Power draw question

[msknight]

I have a zeddy which runs off the 9v supply. It was measured drawing 330mA.
To display it on HDMI, I'm using a PiZero which is pulling 190mA off 5v.
To deal with storage, I have a ZX-Wespi on an ESP32 board which was pulling 90mA on 5v.

I wanted to put them all in the same case, and be sure I had enough juice to power everything.

In theory, according to Ohms law, if my maths were correct, (which is probably isn't) something pulling 250mA on a 5V supply, would be pulling 450mA on a 9v supply.

To my newbie brain... more electronics equals drawing more "power" ... every mouth has got to be fed.

The 7805 was replaced with an I50115 with 5v at 1.5A max, so I believed I had headroom from the regulator, but when the bench power supply said that everything... at 9v... was only pulling 385mA... I knew my maths was way out, and that if resistance in parallel is at play here... how do I ensure that I'm not going to overload something.

At the moment, the only thing that seems to be pulling with any consistency... is my hair!

[1024MAK]

So, we have to be clear here. There is a difference between electrical current and electrical power.

With a 7805, the input current will be almost exactly the same as the output current (the 7805 uses a small amount of current itself, typically less than 8mA).

So if the load on the 5V output is:

• ZX81 using 330mA
• PiZero using 190mA
• ZX-Wespi using 90mA

Then the total output current of the 7805 will be 610mA (or thereabouts).
That is 330mA+190mA+90mA.

The input current to the 7805 would be 618mA (or thereabouts). The input current is also the current from the "9V" PSU. This current will be similar regardless of the actual input voltage (as long as it's in a sensible range, say between 7V and 12V).

DC/DC converters / Switching regulators (like the I50115) work differently. Now rather than thinking about just current, you also have to think in terms of power. Power in Watts (W) for electricity is Voltage X Current.

If the output load is the same as the example above for a 7805, that output power is 5V x 610mA, which is 3.05W.

If the DC/DC converter / Switching regulator is 85% efficient, then it will take 3.59W from the "9V" PSU.
That is 3.05/0.85 = 3.59.

If the output from the "9V" PSU is really 9.0V DC, for a 3.59W input, the DC/DC converter/ Switching regulator will draw 399mA.
That is 3.59 / 9 = 0.399.

Yes, that's correct, the input current is SMALLER than the output current.

If the "9V" PSU is a Sinclair UK700 or indeed, any of the Sinclair 9V PSUs, then because these are of the unregulated type, the actual output voltage will be somewhere between 9.5V and 12V. Typically for a UK700, it will be around 11V to 11.5V for output currents of 300mA to 400mA.

If the actual output voltage of a Sinclair PSU is 11V, then the current drawn by the DC/DC converter / Switching regulator would be 326mA.

Remember, energy cannot be made or destroyed, it can only be converted from one form to another. The 7805 converts the unwanted electrical power into heat (which is why they get hot). A DC/DC converter / Switching regulator produces much less heat, because its much more efficient. Hence as the input voltage goes higher, the input current reduces.

Mark

[msknight]

I'm going to have to re-read this slowly in the morning.... but the 330mA was measured at 9v. So at 5V wouldn't that make it less according to ohms law, or am I barking up the wrong tree?

[1024MAK]

I'm going to have to re-read this slowly in the morning.... but the 330mA was measured at 9v. So at 5V wouldn't that make it less according to ohms law, or am I barking up the wrong tree?

Ohms law is not the law you want...

Ohms law is applicable if we were only dealing with resistance. But with semiconductors like a 7805 it's more complex.

With a DC/DC converter / switching regulator, inductors and magnetic fields and capacitors come into it. Energy is stored in the magnetic field of the inductor. Then transferred to a capacitor, then to the load. Yes, this makes it very complex. Far more complex than Ohms law can explain.

Mark

[msknight]

OK - I get where I went wrong, and I noted that you converted the power to watts to come up with the total.

The zeddy was supplied by my desktop power supply when it was fitted with the 7805, while the Pi and ESP32 were measured by a USB device which was positioned after the transformer.

The final reading of everything together, was done by the desktop power supply, but after I'd fitted the I50115.

What I don't get is this...

The 7805 is generating heat from the "unwanted" electrical power. I believed that circuits only draw what they want, so what is this unwanted power?

When I do the next zeddy, I'll put everything in on the 7805 and run it from the bench, and then replace with the I50115 and run that from the bench, and see what the difference in figures is.

