# Ramp-up rise time?

Discussion in 'General Electronics Chat' started by tracecom, Jun 9, 2012.

1. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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How does one calculate the rise time of the ramp-up circuit at PH1 in the attachment? I think the output voltage will be approximately 11.3 V; is that correct? Thanks.

• ###### BC547B Ramp-up Circuit.png
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2. ### #12 Expert

Nov 30, 2010
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Look at my blog #3

might be as high as 11.7 volts.

3. ### ErnieM AAC Fanatic!

Apr 24, 2011
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11.3V is a good estimate.

A useful equation for cap charge or discharge times is:

Code ( (Unknown Language)):
1.                  (Vss-Vi)
2. time = RC * ln  ----------
3.                  (Vss-Vf)
4.
5. Where:
6. RC   = Resistance times Capacitance (the 'tau')
7. Vss  = final steady state voltage
8. Vi   = initial voltage on cap
9. Vf   = final voltage on cap
10. ln   = natural log
11.
12.
The significant voltages in your circuit are
zero: where the cap starts off
0.7V: when Q1 turns on and the output starts to rise ground

Note there is a delay time between 0-0.7V, and the rise time between 0.7 and 12.

Also note it theroetically takes an infinite time to get to 12V, and the equation will predict exactly that (so cheat a little on Vf).

Last edited: Jun 10, 2012
4. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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Thanks for the explanation, however I am missing something. I don't see how the component values in my circuit are taken into account by the equation.

I have breadboarded the circuit, and, using my DMM, I can watch the emitter voltage rise when the switch is closed. I can also see a change in the rise rate based on the size of C1, i.e., the larger C1 is, the slower the voltage rises.

This little circuit is intended to provide a turn-on signal to an IC that stays inactive until a certain pin reaches 8.5 V. So, what I am actually looking for is the time it takes for the emitter of Q1 to reach 8.5 V. When I use the equation you provided, the answer seems to be 1.23 seconds, which would be okay for my practical application. However, from a learning perspective, I also want to really understand the ramp-up circuit operation, and how to do the calculations based on the component values.

Any additional insight you can give me would be much appreciated.

Last edited: Jun 10, 2012
5. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Oops, my bad. I left out the RC term in that equation. I corrected my post above to reflect that.

When you get that 1.23 seconds, it is actually 1.223*R*C. So you have to include those factors to predict your desired response.

tracecom likes this.
6. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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OK, so in my schematic, the time is still 1.223 seconds because 10k times .0001 (100μF) is 1. Correct?

And my final (maybe) question. Would a 2N2222A work just as well? I substituted one for the BC547B in my breadboard and didn't detect any change.

7. ### #12 Expert

Nov 30, 2010
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Vo = dV (e^-t/RC)
t = -RC Ln Vo/dV
RC = -t/(Ln Vo/dV)
R = -t/C (Ln Vo/dV)
C = -t/R (Ln Vo/dV)
dV = Vo/(e^-t/RC)
e = 2.71828183

Glad you could use one of these equasions.

8. ### ErnieM AAC Fanatic!

Apr 24, 2011
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OK, 10K * 100 uF = 1.0, so in this case my on omission didn't matter.

I didn't consider the transistor in my answer. A 2N2222A should be OK.

One point: this circuit take just as long to discharge as it does to charge, so there is a lower limit as to how fast you can turn the switch off then back on again. You could change that by adding a series diode and a (smaller) resistor across R1.

9. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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I appreciate you responding to my question, but I wasn't knowledgeable enough to use any of your equations without more explanation.

10. ### #12 Expert

Nov 30, 2010
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Darn. Maybe I sould put definitions of the letters in that blog. Would that fix it for you?
Rephrasing: What would make it useful for you?