# Has anyone used a capacitor to provide the Vbe for an NPN BJT.

#### crutschow

Joined Mar 14, 2008
25,664
The problem is it's not enabling CE which is inconsistent with that chart which shows Ic > 0.
What is the circuit and the actual collector current when you measure this?
The collector current in my simulation is less than 1mA for a Vbe of 650mV.
0.7V is not a fixed voltage, it's just the approximate Vbe voltage for a typical BJT transistor with a few mA of collector current.

Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
Yes, I am giving you the steady state Vbe. The time it takes to get there varies from 0.28 s for 1K and 2.6 at 10K, which is again as I expect.

Interestingly the voltage rise is pretty close to linear, which makes sense because the exponential behavior of the diode and the logarithmic behavior of and RC counteract each other.

BobView attachment 198637
What is the difference between V(n003) and V(n004) in that graph. My peak is more like V(n003). Attached is the charging and discharging profile that I expect with the particular RC combo I am using at the moment. 3300 uF and 10kOhm with 9V supply. It gives me a 33 second time constant and a capacitor voltage of 0.7V after 2.64 seconds. These are the prevailing condition at present. In short I am expecting the CE to start flowing 2.64 seconds after the 9V is applied. That's not happening.

What is the circuit and the actual collector current when you measure this?
The collector current in my simulation is less than 1mA for a Vbe of 650mV.
0.7V is not a fixed voltage, it's just the approximate Vbe voltage for a typical BJT transistor with a few mA of collector current.
Ic is 0 but as you pointed out before it may be < 1 mA which I can't detect on my equipment. So I will put my DSO on it tomorrow and see if I can see a pulse.

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#### crutschow

Joined Mar 14, 2008
25,664
which I can't detect on my equipment.
If you put a 1kΩ resistor in series with the collector, you can measure the voltage across that to determine the current (1mA per volt).

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#### BobTPH

Joined Jun 5, 2013
2,587
My simulation is showing Vbe for the 10K and the 1K resistors. The steeper one is, of course, with the 1K.

Bob

#### crutschow

Joined Mar 14, 2008
25,664
In short I am expecting the CE to start flowing 2.64 seconds after the 9V is applied. That's not happening.
Here's the simulation showing the transistor starting to turn on at around 2.5s when the Vbe reaches about 650mV.

View attachment 198660

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#### crutschow

Joined Mar 14, 2008
25,664
Interestingly the voltage rise is pretty close to linear, which makes sense because the exponential behavior of the diode and the logarithmic behavior of and RC counteract each other.
That's not the reason.
It look linear because you are looking at only about the first 8% of the RC exponential rise.

#### crutschow

Joined Mar 14, 2008
25,664
In short I am expecting the CE to start flowing 2.64 seconds after the 9V is applied. That's not happening.
Here's a simulation showing the collector current (yellow trace) versus the base-emitter voltage (blue trace).
The current reaches 1mA after 2.5s where Vbe is 655mV.

Note the expected exponential increase in collector current with Vbe (vertical current scale is logarithmic) cause by the logarithmic increase in base current.

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#### ebeowulf17

Joined Aug 12, 2014
3,274
Here's a simulation showing the collector current (yellow trace) versus the base-emitter voltage (blue trace).
The current reaches 1mA after 2.5s where Vbe is 655mV.

Note the expected exponential increase in collector current with Vbe (vertical current scale is logarithmic) cause by the logarithmic increase in base current.

View attachment 198665
A key difference between your sim and the thread starter's sketch is that you've got R1 controlling base current, but nothing limiting collector current. In the original sketch, both the base and the collector are being fed through the resistor.

With that setup, and using 9V and 10k, there will be less than 1mA of current flowing anywhere! I can't imagine what purpose that circuit is meant to accomplish, but that's how it's drawn, and that would explain never getting measurable current of 1mA or greater!

#### Alec_t

Joined Sep 17, 2013
11,744
Vbe usually needs to be a minimum of 0.7V
It all depends on how much collector current you want. Even 0.1V would likely result in a tiny collector current, whereas 0.8V would result in a much higher current.

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Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
A key difference between your sim and the thread starter's sketch is that you've got R1 controlling base current, but nothing limiting collector current. In the original sketch, both the base and the collector are being fed through the resistor.

