There have been a couple of threads lately about solid state relays that got me thinking. Here is my latest crazy idea...
I have always been interested in trying to use a lower VA, higher voltage transformer in a linear power supply compared to what would normally be used. By higher I mean secondary voltage that is maybe 20Vac higher and VA is half. Of course a smaller transformer will be lighter, cheaper, but less well regulated. This is why you need a higher secondary voltage - to make up for the increased sag at higher current draw. A regular old linear power supply is made from the transformer, a rectifier, and some smoothing caps. This will have a much higher voltage than what is actually needed (e.g. for the amplifier rails) so the question is how to reduce the DC voltages down to what is practical.
The latest idea for this is a PWM driven SSR. "Fast" SSRs exist that have of and off switching times that are 1msec or less. At 1ms you could just get away with a 1kHz switching frequency. The advantage of this is that you can filter out the switching pulses with a simple downstream RC (the C is needed anyway). As the switching frequency rises, the required R value falls for a given attenuation, and smaller R means lower power loss in the resistor. Compared to CRC filtering at 50/60Hz, this would be a big improvement.
But how to you drive the PWM to control the SSR "on" to "off" ratio? This can easily be done with a microcontroller that monitors the downstream rail voltages and adjusts the PWM to "regulate" the downstream cap voltage. The PWM must be able to have a very low or off mode so that under no-load conditions the downstream charge (and this voltage) in the cap will not slowly rise. A bleed resistor will provide a minimum required current draw. Since microcontrollers typically output a 5V digital signal natively and SSRs can be made to turn on with as little as 3V and are optically isolated, you can just directly drive the SSR with the microcontroller.
There will be some voltage drop across the SSR, but this is usually about 1V and SSRs are designed with large heat transfer surfaces so any dissipated power can be easily removed.
Using this approach, I can envision a power supply that would have a very stable output (due to SSR regulation and RC filtering). The output voltage could be tailored to the particular needs of the application over a pretty wide voltage range while using a single transformer just by adjusting the "setpoint" in the microcontroller.
One real advantage of this approach is that under high current draw the conduction angle thru the rectifier will increase and this might increase efficiency a bit and help in other ways, too.
Does it make sense to build something like this, even if just for amusement purposes?
I have always been interested in trying to use a lower VA, higher voltage transformer in a linear power supply compared to what would normally be used. By higher I mean secondary voltage that is maybe 20Vac higher and VA is half. Of course a smaller transformer will be lighter, cheaper, but less well regulated. This is why you need a higher secondary voltage - to make up for the increased sag at higher current draw. A regular old linear power supply is made from the transformer, a rectifier, and some smoothing caps. This will have a much higher voltage than what is actually needed (e.g. for the amplifier rails) so the question is how to reduce the DC voltages down to what is practical.
The latest idea for this is a PWM driven SSR. "Fast" SSRs exist that have of and off switching times that are 1msec or less. At 1ms you could just get away with a 1kHz switching frequency. The advantage of this is that you can filter out the switching pulses with a simple downstream RC (the C is needed anyway). As the switching frequency rises, the required R value falls for a given attenuation, and smaller R means lower power loss in the resistor. Compared to CRC filtering at 50/60Hz, this would be a big improvement.
But how to you drive the PWM to control the SSR "on" to "off" ratio? This can easily be done with a microcontroller that monitors the downstream rail voltages and adjusts the PWM to "regulate" the downstream cap voltage. The PWM must be able to have a very low or off mode so that under no-load conditions the downstream charge (and this voltage) in the cap will not slowly rise. A bleed resistor will provide a minimum required current draw. Since microcontrollers typically output a 5V digital signal natively and SSRs can be made to turn on with as little as 3V and are optically isolated, you can just directly drive the SSR with the microcontroller.
There will be some voltage drop across the SSR, but this is usually about 1V and SSRs are designed with large heat transfer surfaces so any dissipated power can be easily removed.
Using this approach, I can envision a power supply that would have a very stable output (due to SSR regulation and RC filtering). The output voltage could be tailored to the particular needs of the application over a pretty wide voltage range while using a single transformer just by adjusting the "setpoint" in the microcontroller.
One real advantage of this approach is that under high current draw the conduction angle thru the rectifier will increase and this might increase efficiency a bit and help in other ways, too.
Does it make sense to build something like this, even if just for amusement purposes?
To be honest, you will probably encounter more problems than fun with that idea.
What you intend to do could be done simply and effective with an unregulated supply as you mentioned followed by some buck converter.
This is not new at all - but it works😉
What you intend to do could be done simply and effective with an unregulated supply as you mentioned followed by some buck converter.
