Skinny thanks for the video as I’m a retired aircraft mechanic and just started teaching my self the electronics side of things and how everything interacts. The resistors kit you mentioned does not exist any longer. Do you know of a good alternative. What supplier/brand would you recommend? Thanks for your time and knowledge Artie 👍👍👍
Thanks for the explanation. There's one thing I don't get. At 5:00 into the video, you explain how the capacitor on the right develops a negative charge which being connected to the base of the transistor on the left, turns it off. What I don't get is - the base of that transistor is also connected via R3 to Vcc (+9v), so - a) why doesn't that keep the base positively biased enough to keep the transistor on? b) why doesn't the aforementioned capacitor dissipate its negative charge through this connection to Vcc (+9v)?
Hi Jason, may be if you have some time to explain a D Flip-Flop using discrete transistors, I've read it's possible to build one using two of them but I find that hard to believe, I'm struggling with this as I want to wire up a 4-bit synchronous binary counter, I don't want to simply go outside and buy a 74HC163, I want to build one. Thank you.
Hello Sir! Although I'm five months late, I think I can answer your question. If you notice carefully, resistor R3 has quite a high value (220K) and resistor R4 has a comparatively smaller value (1K). Since R3 and R4 have different values, the potential drops across R3 and R4 are different (we can calculate potential drop using Ohm's Law: V = I × R. Higher resistance means greater potential drop.). The supply tries to maintain this potential difference across R3 and R4. If we assign V3 to be the potential at the node(node to which the capacitor is attached) after resistor R3 and V4 to be the potential at the node (node to which the capacitor is attached) after resistor R4, the supply tries to maintain a constant value for V4 - V3, that is a constant potential difference. However, when the transistor turns on, V4 becomes zero as it gets directly connected to ground. Since the supply wants to maintain a constant value for V4 - V3, it turns potential V3 into negative. That is how you get a negative potential! I hope I've answered your question properly sir.
The capacitors will never be reverse charged more than a few tenths of a Volt. Hook up a multimeter and check it for yourself. If they were to be reverse charged to 8 Volts, they would probably be destroyed and possibly explode if they are electrolytics.
So the on time relates to one R and the off time relates to the other R. I guess that makes sense. How do I determine turning the LED on and off 10 times in one second equal on and equal off time. Vs 5 on time but the off time be 1.5 portion longer. All in that one second for each output. What is the frequency of this set up? 2 HZ?
Nice video. I am looking for a simple timing circuit with a 3 minute off and a quick pulse on to activate another low current circuit in place of a cheap motion sensor on my owl decoy(to keep rabbits out of my yard) . I presume I can adjust duty cycle by the combination of resistor /capacitor on each side? I wanted to ask before I invest the time to experiment.
I'm looking closely at your infrared led flasher and it doesn't seem to be the same circuit i see only 2 resistors. Do i miss something. Thanks. I would love to make it !!!
Hi, tnx for the video. Where will the current flow if c2 is discharging? Is it flowing thorugh r3 to the diode on the right and then through the transistor on the right and filnally to the ground?
I dont understand how 9v kan get to be only 0,95 volt after R2 and R3? i thought the resistors limited the current and not the voltage. (i am green in this)
The resistors do not limit the voltage. The 'turn-on' voltage of the transistor does. If the voltage ever reaches above 0.95 volts, that voltage would turn on the transistor at the base. Once that transistor is turned on, that base voltage will stay at 0.95V.
I have a number of beginner electronics books and they all use bipolar transistors in the astable multivibrator build. So I looked up youtube videos about the same topic and they all use bipolar transistors. So why does everyone use them? Why not use mosfets or I.G.B.T.s or is there a reason why they are never used?
I would not suggest to operate this kind of circuit at operating voltages higher than 5 V. The issue with this circuit is, that in the switching moment the voltage at the base of the transister gets lower than the avalanche break down voltage of typical silicon BJT, with is in the range of -6 to -8 V. Especially in the case of larger capacitors the power of the excess charge can be high enough to shoot through the base emitter diode of the transistor. In the typical case of lower capacitances the time constant is lower than expected (0.7*R*C). If you need to apply this circuit for higher operating voltages than 5 V (which is almost never the case) you have to select BJT with larger avalanche break down voltages or you can chose low power MOSFETs, which can be operated usually at gate voltages of -10 down to -20 V. In the literature you often find circuit designs in which the BJT are operated beyond their save specification range for their avalanche break down voltage even with capacitors in the range of multiple 100 µF. So they often fail after a while.
I do not understand for the life of me, why 0.95v, why?????????? Isn't the forward voltage of a BJT BE or Vbe, isn't it 0.65v or 0.7v and it being more than that value, doesnt it signal a bad bjt, seriously I'm lost here
Just great. Well put together. I have been searching like forever for a clear and understandable explanation. Thank you!!
Thank you for the video - it was very well explained.
Your're welcome!
Skinny thanks for the video as I’m a retired aircraft mechanic and just started teaching my self the electronics side of things and how everything interacts. The resistors kit you mentioned does not exist any longer. Do you know of a good alternative. What supplier/brand would you recommend? Thanks for your time and knowledge Artie 👍👍👍
Great video & explanation. Please do more ( time permitting of course 👍 )
Thanks for the explanation. There's one thing I don't get. At 5:00 into the video, you explain how the capacitor on the right develops a negative charge which being connected to the base of the transistor on the left, turns it off. What I don't get is - the base of that transistor is also connected via R3 to Vcc (+9v), so -
a) why doesn't that keep the base positively biased enough to keep the transistor on?
b) why doesn't the aforementioned capacitor dissipate its negative charge through this connection to Vcc (+9v)?
