Help with voltage divider

Thread Starter

sparkie78

Joined Nov 5, 2016
19
I put together this common astable multivibrator circuit and powered it from my power supply at 5v and it worked fine. I wanted to use it by plugging it in to a wall socket. So I used the voltage divider to drop the voltage to around 10 volts and utilized a zener to provide a steady voltage. With the center tap not connected to anything I read 9.9 volts. But when I connected it to the zener resistor, it immediately dropped to 4.85v with the zener output at 4.53v. And when I connected it to the multivibrator, the zener output dropped to 3.59v and the multivibrator wouldn't work. I've had this kind of trouble before when trying to use a voltage divider. What am I doing wrong? I had to draw the circuit in the 'Paint' program, I hope you can read it allright.Vtg Dvdr ckt.jpg
 

dl324

Joined Mar 30, 2015
12,210
The problem is that your voltage divider isn't "stiff" enough.

Get rid of the 1.5k and 80 ohm resistors and make the 15k resistor 3W or more. Put an electrolytic cap across the zener to reduce ripple.

The 109V you're measuring isn't correct. Peak voltage will be closer to 170V. 15k will give you a zener current of about 10mA
 

Audioguru again

Joined Oct 21, 2019
2,353
Your circuit is missing an important transformer that will prevent you and anybody nearby from being electrocuted.
I show the output of a bridge rectifier as being humps of voltage that go from 0V to 160V over and over. The zener diode does not prevent the voltage from dropping to 0V but a large value filter capacitor parallel to the zener diode will.
Your 1k5 and 80 ohms resistors are not needed.

Where did you find the completely wrong multivibrator circuit? Look here:
 

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xox

Joined Sep 8, 2017
517
I put together this common astable multivibrator circuit and powered it from my power supply at 5v and it worked fine. I wanted to use it by plugging it in to a wall socket. So I used the voltage divider to drop the voltage to around 10 volts and utilized a zener to provide a steady voltage. With the center tap not connected to anything I read 9.9 volts. But when I connected it to the zener resistor, it immediately dropped to 4.85v with the zener output at 4.53v. And when I connected it to the multivibrator, the zener output dropped to 3.59v and the multivibrator wouldn't work. I've had this kind of trouble before when trying to use a voltage divider. What am I doing wrong? I had to draw the circuit in the 'Paint' program, I hope you can read it allright.Vtg Dvdr ckt.jpg
The voltage divider and zener produces a stable voltage REFERENCE. That implies high impedance loads. For normal loads you should use an actual power supply. Or at the very least an op amp driver set up in a follower configuration.
 

DickCappels

Joined Aug 21, 2008
6,776
One problem is that the output resistance of the voltage divider is roughly equal to the parallel combination of the two resistors. By the time you bring the output resistance of the voltage divider down to something that you would be happy with the values of the resistors would be much lower than they are now and they would dissipate a lot more power than they do now.

A small transformer can get you about 9V at a much lower impedance, and it might even save you from a painful shock or worse.

By the way, driving the LEDs in series with the capacitor in this circuit with these voltages pretty much assures that they will be subject to reverse voltage, something that many LED manufactures warn us not to do. You can fix this by putting a diode facing in the opposite direction in parallel with the LED.

Along the same lines, many multivibrator circuits on the internet show "normal" transistors in this circuit with power supplies of 9V or more. the problem is that with power supplies above about 6V when a transistor is switched off the reverse base-emitter voltage exceeds that specified with many general purpose bipolar transistors. This may shorten the life of your circuit by making the low current gain of the transistor less over time.
 

Thread Starter

sparkie78

Joined Nov 5, 2016
19
Your circuit is missing an important transformer that will prevent you and anybody nearby from being electrocuted.
I show the output of a bridge rectifier as being humps of voltage that go from 0V to 160V over and over. The zener diode does not prevent the voltage from dropping to 0V but a large value filter capacitor parallel to the zener diode will.
Your 1k5 and 80 ohms resistors are not needed.

Where did you find the completely wrong multivibrator circuit? Look here:
I copied the multivibrator verbatim from this youtube video
At the risk of sounding stupid, I don't understand how the capacitors function in that last drawing. It seems to me that the negative side of the capacitor should be tied to ground, yet it looks like it's being fed through the 47K resistor, which is connected to the positive rail. And the base of the transistor must have a positive voltage to work, yet it is also connected to the capacitor negative. It appears that the capacitor is connected to the positive rail on both sides.
 

