## Riddle me this... where's the current going?

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A straight-forward oscillator as above, 4MHz. The two gates are the only ones used on the package; all other inputs are tied high or low as most convenient for the wiring, about half each way. The feedback resistor is 1M, the series resistor 680.

It gives a nice stable 4MHz as might be expected and takes around 800uA... but hanging a 10M scope probe on it increases the standing current to over 1mA.

Since V=IR, and the scope resistance is 10M, that implies that 2000v is being generated somewhere :shock: which it obviously isn't. A 10M resistor across the output pin and ground makes no difference to the current... so it must be the capacitance. But that's only picofarads.

Where on earth is that current going?

You're loading the signal with another 50pf of capacitance from the scope probe...the 10M is not affecting it in any significant way

Agree, though it might not be that much capacitance with a "real" 10X scope probe. If the "probe" is one of those (expletive deleted) things that is just a length of coax with clips on the end, it could be even higher.

Here is what happens:

1) How much charge does it take to charge up 50pf to 5V (assumed Vcc)?
Q = C*V = 50pf * 5V = 250pC (picoCoulombs)

2) With a 4MHz signal, the capacitance is charged every 250ns, then the charge is "thrown awaty" so that, 250ns later, it has to be charged again. This is the current that is required to charge the probe capacitance at 4MHz.
I = dQ/dT = 250pC/250ns = 1mA

Chartman must have done exactly this calculation to come up with 50pf. That is a pretty miserable "probe"!

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

You really should not use HC04 because it is a buffered inverter, but a HCU04 in this application. It could have a square wave at pin 12 because of excess gain.

You can also find a two-gate chip with single unbuffered inverter and single schmitt trigger inverter.

Probing pointers, Take 2 by Jack Ganssle (embedded.com; March 25, 2012) shows what happens when using a cheap probe, a much better alternative, and CMOS's latchup.

"Dare to be naïve." - Buckminster Fuller

Certainly, a 10 MÎ© // 50 pF scope probe is not considered to be a 'good' probe.

But, that particular probe actually raised the total oscillator dissipation from 800 Î¼A 1 mA; it is not responsible for the whole 1 mA but for the difference of 200 Î¼A (= 1 mA - 800 Î¼A) only. In that regard, dQ becomes equal to 200 Î¼A * 250 ns = 50 pC and the effective capacitance equal to 50 pC / 5.0 V = 10 pF, assuming that Vcc = 5.0V.

-George

I hope for nothing; I fear nothing; I am free. (Nikos Kazantzakis)

Agree - the capacitance (and it is a proper probe) must be about 10pF.

With regard to the chip selection: this is an attempt to free up a pin on a (EXPLETIVE DELETED) processor by using an external oscillator with no extra chips; it's using the last two spare gates.

It's been nuked as an idea; too much current. Instead I'm using those two inverters and a few discretes to detect the difference between a 100us and a 500us input at much lower current.