[SOLVED] RC differentiator doubts and queries

[d123]

Hi,

iC = C*(dVc/dt)

1)
(i) a) Is dt instantaneous or conditioned by RC selection?
b) How is Rsource relevant, or not, to dt?
(ii) In schematic A, what would dt be: >0s or 10ms?

2)
(i) Does ib = iC?
(ii) If so, presumably iC must be calculated for desired ib?

3)
At t = >0s (when pushbutton is initially pressed), does VR = Vin immediately? I thought so.

4)
If ib = iC, which schematic is most helpful for protecting the BJT base from excessive input currents: A, B or C?

5)
(i) As the differentiator swings below ground at turn-off of a square wave input signal as far as it rises above ground at turn-on of input signal, what would happen at the 'Vin' node at input turn-off?
(ii) How can the negative-going excursion be limited (in the 'A' schematic) - presumably/possibly a diode from ground to tbe Vin node, anode to ground and cathode to Vin?
(iii) How would the negative-going excursion affect other components/voltages/ground/voltage references on a pcb with a solid ground plane?

Children's doodles attached...

Thanks.

 

[barry]

Some random thoughts:

Basically, you need to learn how to do circuit analysis. Loop equations. Mesh equations. Laplace transforms. Kirchoff's laws. Or, just cheat and use a circuit simulator program. I think that would go a long way towards helping you understand what's going on here.

dv/dt for the cap changes over time. dt is not just some constant value.

Your formula is the relationship between instantaneous current, capacitance and dv/dt. It has nothing to do with resistance.

2. NO. Didn't you notice you have a resistor in your circuit? Doesn't some current flow through that?

 

[BradtheRad]

It's easy to forget about unseen or unknown input resistance in the incoming source at left. Where there's a capacitor it's necessary to account for this. Or else put in a neighboring resistor so that a sensible RC time constant can be calculated.

One of your schematics has what looks like a resistor providing input resistance. Without it the simulator might allot some internal resistance in the bias of the transistor. However it's unpredictable and the simulator may halt with an error.

Likewise the simulator may have an easier time if you put in a resistor where you have two capacitors in series.

 

[d123]

Hi,

Some random thoughts:

Basically, you need to learn how to do circuit analysis. Loop equations. Mesh equations. Laplace transforms. Kirchoff's laws. Or, just cheat and use a circuit simulator program. I think that would go a long way towards helping you understand what's going on here.

You know what, I completely agree with you, Barry, thanks for saying so. I frequently see huge gaps in my basic understanding and know I need to study the maths behind it. I'm trying to take my circuits more seriously and doing more calculations, not just simulating (even though that can help a lot) but have a very long way to go in this respect.

dv/dt for the cap changes over time. dt is not just some constant value.

Your formula is the relationship between instantaneous current, capacitance and dv/dt. It has nothing to do with resistance.

That I understand, just wasn't sure how to ascertain dt for peak current and therefore peak Vout in a differentiator. The rising slope will have a rise time ftom 0 to peak value that is the dt I was unsure of how to obtain. Brad's reply has answered my doubt.

2. NO. Didn't you notice you have a resistor in your circuit? Doesn't some current flow through that?

Okay. I did my sums from this tutorial, it says iR = iC, and iC = C/(dVin/dt). So, in my ignorance I would have assumed ib =iC also.

Thanks a lot for your timely reminder that I need to also grapple with the basics, actually does help.

--- Updated Oct 3, 2020 ---

Hi Brad,

It's easy to forget about unseen or unknown input resistance in the incoming source at left. Where there's a capacitor it's necessary to account for this. Or else put in a neighboring resistor so that a sensible RC time constant can be calculated.

That's what I dimly gleaned, just wasn't sure, thanks so much for clarifying. So, e.g. source resistance (and to be pedantic, even pushbutton contact resistance) will shape the dt in that formula for when the peak is reached.

One of your schematics has what looks like a resistor providing input resistance. Without it the simulator might allot some internal resistance in the bias of the transistor. However it's unpredictable and the simulator may halt with an error.

Likewise the simulator may have an easier time if you put in a resistor where you have two capacitors in series.

Thanks for those helpful hints, simulation quirks like that can be a real problem for an assortment of reasons, much appreciated.

