26-05-2020, 11:51 AM
Applies to anything, even the mains 
![[Image: equivalent.gif]](https://www.markhennessy.co.uk/articles/equivalent.gif)
Even a low source impedance (Rint) of 0.1 ohms will drop a volt for every 10 amps taken.
A volt or two might not sound like much in the context of a nominal 230V, but go back to rectifier theory:
Remember that when you rectify and smooth an AC waveform, you only draw current at the peaks of the waveform. At other times, no current flows.
So when the current is flowing, it has to be much larger than the average DC current delivered to the load. Remember, it has to charge the capacitor sufficiently so it can supply the DC current until the next peak.
As a rule of thumb, a full-wave rectifier draws peak currents of around 4 times the average DC current.
Whereas a half-wave rectifier might draw closer to 10 times the average DC current!
Obviously, the exact values depend on many variables - don't bother trying to calculate them as you'll go mad! Just build and measure.
This, by the way, explains why the peaks of the AC waveform always appear to be slightly "squashed" when you look on an oscilloscope. Many of the devices connected to the mains only draw current at the peaks, and because of the non-zero source impedance, the voltage drops slightly during that time. That is especially noticeable at work, but might be less so domestically, where the biggest consumers are things like showers and cookers, which are obviously resistive, so draw current over the whole cycle.
This is obviously a bit of a pain, but has got better in recent times because switched-mode power supplies are increasingly fitted with PFC (power factor correction), which attempts to draw current over the whole AC cycle like a simple resistive load.
Much more about power supplies (from an audio POV) here: https://www.markhennessy.co.uk/articles/...pplies.htm
Anyway, back to DC on the mains. A full-wave rectifier draws current in "gulps" on every half-cycle, but although it's not ideal, at least the current draw is symmetrical. So it does not have a significant DC component.
But a half-wave rectifier only draws current pulses in one direction. So the current waveform is not symmetrical. That means the waveform has a DC component. Combine that with the source impedance inherent in the mains supply, and you can see how the voltage waveform can be slightly shifted by these unsymmetrical currents.
A colour TV with half-wave rectification might draw 230 watts (made up value to keep the numbers simple!). That means the average DC current taken by the set, after the rectifier, is obviously 1 amp. But the peak currents taken by this set might be around 10 amps. For an impedance of 0.1 ohms, that's obviously a volt. The DC component of the current waveform is hard to predict, but might be 0.1V, perhaps. Not a big deal, but that's just 1 TV, and the 0.1 ohms source impedance is generously low - it might be more like 0.5 ohms, depending on where you live. As an exercise, consider a row of 10 cottages on the end of long overhead supply with a source impedance of 1 ohm, and all of them with a colour TV!
![[Image: equivalent.gif]](https://www.markhennessy.co.uk/articles/equivalent.gif)
Even a low source impedance (Rint) of 0.1 ohms will drop a volt for every 10 amps taken.
A volt or two might not sound like much in the context of a nominal 230V, but go back to rectifier theory:
Remember that when you rectify and smooth an AC waveform, you only draw current at the peaks of the waveform. At other times, no current flows.
So when the current is flowing, it has to be much larger than the average DC current delivered to the load. Remember, it has to charge the capacitor sufficiently so it can supply the DC current until the next peak.
As a rule of thumb, a full-wave rectifier draws peak currents of around 4 times the average DC current.
Whereas a half-wave rectifier might draw closer to 10 times the average DC current!
Obviously, the exact values depend on many variables - don't bother trying to calculate them as you'll go mad! Just build and measure.
This, by the way, explains why the peaks of the AC waveform always appear to be slightly "squashed" when you look on an oscilloscope. Many of the devices connected to the mains only draw current at the peaks, and because of the non-zero source impedance, the voltage drops slightly during that time. That is especially noticeable at work, but might be less so domestically, where the biggest consumers are things like showers and cookers, which are obviously resistive, so draw current over the whole cycle.
This is obviously a bit of a pain, but has got better in recent times because switched-mode power supplies are increasingly fitted with PFC (power factor correction), which attempts to draw current over the whole AC cycle like a simple resistive load.
Much more about power supplies (from an audio POV) here: https://www.markhennessy.co.uk/articles/...pplies.htm
Anyway, back to DC on the mains. A full-wave rectifier draws current in "gulps" on every half-cycle, but although it's not ideal, at least the current draw is symmetrical. So it does not have a significant DC component.
But a half-wave rectifier only draws current pulses in one direction. So the current waveform is not symmetrical. That means the waveform has a DC component. Combine that with the source impedance inherent in the mains supply, and you can see how the voltage waveform can be slightly shifted by these unsymmetrical currents.
A colour TV with half-wave rectification might draw 230 watts (made up value to keep the numbers simple!). That means the average DC current taken by the set, after the rectifier, is obviously 1 amp. But the peak currents taken by this set might be around 10 amps. For an impedance of 0.1 ohms, that's obviously a volt. The DC component of the current waveform is hard to predict, but might be 0.1V, perhaps. Not a big deal, but that's just 1 TV, and the 0.1 ohms source impedance is generously low - it might be more like 0.5 ohms, depending on where you live. As an exercise, consider a row of 10 cottages on the end of long overhead supply with a source impedance of 1 ohm, and all of them with a colour TV!







