Without the simulator, is there any easy way to calculate the power dissipation in R1 ?

[richsf]

SiliconWizard, that is what I did first.  Current for elements in series is all the same, so current through R1 is about 100 mA.  So that would work out to 0.75 watts.  But, in a real circuit, that resistor gets way hotter than 3/4 watt.  The simple i squared R isn't right due to the waveform shape.

thanks for the comment.

[Wallace Gasiewicz]

First of all the diagram is wrong.  A 1/2 wave rectifier will put out V max / 3.14 .   
So the Volts is 51.6. Close to 1/2 of the AC V RMS, since it only rectifies half the wave.  According to some guy named Ohm, 51.6 divided by resistance , 1150 Ohms  equals current which comes out to .045 (45 mA) the diagram has 101 mA, which I think is quite wrong  and misleading. They are using the RMS input as the DC volts, which is entirely wrong.
Anyone agree?

[SiliconWizard]

SiliconWizard, that is what I did first.  Current for elements in series is all the same, so current through R1 is about 100 mA.  So that would work out to 0.75 watts.  But, in a real circuit, that resistor gets way hotter than 3/4 watt.  The simple i squared R isn't right due to the waveform shape.

thanks for the comment.

I edited my post, sorry for the shortcut, and your post raises a more valid question than initially looked.
Yes, you need to use RMS values here, as this is AC. Only for pure DC would the P = R*I^2, with I = average current, be correct.

So, the RMS measurement is the one that you need here, and yes, this gives a much higher power dissipation in R1 (while the difference for R3 is much smaller, as it's "almost" DC).

Note that if this is a practical circuit, powered say, from mains, you may want to be careful with your components. The capacitor, in particular, should be rated appropriately. And, as you saw, it's not a particularly efficient way of rectifying a high voltage, with these values.

[Andy Chee]

This is a common circuit in old antique radios,

This might shed some light on rectifier circuit topologies, including your half-wave circuit.

https://www.sound-au.com/articles/rectifiers.htm

[Harry_22]

The task is complicated by the capacitor. It quickly charges through a 75 Ohm resistor during the active half-wave and passes energy further during the second part of period.

It should be noted that the values ​​in this task are specifically selected for P= 4 W answer.

Microcap file in attachment.

[Stray Electron]

Harry,

  Should the second trace (the green one) begin at 0.00 seconds and not 0.20 seconds?

   Oh wait,  I guess that it's off of the top of the scale so never mind. 

[floobydust]

It's two things to worry about.
Average power dissipation I simulate to ~4.1W at steady state.
But the massive inrush current IF you happen to plug in when mains is at peak voltage is a concern.
Over 200W peak (1.6A) for the first half-cycle. This is hard on the resistor until the filter cap charges up.
I have forgotten the wirewound resistor element time-constant, would have to dig for that. You don't want the wire to fuse open-circuit.
Reasonable overload for wirewound resistors is 10x, so a 20W part would be fine I think.
Unless you like to gamble, then run a 10W part.

Is there a mains fuse, for safety coverage? It should be a fusible, flameproof part. MOX can be purchased like that but they take much less inrush compared to wirewound.

[richsf]

Thanks for the help.  Especially taking time to run the simulations.

History:  I should say, this is not a new design.  It is typical of many vacuum tube radios in the 1940s - 1960s where making a line operated power supply without a transformer was very common and cheap -- many millions were made, they were 30-40 watts, it would not pass safety rules today.  There was a point in time where the industry made a transition from using vacuum tube rectifier to using selenium rectifiers.  That was a big advantage since there was no power wasting rectifier filament.  The selenium rectifiers had a forward voltage of 5-7 volts for a 100 mA device, but the vacuum tube rectifiers had 20 or more volts lost.  1N400x had not yet been invented.

Clearly R1 dissipates much more power than R2, even though at first glance they both are in the same current path. 

I am still somewhat confused about one thing.  In the simulation I get the power pulse each cycle as you guys did, and when I measure as average I get about 4 watts which several of you did. 

* The question  now is whether AVERAGE is the right measurement for calculating the dissipation of the resistor, or should I use RMS power calculation which I get 9.6 watts ?

Also, in my start time and end time I used start = 400ms (integral cycle aligned) by which time things are stable, and end = 483.333 ms which is also an even 1/60 second frame.  RMS calculations require a window of integral half cycles I think.