 

[betwixt]

adding to the above... where does your negative voltage come from?
Also note that opening the switch does not discharge the input capacitor, it just isolates it. There will only be a single pulse as the switch closes and the voltage on the transistor base will be subject the characteristics of the B-E PN junction.

Brian.

 

[d123]

Hi Brian,

adding to the above... where does your negative voltage come from?

The tutorial shows the differentiator output at Vin input signal falling edge swinging below ground on a single-supply circuit. Attached image shows copied tutorial stuff, etc.

Also note that opening the switch does not discharge the input capacitor, it just isolates it.

Yes, but capacitor voltage will decay at whatever RC rate is chosen.

There will only be a single pulse as the switch closes and the voltage on the transistor base will be subject the characteristics of the B-E PN junction.

Okay, you mean regardless of Vin and differentiator Vout, Vbe will only be e.g. 0.7V?
So long as it wouldn't be one of those epic fails of 5 Vin and 0.1 Vout such as may happen with an RC LPF when amateurs such as myself use circuits they neither understand nor even attempt to calculate .

Besides question(s) about possible circuit side-effects caused by the differentiator negative-going pulse, I'm trying to ascertain that dt in the iC formula is not 0.000001 seconds, but due to source resistance would be more like e.g. 0.000100 seconds. 650mA is really bad, 6.5mA is acceptable, especially if that's what a BJT base would also 'see'/how that iC affects ib of a BJT with no series resistor into the base.

Thanks.

 

[FvM]

The tutorial shows the differentiator output at Vin input signal falling edge swinging below ground on a single-supply circuit. Attached image shows copied tutorial stuff, etc.

If the tutorial actually uses a pushbutton (single throw switch) as input voltage, it's completely wrong. Can you show a screenshot or give a link?

Yes, but capacitor voltage will decay at whatever RC rate is chosen.

Did you notice that the capacitor is disconnected as long as the switch is open? How do you calculate a RC time constant in this case?

I'm trying to ascertain that dt in the iC formula is not 0.000001 seconds, but due to source resistance would be more like e.g. 0.000100 seconds. 650mA is really bad, 6.5mA is acceptable, especially if that's what a BJT base would also 'see'/how that iC affects ib of a BJT with no series resistor into the base.

Excurse: dt is a symbol in differential calculus, an infinitisimal time interval. If you refer to finite time intervals, you write \[ \Delta t \]

Returning to practical circuit analysis. Without current limiting elements, the pulse current in your circuit would be inifinite. Fortunately the real circuit has some series resistance and inductance. Also the switch closure takes some time. If we assume ideal 6.5 V supply, switch and capacitor, the transistor can't absorb more than 0.5 C U² = 2 µJ, which most likely won't kill it. More likely, the switch will fastly degrade.

Don't design the circuit without a small series resistor, e.g. 100 ohms. And add a pull-down resistor after the switch to discharge the resistor, if you want the circuit working for more than one button press.

 

[d123]

Hi FvM,

If the tutorial actually uses a pushbutton (single throw switch) as input voltage, it's completely wrong. Can you show a screenshot or give a link?

It doesn't show a pushbutton, it's this tutorial from Electronics Tutorials called: 'RC Differentiator'.

Did you notice that the capacitor is disconnected as long as the switch is open? How do you calculate a RC time constant in this case?

. I have the right to remain silent, so as to not make even more of a fool of myself.

Excurse: dt is a symbol in differential calculus, an infinitisimal time interval. If you refer to finite time intervals, you write \[ \Delta t \]

Okay, you've just cleared up a misconception I had from trying to parse formulas where each symbol wasn't always defined over the past year or so... t can be anywhere on an endless time line, delta t is a specific point on the timeline - I had thought they meant the same thing.

Returning to practical circuit analysis. Without current limiting elements, the pulse current in your circuit would be inifinite. Fortunately the real circuit has some series resistance and inductance. Also the switch closure takes some time. If we assume ideal 6.5 V supply, switch and capacitor, the transistor can't absorb more than 0.5 C U² = 2 µJ, which most likely won't kill it. More likely, the switch will fastly degrade.

Thanks, I wouldn't have known that - never done any Joule calculations, only Watts, and hadn't considered switch closure time, nor switch degradation.

Don't design the circuit without a small series resistor, e.g. 100 ohms. And add a pull-down resistor after the switch to discharge the resistor, if you want the circuit working for more than one button press.

Thank you, that is of practical help.