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Q&A Change of pins in monostable multivibrator

Your circuit will make sense if you do two things: significantly increase both R3 and R4 reverse D1 or move it to Q1's base Let's first see the role of the resistances R3 and R4. These are ...

posted 3y ago by Circuit fantasist‭  ·  edited 3y ago by Circuit fantasist‭

Answer
#7: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T14:44:56Z (over 3 years ago)
Minor edit
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in these cases - reversed in the first (OP's) case and connected in this direction in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode since the driving circuit is disconnected when the transistor is off.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is "inactive" (zero for the first case and high for the second case). So it should be backward biased in these cases - reversed in the first (OP's) case and connected in this direction in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode since the driving circuit is disconnected when the transistor is off.
#6: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T12:36:04Z (over 3 years ago)
Refining
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in these cases - reversed in the first (OP's) case and connected in this direction in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in these cases - reversed in the first (OP's) case and connected in this direction in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode since the driving circuit is disconnected when the transistor is off.
#5: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T12:33:48Z (over 3 years ago)
Minor edit
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in these cases - reversed in the first (OP's) case and connected in this direction in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
#4: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T12:28:54Z (over 3 years ago)
Minor edit
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2's base (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
#3: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T12:27:16Z (over 3 years ago)
Minor edit
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • They are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • These are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
#2: Post edited by user avatar Circuit fantasist‭ · 2021-09-19T12:26:00Z (over 3 years ago)
Minor edit
  • Your circuit will make sense if you do two things:
  • * significantly increase R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • They are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
  • Your circuit will make sense if you do two things:
  • * significantly increase both R3 and R4
  • * reverse D1 or move it to Q1's base
  • Let's first see the role of the resistances R3 and R4.
  • They are base resistors; so their resistances should be higher than the collector resistances R1 and R2.
  • For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).
  • Let's now see what the role of the diode D1 is.
  • You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.
  • If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.
#1: Initial revision by user avatar Circuit fantasist‭ · 2021-09-19T12:25:14Z (over 3 years ago) 
Your circuit will make sense if you do two things:
 * significantly increase R3 and R4
 * reverse D1 or move it to Q1's base

Let's first see the role of the resistances R3 and R4.

They are base resistors; so their resistances should be higher than the collector resistances R1 and R2.

For example, Q1's base current is roughly IB1 = V1/R4 = 10/100 = 100 mA which is too much. R3 is a heavy load for the circuit output (the common point between R2 and Q1's collector) since R2 and R3 form a voltage divider with ratio of R3/(R2 + R3) = 0.1; so the output voltage will be only 1 V. Also, R3 would be a heavy load for the positive input voltage source connected to Q2 (if you decide to control the circuit this way).

Let's now see what the role of the diode D1 is.

You can toggle this circuit in two ways - by applying a positive voltage to Q2's base (through a resistor) or by applying a zero voltage (ground) to Q1's base. In both cases, the role of the diode is to separate (if needed) the input voltage source from the circuit when its voltage is inactive (zero for the first case and high for the second case). So it should be backward biased in the first (OP's) case and forward biased in the second (Olin's) case.

If you toggle the circuit by a circuit with the so-called *open collector* output (a humble transistor), you do not need a diode.