[1024MAK]

What I don't get is this...

The 7805 is generating heat from the "unwanted" electrical power. I believed that circuits only draw what they want, so what is this unwanted power?

Ah, be careful what you believe!

The job of a voltage regulator like a 7805 (and this applies to all series pass voltage regulators no matter what type), is to produce a steady DC output voltage, within a tight voltage range and with the minimum amount of noise and ripple.

To do this, the input voltage must always be higher than the rated output voltage. Plus, an allowance for the voltage "drop" across the regulators internal circuitry (this is also known as the "drop out" voltage). With a 7805, this figure is 2V. Hence why you see the minimum input voltage for a 7805 given as 7V.

Hence, circuit designers normally allow plenty of margin above 7V. As normally a 7805 is fed from an unregulated DC supply, which comes from a mains transformer, a bridge rectifier and a smoothing (filter) capacitor. So, have to consider mains voltage variations and the ripple level on the smoothing capacitor. Ripple is the voltage falling in-between each half cycle of the mains sine wave.

That's why the Sinclair PSUs have an output rated at 9V.

Now, as I said earlier, with a 7805, the output current is almost the same as the input current (I'm ignoring the small amount of current used by the 7805 to keep the calculations simple). But there is a difference in voltage between the input and output.

This voltage can't magically just disappear. Let's look at it from the point of view of the power.

If a 7805 has an input voltage of 9V, an output voltage of 5V, and 400mA (0.4A) of current is passing through it to the load, then:

the input power is 9V x 0.4A = 3.6W

the "output" power (or the power 'used' by the load) is 5V x 0.4A = 2W

But wait! Where did the other 1.6W go?

We know that the current is 0.4A, so we can do another calculation:

V = P/I

1.6W / 0.4A = 4V

Note that 9V - 5V = 4V.

As energy can't just disappear, it has to go somewhere. In the case of a 7805, this 1.6W is being converted to heat, hence the need for a metal heatsink to be fitted to the 7805. That's the "unwanted power".

The statement that "I believed that circuits only draw what they want" is not entirely correct. For circuits that are designed or intended to operate from a 'fixed' and defined supply voltage, (in a perfect world) the only variable is the amount of current that they use (in practice temperature also has an effect). The amount of power follows the current. As different parts change state, the current will vary. Then "circuits only draw what current they want" is usually true.

But with a voltage regulator, within it's normal range of output current, as its the load that determines the current, the amount of power that is converted to heat is outside the control of the 7805 (unless the internal circuitry senses the output current is too high or the temperature gets too hot in which case it will restrict the output voltage to reduce the output current and its temperature). Hence for a 7805, "I believed that circuits only draw what they want" is incorrect.

For anyone confused about electrical power for DC systems:

The power, P (in Watts, W) is voltage (V or U depending on which you prefer) x current (I)

Or P = V x I

This can be rearranged:

V = P/I

or

I = P/V

Mark

[msknight]

OK - I think I'm getting it.

You calculated the figures as watts because that was the true draw.

In the case of the Pi and ESP32, this was measured after the transformer which was probably reasonably accurate.

In the case of the zeddy, however, this measurement was before the regulator so it was the consumption of the zeddy and the "waste" of the 7805.

When I changed the regulator and put everything inside, it was not as high as I believed it would be, because I don't have as much "waste" because I'm not using the 7805 any more... and thus I can still rely on the amp figure reported by the desktop power supply and I can relax as I'm well within limits.

Do I have that right please?

By the way... thank you for the time you're putting in to these explanations. I really appreciate it.

[1024MAK]

OK - I think I'm getting it.

You calculated the figures as watts because that was the true draw.

True power consumption is in Watts, yes.

In the case of the zeddy, however, this measurement was before the regulator so it was the consumption of the zeddy and the "waste" of the 7805.

Well, as you were measuring current, the current reading is approximately the same as the current used by the Zeddy. But the power figures are very different, because the voltages are different.

When I changed the regulator and put everything inside, it was not as high as I believed it would be, because I don't have as much "waste" because I'm not using the 7805 any more... and thus I can still rely on the amp figure reported by the desktop power supply and I can relax as I'm well within limits.

Do I have that right please?

Yes

By the way... thank you for the time you're putting in to these explanations. I really appreciate it.

No problem. As long as what I write makes sense

If it doesn't, I'll have to start talking about buying and cutting bits of wood...

Mark