This is not new at all - but it works😉
You can make an AC buck converter with two ac mosfet SSRs and then feed the transformer through an LC filter.
However there is little point in doing this.
If you want a more reasonable operation, build a buck converter after the 60hz transformer and rectifier.
you don't need a micro controller, and there are hundreds of half bridge drivers for buck and other topologies.
no need for a micro controller.
However there is little point in doing this.
If you want a more reasonable operation, build a buck converter after the 60hz transformer and rectifier.
you don't need a micro controller, and there are hundreds of half bridge drivers for buck and other topologies.
no need for a micro controller.
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I'm just using what I know.
I'm not implying in any way, shape, or form that this is new or "better". Otherwise companies would be building products this way!
I don't want to make something that is using an inductor to store charge... My concept PS doles out charge in "chunks" controlled by the "on" time of the SSR as needed to maintain the voltage and charge downstream of the SSR on the secondary bank of caps as the demand on the PS (from the amplifier) varies. Seems very simple...
I am more interested in knowing if it would "work", of what the pitfalls might/will be.
I'm not implying in any way, shape, or form that this is new or "better". Otherwise companies would be building products this way!
I don't want to make something that is using an inductor to store charge... My concept PS doles out charge in "chunks" controlled by the "on" time of the SSR as needed to maintain the voltage and charge downstream of the SSR on the secondary bank of caps as the demand on the PS (from the amplifier) varies. Seems very simple...
I am more interested in knowing if it would "work", of what the pitfalls might/will be.
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Charging a capacitor through a pwm modulated switch will be quite lossy if you omit any inductors.
I don't see how it would be all that lossy... There is only dissipation in the SSR, which would be 1V drop times the time-averaged current drawn, which won't be any more than the amplifier is using... Let's say that we have 50V rails and the amplifier is operating into a 4 ohm load. I figure that would be a maximum current draw of (50V-3Vdrop)*0.707 = 33Vrms, P=Vrms^2/R = 275W, Irms = Vrms/R ~ 8.5 rms amperes. So the power lost in each SSR if the amp was operating at full power would be about 1V*8.5A=8.5 Watts.
The transistors in the SSR are either "on" or "off" which are both low-loss states. It's only the voltage drop across the device during the conduction pulses that I see as leading to power loss and heat generation. It seems pretty insignificant to me, unless I am not including some other losses.
You are missing something critical.
Without an inductor or cap storing energy in the awitching action you cannot get more rms current out then you put in, so if you are running at say 8.5A and the input is 20V higher then the output you will be dissipating 170W in the resistances of your switches caps and traces.
Smaller R means a lower power loss AT A GIVEN CURRENT, but as you reduce R the rms current rises rapidly, and that squared term becomes a problem, inductors beat resistors for filtering anyway, and can flywheel which a resistor cannot.
Consider, if we are switching and the input is say 70V and we have 50V on the output cap and 1 ohm series, load is 1A, then the moment that switch closes the resistor has 20V across it and 20A flows, power in the resistor is 400W falling as the voltage on the cap bank comes up (But we dont want too much ripple, so we cannot allow the output cap to charge too far, if we open the switch when the cap reaches say 55V (10% ripple) then the current in that resistor as we open the switch is still 15A and the dissipation in that resistor is still over 200W.
Now our load is 1A, so 1C/second, and our average charging current is ~17A, so the duty cycle for steady state is trivially about 6 percent or so.....
If the average dissipation during a conduction pulse is say 300W, then the average power loss in our resistor is 300 * 6/100 = 18W.
This not coincidentally is exactly the same losses you would get with a linear regulator!
If you want a switcher, just build a switcher.
Regards, Dan.
Without an inductor or cap storing energy in the awitching action you cannot get more rms current out then you put in, so if you are running at say 8.5A and the input is 20V higher then the output you will be dissipating 170W in the resistances of your switches caps and traces.
Smaller R means a lower power loss AT A GIVEN CURRENT, but as you reduce R the rms current rises rapidly, and that squared term becomes a problem, inductors beat resistors for filtering anyway, and can flywheel which a resistor cannot.
Consider, if we are switching and the input is say 70V and we have 50V on the output cap and 1 ohm series, load is 1A, then the moment that switch closes the resistor has 20V across it and 20A flows, power in the resistor is 400W falling as the voltage on the cap bank comes up (But we dont want too much ripple, so we cannot allow the output cap to charge too far, if we open the switch when the cap reaches say 55V (10% ripple) then the current in that resistor as we open the switch is still 15A and the dissipation in that resistor is still over 200W.
Now our load is 1A, so 1C/second, and our average charging current is ~17A, so the duty cycle for steady state is trivially about 6 percent or so.....