This result, even 4 watts, is surprising to me -- initially I was thinking in terms of I^2 * R .  It was shocking, pun intended.

[Harry_22]

And it’s not for nothing that you installed two 75 ohm resistors .

Look at the paradoxical situation:

Two identical resistors are placed in series.
Both can be referred to internal source resistance.
But the dissipated power differs to five times! 

[Stray Electron]

Thanks for the help.  Especially taking time to run the simulations.

History:  I should say, this is not a new design.  It is typical of many vacuum tube radios in the 1940s - 1960s where making a line operated power supply without a transformer was very common and cheap -- many millions were made, they were 30-40 watts, it would not pass safety rules today.  There was a point in time where the industry made a transition from using vacuum tube rectifier to using selenium rectifiers.  That was a big advantage since there was no power wasting rectifier filament.  The selenium rectifiers had a forward voltage of 5-7 volts for a 100 mA device, but the vacuum tube rectifiers had 20 or more volts lost.  1N400x had not yet been invented.

Clearly R1 dissipates much more power than R2, even though at first glance they both are in the same current path. 

I am still somewhat confused about one thing.  In the simulation I get the power pulse each cycle as you guys did, and when I measure as average I get about 4 watts which several of you did. 

* The question  now is whether AVERAGE is the right measurement for calculating the dissipation of the resistor, or should I use RMS power calculation which I get 9.6 watts ?

Also, in my start time and end time I used start = 400ms (integral cycle aligned) by which time things are stable, and end = 483.333 ms which is also an even 1/60 second frame.  RMS calculations require a window of integral half cycles I think.

This result, even 4 watts, is surprising to me -- initially I was thinking in terms of I^2 * R .  It was shocking, pun intended.

   The difference in power in R1 vs R2 is due to the fact that C1 pulls more current through R1 on the positive half cycle as compared to R2.   Large AC powered transformer based welders have a similar dilemma when attempting to use Power Factor Correction Capacitors in the input.  The PFCCs are very desirable when the welder is under load but then the welder is unloaded or lightly loaded, the PFCCs attempt to pushes power back into the AC grid. That results in power flowing both ways and very high current levels in the wiring that's powering the welder.  IIRC the manual for the Miller Syncrowave 300 welder talks about it in slight detail but Harry's simulation demonstrates it much more clearly. 

[Harry_22]

* The question  now is whether AVERAGE is the right measurement for calculating the dissipation of the resistor, or should I use RMS power calculation which I get 9.6 watts ?

We have to use RMS voltage or current measurements in AC for power calculation. This is what they were invented for.

The power for DC was                    U²/R                             or    I²*R
and the power for AC has become   AVG(U²))/R = U_RMS²/R   or    AVG(I²)*R = I_RMS²*R

Please see the power calculation via RMS:

[Harry_22]

   The difference in power in R1 vs R2 is due to the fact that C1 pulls more current through R1 on the positive half cycle as compared to R2. 

Ok, we disconnect R3 load and there is no current via C1 (R1).

So no small current via load no big current via R1  .

[3roomlab]

using idt() ~ 2.7w

[Wallace Gasiewicz]

I don't use simulators.  Is it possible that someone could simulate the circuit without the Ground??  I just bread boarded the circuit and applied 10V P-P sine. A 1/4 W R1 does not even get warm. I have lots of measurements also, if anyone interested.

I get exactly what Harry got with an open circuit.

[richsf]

Yes Harry22, I noticed that a resistor in a radio was hotter than 'made sense' so I knew there was something wrong with my understanding.  Thank you and others for helping me get my mind right !  Picking two same value resistors demonstrates that.  Easy to 'under-think' it.

Two identical resistors are placed in series.
Both can be referred to internal source resistance.
But the dissipated power differs to five times!

I still have to review a couple more times the posts in these threads, but I have a much better understanding now.

Thanks again.  And, what a great calculator spice type simulators are !  It can give answers even when you don't believe it.

And Wallace, what you did is 'real world simulation'.  I'm not sure what you mean by simulate without a ground, but the simulator requires a ground reference to make the math work.  Even if you have an isolated primary and secondary in a transformer, you have to connect the secondary ground to the primary ground, though you can use a 10 Meg resistor to do it so there is no consequence in the results.