With that setup, and using 9V and 10k, there will be less than 1mA of current flowing anywhere! I can't imagine what purpose that circuit is meant to accomplish, but that's how it's drawn, and that would explain never getting measurable current of 1mA or greater!
That's correct. The purpose of the circuit is to charge and discharge the capacitor through CE once every 2.64 seconds, with this RC config, but the frequency can be changed by varying the R in the RC. Just in case anyone was wondering. It works well with other components. I just need to figure out what's going on with these transistors.

It all depends on how much collector current you want. Even 0.1V would likely result in a tiny collector current, whereas 0.8V would result in a much higher current.
I need to take a closer look at that dependency. I don't fully understand the internal workings of the transistor. Maybe that's my problem. Too small an Ic with my current set up.

Here's a simulation showing the collector current (yellow trace) versus the base-emitter voltage (blue trace).
The current reaches 1mA after 2.5s where Vbe is 655mV.

Note the expected exponential increase in collector current with Vbe (vertical current scale is logarithmic) cause by the logarithmic increase in base current.

View attachment 198665
What's that software you are using? Is it open source of does it cost money?

Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
Thanks for all your input folks. Just a quick update. I replaced the 2N2222s with a couple of TIP120s (Darlingtons) just because they are a bigger package and twice the Vbe (1.4V). Tested them first. The RC cap now reaches 1.24V which is still too little to open the transistor CE. So it's the issue remains. No change really. Stumped.

#### crutschow

Joined Mar 14, 2008
25,664
What's that software you are using?
It's LTspice, free download from Analog Devices.
The RC cap now reaches 1.24V which is still too little to open the transistor CE. So it's the issue remains. No change really. Stumped.
You are connecting the transistor(s) as a diode, so they act like a diode.
It will never "open" the transistor CE beyond what you see.

To act like a transistor, the collector needs to be separate from the base, as my post #27 shows.

Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
Here is what I have found when taking a few measurements.

When the collector is NOT connected to the capacitor +ve pin there is a voltage of 0.723V across the C and B pins and a voltage of 0.6V between the C and E pins which adds up to the voltage of 1.323V across the B and E pins.

When the collector IS connected to the capacitor +ve pin there is NO a voltage (0V) across the C and B pins and a voltage of 1.246V between the C and E pins which adds up to the voltage of 1.246V across the B and E pins.

So the problem seems to be that I am connecting the collector is connected to the +ve pin of the capacitor because it's not going to work with a CB voltage of 0V is it.

Is this a correct interpretation of this data?

I am not sure how I can get around this of course. Back to the dawing board.

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Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
It's LTspice, free download from Analog Devices.
You are connecting the transistor(s) as a diode, so they act like a diode.
It will never "open" the transistor CE beyond what you see.

To act like a transistor, the collector needs to be separate from the base, as my post #27 shows.
Thanks for this. I didn't receive a notification so of this response. I have just come to it after posting further findings which would seem to confirm what you say here.

#### crutschow

Joined Mar 14, 2008
25,664
I'm still not clear as to exactly what you want the complete circuit to do.
What is supposed to happen after the delay?

Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
I'm still not clear as to exactly what you want the complete circuit to do.
What is supposed to happen after the delay?
I want it to discharge the capacitor. The capacitor then recharges again and repeats. You can control the frquency by altering the time constant by varying the resistor. You basically have an on/off switch.

#### djsfantasi

Joined Apr 11, 2010
6,962
I want it to discharge the capacitor. The capacitor then recharges again and repeats. You can control the frquency by altering the time constant by varying the resistor. You basically have an on/off switch.
On/off switch or a repeating on/off cycle?

Thread Starter

#### RAMBO999

Joined Feb 26, 2018
247
On/off switch or a repeating on/off cycle?
Have a look at my posting #15. There's a video link. That's how it works.

#### Audioguru again

Joined Oct 21, 2019
2,067
You want an oscillator circuit that has a very low on-off frequency. Your extremely simple circuit does not do that.

I would use a logic Schmitt-Trigger inverter to make an oscillator and a Cmos Logic IC can make a very low frequency and use a very low power supply current powering it. Some people will use a 555 timer or a Cmos 555 timer IC to make an oscillator.
Look at them in Google.

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#### RAMBO999

Joined Feb 26, 2018
247
Alas, I am just reaching the conclusion that it's not going to work with transistors although it works happily with the TQ2-5V relay. I will take a look at those now actually and see what they are about. I won't completely give up on the BJTs. I may have a flash of inspiration. It's happened before.

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