If the average dissipation during a conduction pulse is say 300W, then the average power loss in our resistor is 300 * 6/100 = 18W.
This not coincidentally is exactly the same losses you would get with a linear regulator!
If you want a switcher, just build a switcher.
Regards, Dan.
I don't see where you are coming up with 170W of dissipation. Let's ignore dissipation in traces and wiring for the moment. When the SSE relay is open there is no dissipation, right? No current flow (or negligible dribble of a couple of mA leakage) means no dissipation. Only when the relay is closed will dissipation occur in the relay and the resistor of the downstream RC filter. Moreover, the voltage drop (at least what I can glean from spec sheets) is on the order of 1V in the SSR so for every amp flowing you get 1W. I used the RMS amplifier current to get the time averaged current flow, and this was 8.5W. That's not much. The other place dissipation will occur is in the resistor. If we are switching at 1kHz and we want 40dB of attenuation at 1kHz we need a corner frequency of 10Hz. This can be achieved with a 1R resistor and 15,000uF cap. If the time-averaged current flow through the resistor is 8.5Arms then by I^2R we are dissipating about 72W. That is a lot, but you can get a bolt on chassis-mount resistor that is rated for this. Also, the current I used in these calculations (8.5Arms) was calculated for full power operation of a 275W amp into 4 ohms. That is not typical of music signals either, and average power will be much less, meaning average current and average dissipation will be much less.
Also, the 1R resistor has another necessary function - it limits the instantaneous current flow that will occur the instant that the SSR closes to 1A per Volt difference between the upstream caps and the downstream ones. This must be less than the rated current limit for the SSR.
Don't follow you here Dan. All that really matters for dissipation is the average dissipation, not the instantaneous values, which will be limited by the max current flow when the SSR closes.
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RMS vs Average current is where you are going wrong.
As the duty cycle falls the RMS current in the resistor (And thus the heat in the resistor) rises for a given final load current.
The current during conduction must average ILoad/df where df is the fractional duty cycle, but the RMS current (And thus the heat load in the resistor) is the root of the mean of the square so intgrated over a cycle it is Sqrt(((ILoad/df)^2)/df )= ILoad/df, thus the heat in the resistor rises linearly as the duty cycle falls exactly as a linear regulator would do.
What you describe is halfway to a buck converter, but never quite gets there, use mosfets, run at 250KHz or so and use off the shelf magnetics, you will be much happier.
Seriously buck converters are not hard, and low RdsOn mosfets are cheaper then SSRs.
Regards, Dan.
As the duty cycle falls the RMS current in the resistor (And thus the heat in the resistor) rises for a given final load current.
The current during conduction must average ILoad/df where df is the fractional duty cycle, but the RMS current (And thus the heat load in the resistor) is the root of the mean of the square so intgrated over a cycle it is Sqrt(((ILoad/df)^2)/df )= ILoad/df, thus the heat in the resistor rises linearly as the duty cycle falls exactly as a linear regulator would do.
What you describe is halfway to a buck converter, but never quite gets there, use mosfets, run at 250KHz or so and use off the shelf magnetics, you will be much happier.
Seriously buck converters are not hard, and low RdsOn mosfets are cheaper then SSRs.
Regards, Dan.
It now makes sense to me (under your assumption of 20V Vin-Vout) that the dissipation in the SSR. In addition to this there will be the 72W dissipation in the resistor that I calculated for that average 8.5A draw. As you said, just like a linear regulator you dissipate more heat as the difference between Vin and Vout increases, but (I believe) this is only for the SSR and not the resistor, which is only dissipating heat when current flows. My mistake was that I was considering the transistor to behave like a resistor in terms of dissipation.
Now that its more clear that there are not any magical advantages of the SSR approach I might as well just stick with my previous idea:
http://www.diyaudio.com/forums/powe...tance-multiplier-over-voltage-protection.html
http://www.diyaudio.com/forums/powe...tance-multiplier-over-voltage-protection.html
I appreciate you accepting the arguments of this discussion and frankly revise your idea. That is what I call a discussion!😉
Ah, not so fast! 😉
I'm still considering building something like this just to try it out. I now better understand the dissipation issues that I would/will face. It would be a fun (for me) project anyway, and I could use a high current high voltage regulator for a couple of applications for which I pretty much already own everything else. So... maybe it will come to life after all.
I'm still considering building something like this just to try it out. I now better understand the dissipation issues that I would/will face. It would be a fun (for me) project anyway, and I could use a high current high voltage regulator for a couple of applications for which I pretty much already own everything else. So... maybe it will come to life after all.
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