# Help with the math? Zener and rectifier for dual powered logic

Discussion in 'The Projects Forum' started by s_mack, Feb 24, 2013.

1. ### s_mack Thread Starter Member

Dec 17, 2011
187
5
Hi. I wouldn't mind some eyes on this in case I'm missing something. See attached.

The idea is that two logic IC's (a demux and a inverter gate) are deciding how signals are routed depending on whether or not USB or battery power, or both, is present. The issue is that the logic IC's themselves need power. So in this diagram, Vaa is the battery and it can range from 7.0 to 12.6 Volts. Vbb is a steady 5V from USB. I use a 4.7V zener to regulate the battery down and use a common cathode rectifier to isolate the two sources. The rectifier has a voltage drop as low as 200mV at very low current so I figure 4.7V takes me down to 4.5V, which is the minimum that the demux can work with* (the gate can go much lower) and also ensures the 5.5V max is respected. 4.7V should also be low enough that the 7V minimum battery voltage is far enough away for the Zener to do its job (right?). Considering this is only powering these two logic chips, I believe... if I'm understanding the datasheets correctly, they'll draw a max of only 13uA. But its the math that I wouldn't mind more eyes on.

Or am I way off because the chips' datasheets both have a ΔIcc figure that I'm not really understanding.

This is eventually (hopefully) for production so component cost is important. I realize I can use a vReg rather than the zener if I have to.

Anyway, hope its clear. Thanks.

- Steven

* Actually the demux datasheet by TI says 4.0V, but availability may be an issue and NXP's replacement says 4.5V so I want to stick with that.

SN74CBT3257BQ
74LVC1G04GV
BAT754C,215/BKN
MMBZ5230B-7-F

• ###### zener-reg.png
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Last edited: Feb 24, 2013
2. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
Your 13μA is under the condition that the output is not drawing any current at all. You need to include whatever current is being drawn by the load on these parts. The ΔIcc appears to refer to the additional current per pin than can be drawn if the input is a diode drop below Vcc.

If your Vaa is 12.6V how much power will the resistor be dissipating? Are you sure you want to push your power ratings that close?

3. ### s_mack Thread Starter Member

Dec 17, 2011
187
5
Well, it would seem to me that the resistor is not dissipating much. But again, math isn't my strongest subject

12.6V - 4.7V = 7.9V * 13μA = 0.1mW or far from the 62.5mW rating.

But, of course, that's if its 13μA... which you say its not. So what is it? I don't really understand the ΔIcc figure. Let's take the inverter... its input is tied to Vcc and its output is going to the demux's OE pin. I can't imagine that uses much, but I suppose its the max ΔIcc figure shown on the demux's datasheet?... so that's a max of 2.5mA (thought I can't imagine its anywhere near that. If so, then that means the max for the gate is 2.51mA. Now the demux I really don't understand. First, ΔIcc says its for "control inputs", which I take to mean OE and S... not the A and B pins, right? Again, max of 2.5mA (do we count OE twice? Already counted it for the gate) so I suppose that means the demux's max draw is 5.03mA for a total of 7.54mA?

I'm so lost I have no idea if that's correct, but if it is... 7.9V * 7.54mA = 59.57mW worst case which is still (barely) ok on the 1/16W resistor.

If I am wrong on the values... if it could be much higher than that... then I'm not sure the resistor is my only issue. Upping the wattage of the resistor is easy of course. But I probably have to take a careful look at the Zener choice again... at least.

If the A and B pins on the demux have to be taken into consideration too... they are MISO, MOSI, SCK and RST of an ISP connection. I can't imagine that draws too much power either, but it doesn't make sense to me that they'd have anything to do with the equation.

4. ### WBahn Moderator

Mar 31, 2012
17,716
4,788
If a 1kΩ resistor has 12.6V on one side of it and 4.7V on the other side of it, how much current is flowing in it? How much power is being dissipated by it?

5. ### s_mack Thread Starter Member

Dec 17, 2011
187
5
yeah, i thought too much, didn't I?

potential diff is 7.9 / 1kΩ = 7.9mA so 62.41mW. Gotcha.