Thread Starter

sparkie78

Joined Nov 5, 2016
19
One problem is that the output resistance of the voltage divider is roughly equal to the parallel combination of the two resistors. By the time you bring the output resistance of the voltage divider down to something that you would be happy with the values of the resistors would be much lower than they are now and they would dissipate a lot more power than they do now.

A small transformer can get you about 9V at a much lower impedance, and it might even save you from a painful shock or worse.

By the way, driving the LEDs in series with the capacitor in this circuit with these voltages pretty much assures that they will be subject to reverse voltage, something that many LED manufactures warn us not to do. You can fix this by putting a diode facing in the opposite direction in parallel with the LED.

Along the same lines, many multivibrator circuits on the internet show "normal" transistors in this circuit with power supplies of 9V or more. the problem is that with power supplies above about 6V when a transistor is switched off the reverse base-emitter voltage exceeds that specified with many general purpose bipolar transistors. This may shorten the life of your circuit by making the low current gain of the transistor less over time.
Geez, I didn't know about any of that. This 'simple' circuit turned out not so simple for me. Thank you for your help.
 

Thread Starter

sparkie78

Joined Nov 5, 2016
19
The problem is that your voltage divider isn't "stiff" enough.

Get rid of the 1.5k and 80 ohm resistors and make the 15k resistor 3W or more. Put an electrolytic cap across the zener to reduce ripple.

The 109V you're measuring isn't correct. Peak voltage will be closer to 170V. 15k will give you a zener current of about 10mA
Thank you for your reply.
 

Audioguru again

Joined Oct 21, 2019
2,353
The base voltage of the transistor is never more than +0.7V because the emitter is at 0V.

In my circuits with the 5V supply, when a transistor is turned off then the collector goes to about +3.7V.
Then the capacitor has 3V across it with the correct polarity.

When this transistor turns on, its collector goes to +0.1V and the LED lights brightly and the capacitor with 3V across it sends the other transistor's base to -3.6V turning off that transistor and LED, and this capacitor is discharged by the 47k base resistor.
 

Thread Starter

sparkie78

Joined Nov 5, 2016
19
The base voltage of the transistor is never more than +0.7V because the emitter is at 0V.

In my circuits with the 5V supply, when a transistor is turned off then the collector goes to about +3.7V.
Then the capacitor has 3V across it with the correct polarity.

When this transistor turns on, its collector goes to +0.1V and the LED lights brightly and the capacitor with 3V across it sends the other transistor's base to -3.6V turning off that transistor and LED, and this capacitor is discharged by the 47k base resistor.
I got it. That makes sense. Thank you for taking time to explain this to me. I really appreciate it!
 

sparky 1

Joined Nov 3, 2018
401
image_2020-12-18_011254.png
edited on 12/20/2020 C4 changed to 0.1uF to reduce initial transient mentioned by Audio Guru. Finding mistakes might make a difference.
Also mentioned that the internet has a lot of errors on this circuit, it can be sensitive to voltage changes. The PNP version had to be corrected.
The simulation shows 7.5mA some of the older circuits may not have enough current, leds today are typically using more current.
 
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Audioguru again

Joined Oct 21, 2019
2,353
You should use circuits from datasheets, not from people who know nuthin, then they post videos with errors on You Tube.
This image from the datasheet of a uA78xx regulator from Texas Instruments shows an output capacitor 10 times the value of yours.
 

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Audioguru again

Joined Oct 21, 2019
2,353
Sparky, the polarity of your capacitors is backwards.
Also, the value of the 100k base resistor is much too high to allow each transistor to saturate. The base current will be only (5V - 0.7V/100k= 43uA. Then the saturated collector current will be only 0.43mA to 0.86mA but the 180 ohms in series with a 2V red LED would try to be 16.6mA.

When you decrease the base resistor value 17.2 times to 34.4 times then for the same timing, the value if each capacitor must be increased that much.
 

DickCappels

Joined Aug 21, 2008
6,776
Or better yet, try SPICE there is a lot to learn about using it but you will find an answer to almost any question you can come up with already of the internet or available from members here.
 
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