Help A Friend Help A Friend (Measuring Voltage Drop)

Earlier today I had a friend show me a picture of a simple electrical schematic. He asked me if I remembered how to calculate voltage drops, and I respond positively. After he showed it to me I discovered I forgot how to figure in amplifiers. So, after about an hour of reeducation I'm getting pretty frustrated that I still can't figure out what the hell to do.

Can you fine people of ATS take a look at this schematic, and not only tell me what the voltage out is, but how to calculate it? I would really appreciate it.

It looks amazing.

originally posted by: FlukeSkywalker
It looks amazing.

Err, thanks I guess.

a reply to: RealityIsAbsurd

If you have +1v input, then you will have -50v on the output. The capacitor results in a gain of minus infinity.

originally posted by: graysquirrel
a reply to: RealityIsAbsurd

If you have +1v input, then you will have -50v on the output. The capacitor results in a gain of minus infinity.

I'm pretty sure the OP is looking for the full circuit equation and not just the drop answer.

a reply to: RealityIsAbsurd

If you are talking pure DC, and 'ideal' components, then the cap is pretty much an open circuit and the diode is reverse-biased and so the feedback side across the op-amp (inverting) is not happening, so I'd expect the V-out to go fully to the positive supply rail of the op-amp.

a reply to: chr0naut

It is an inverting amplifier. So, the output will go the the negative supply rail or -50v.

Graysquirrel, my old friend said it's not positive or negative 50VDC. He gave me a hint saying that it wouldn't be any more than positive or negative 10 VDC. I'm currently going over all my calculations again, and will post them so that you all can point out where I'm going astray.

Chr0naut, yeah this is all theoretical in regards to components operating perfectly and all that. You mentioned cap; I take it you mean capacitor? That symbol is for a contact that is normally closed. If there were no line through it that would symbolize a contact that is normally open.

If it were a capacitor the line going through it would be an arrow symbolizing a variable capacitor.

Since it is a normally closed contact it we can disregard it or regard it as a short. It's not a relay with any electromagnetic component.

Am I wrong?

So, there is 1 VDC going across a 10k ohm rated resistor into the negative leg of the amplifier. The amplifier also has a positive and negative 50 VDC going into it. For the life of me I can't remember how the diode, 100k ohm resistor, and normally closed contact play into the equation. I took a break from it and watched a movie, but coming back to it I still can't get it. I've forgotten some fundamental.


Ohm's law is V=I*R. So, 1 VDC= (?)I x 10K ohm. So, I=.0001

One can get Power(P) by:
V*I, I squared x R, or V squared / R

I just multiplied the voltage by the amperage and got .0001 for the power.

Therefore, I've got .0001 amperage(power) and voltage going into the amplifier.

You know I think I just mixed up some calculations when I got 10 VDC. I'm fairly certain that it's -10 VDC.

a reply to: RealityIsAbsurd

If the capacitor is really a short then the out put will -10v for the +1v input.

a reply to: graysquirrel

That's what I'm thinking the Vout is. I guess I got smirked at because I told him that it was positive 10 VDC out instead of -10 VDC.

Ten years ago I wouldn't have had this problem. It's just like all math. If I don't do it everyday my brain tosses it out. I have to reteach myself math about once a year since I don't work with it on a daily basis. I guess I have the gift of modern day computers to thank for that. That's why I get frustrated when I have to actually do the work in my head. Even when I worked with schematics and these kinds of problems in every day work situations troubleshooting was a breeze because I could just take some readings, punch in some values, have a machine do all the brainpower for me and call it a day.

Yes it's an inverting amplifier with the gain set by the resistors at 10x so the output is -10V

The diode is a red herring in this example and does nothing with the voltages quoted. What it does is clamp the maximum positive output at roughly the forward voltage drop of the diode so dependant on the type (silicom, germanium, shottky). So a typical silicon diode would limit the output to a range of about +0.7V to -50V (negative supply rail) but you can only get a positive output if the Vin is negative.

a reply to: Pilgrum

Appreciate it. I thought along the same lines when I told the guy it was 10V. But when he laughed and told me I was wrong it shot me in the foulest of moods. I thought it was pretty ridiculous that I couldn't get something that used to be so simple. So, I guess in my frustration I made it over complicated and started focusing on the wrong things. But every time I did my math I kept getting the same thing over and over. That's when I realized I wasn't inverting it.

a reply to: RealityIsAbsurd

The diode has a Vfd. That Vfd will the be same across the 100K resistor and the...cap? If this is strictly DC then Ic +I(100K) is zero. Vout is Vout = Vin + Vfd (inverted); when f = 0. Unless the question is at time = 0, then the cap is a short and this is a 10x inverter. At some frequency the cap has an impedance, Z = 1/wC where w = 2(PI)f.