Perfect 😉 Always perfect 😎
How do you calculate the resistance if frequency capacitance are given
Hi Jason, may be if you have some time to explain a D Flip-Flop using discrete transistors, I've read it's possible to build one using two of them but I find that hard to believe, I'm struggling with this as I want to wire up a 4-bit synchronous binary counter, I don't want to simply go outside and buy a 74HC163, I want to build one. Thank you.
Hello there, i really like your videos. Can you explain how to understand the minus 8 volts at the capacitor. Thanks!
Hello Sir! Although I'm five months late, I think I can answer your question. If you notice carefully, resistor R3 has quite a high value (220K) and resistor R4 has a comparatively smaller value (1K). Since R3 and R4 have different values, the potential drops across R3 and R4 are different (we can calculate potential drop using Ohm's Law: V = I × R. Higher resistance means greater potential drop.). The supply tries to maintain this potential difference across R3 and R4. If we assign V3 to be the potential at the node(node to which the capacitor is attached) after resistor R3 and V4 to be the potential at the node (node to which the capacitor is attached) after resistor R4, the supply tries to maintain a constant value for V4 - V3, that is a constant potential difference. However, when the transistor turns on, V4 becomes zero as it gets directly connected to ground. Since the supply wants to maintain a constant value for V4 - V3, it turns potential V3 into negative. That is how you get a negative potential! I hope I've answered your question properly sir.
Thank you very much for answering me.
You're welcome😊😊
The capacitors will never be reverse charged more than a few tenths of a Volt. Hook up a multimeter and check it for yourself. If they were to be reverse charged to 8 Volts, they would probably be destroyed and possibly explode if they are electrolytics.
So the on time relates to one R and the off time relates to the other R. I guess that makes sense.
How do I determine turning the LED on and off 10 times in one second equal on and equal off time.
Vs 5 on time but the off time be 1.5 portion longer. All in that one second for each output.
What is the frequency of this set up? 2 HZ?
Didn't know you can use just 2 resistor 5kohm 3.3 v
Nice video. I am looking for a simple timing circuit with a 3 minute off and a quick pulse on to activate another low current circuit in place of a cheap motion sensor on my owl decoy(to keep rabbits out of my yard) . I presume I can adjust duty cycle by the combination of resistor /capacitor on each side? I wanted to ask before I invest the time to experiment.
I'm looking closely at your infrared led flasher and it doesn't seem to be the same circuit i see only 2 resistors. Do i miss something. Thanks. I would love to make it !!!
Hi, tnx for the video. Where will the current flow if c2 is discharging? Is it flowing thorugh r3 to the diode on the right and then through the transistor on the right and filnally to the ground?
Yes
I thought the capacitor would go kaput if both legs were led to voltage?
I dont understand negative voltage
I dont understand how 9v kan get to be only 0,95 volt after R2 and R3? i thought the resistors limited the current and not the voltage. (i am green in this)
The resistors do not limit the voltage. The 'turn-on' voltage of the transistor does. If the voltage ever reaches above 0.95 volts, that voltage would turn on the transistor at the base. Once that transistor is turned on, that base voltage will stay at 0.95V.
Can this circuit work without R1 and R2?
If I remove R1 and R2 what will happen?
If you remove R1, you'll probably destroy a diode. If you remove R2, it will drastically change the frequency and might not work.
Okay. Thanks for the information!
Nice video
Thanks. Glad you liked it.
I have a number of beginner electronics books and they all use bipolar transistors in the astable multivibrator build. So I looked up youtube videos about the same topic and they all use bipolar transistors. So why does everyone use them? Why not use mosfets or I.G.B.T.s or is there a reason why they are never used?
How to select R1 and R4?
If the two LEDs are replaced by two 3v DC motors, will the motors spin alternately? Thanks.
make a video about a simple radio ... i'll be viewing ...
What do you mean by radio? An AM/FM radio receiver, an AM/FM radio transmitter, or something different?
I want you to do a fm transmitor
i mean receiver
I would not suggest to operate this kind of circuit at operating voltages higher than 5 V. The issue with this circuit is, that in the switching moment the voltage at the base of the transister gets lower than the avalanche break down voltage of typical silicon BJT, with is in the range of -6 to -8 V. Especially in the case of larger capacitors the power of the excess charge can be high enough to shoot through the base emitter diode of the transistor. In the typical case of lower capacitances the time constant is lower than expected (0.7*R*C).
If you need to apply this circuit for higher operating voltages than 5 V (which is almost never the case) you have to select BJT with larger avalanche break down voltages or you can chose low power MOSFETs, which can be operated usually at gate voltages of -10 down to -20 V. In the literature you often find circuit designs in which the BJT are operated beyond their save specification range for their avalanche break down voltage even with capacitors in the range of multiple 100 µF. So they often fail after a while.
I do not understand for the life of me, why 0.95v, why?????????? Isn't the forward voltage of a BJT BE or Vbe, isn't it 0.65v or 0.7v and it being more than that value, doesnt it signal a bad bjt, seriously I'm lost here
Are your teachers busy? Spot androidcircuitsolver on google
pov ur teacher sent you this
make car o2 simulator