# PNP pull-down circuit

Hi I've build this simple circuit : I believed that closing the S3 switch (after the C1 was fully charged) the PNP transistor instantaneously would discharged the C1 capacitor. Instead the discharging time is affected by R1 value (as you see on the scope screen). I don't understand why. Any idea? Thanks in advance.

Update

Hi thanks to all for replies. Yes I've understand the issue. The only thing that is not yet clear to me is why, if I remove completely the R1 resistor, the discharging time goes to zero (and this is comprehensible) but I see a very high Ipp (not I)

• R1 limits the transistor's emitter-base current, which in turn limits the emitter-collector current, which is Ib * hFE. Apr 26, 2021 at 14:21
• It might be beneficial to explain why you think the resistor does not affect the discharging time, to explain some misunderstandings better. Apr 26, 2021 at 14:26

I believed that closing the S3 switch (after the C1 was fully charged) the PNP transistor instantaneously would discharged the C1 capacitor.

The transistor is not a perfect conductor. It has some "resistance" (although it is not linear i.e. it doesn't obey Ohm's law). That resistance keeps the capacitor for discharging instantaneously.

Instead the discharging time is affected by R1 value

Yes, the current through the transistor is roughly regulated by the current through the transistor's base. The smaller the base resistor, the more current will flow through the transistor. Typically, somewhere in the neighborhood of 50-200 times more current will flow through the collector than through the base. However, that number drops when the transistor is "saturated".

The bipolar junction transistor is a current amplifier; its collector current is a function of its base current. The resistor limits the base current, therefore the collector current. Check its data sheet for its dc gain characteristic.

Here's a simulation showing how the discharge current into Q1's emitter I(Q1.nE) varies with R1:

simulate this circuit – Schematic created using CircuitLab

Here's the simulation plot result:

This compares R1 = 10k (in blue) vs R1 = 20k (in orange) using CircuitLab's Parameter Sweep simulation option. The top subplot shows V(out), and the bottom subplot shows I(Q1.nE).

When R1 is larger, the base current I(Q1.nB) is smaller for the same voltage difference, resulting in smaller discharge current. A smaller discharge current causes it to take longer to discharge the same initial charge $$\Q = C \cdot V = 12 \ \text{V} \cdot 47 \ \mu \text{F} = 564 \ \mu \text{C}\$$ stored on C1.

• The transistor can be considered as a current sink whose (collector, discharging) current is programmed by the constant base resistor R1 and the varying Vce (Vc1). As a result, the capacitor slowly discharges. The remedy is to swap the emitter and collector... and to connect the resistor through S1 to V1. Then the capacitor would be discharged by a (high enough) constant current. Apr 26, 2021 at 18:26

The base to emitter is the control element. collector to emitter is the load path. When the base receives its voltage from the emitter, a forward drop voltage develops on the base, and when this trans-resistive element (base-emitter junction) establishes current flow, then current will flow through the collector to emitter trans-resistive element. the amount of current in the base to emitter current is proportionally to the current flow of the emitter to collector transresisitive element. This is why they were first called trans-resistors in their beginnings.

Here is a physical construction picture of a transistor: picture from: https://colinjs.com/elec/basicfab/fab.htm

The base to emitter forward bias, changes the resistive barrier from emitter to collector. Unlike FETS, where a trans-conductive, or electron valve device that controls the flow of a layer in constant conduction:

• I wonder what "When the base receives its voltage from the emitter" means... The single base-emitter junction is a "resistive" element (without "trans")... collector-emitter also... All kinds of transistors are transconductive devices (input voltage, output current). Apr 26, 2021 at 15:42
• @Circuitfantasist The base-emitter junction is a diode junction, not a resistive element. Apr 26, 2021 at 15:50
• The base to emitter junction is a diode that shares one of its elements with the collector. So as the base gets forward biased the emitter to collector is getting forward biased. that is why always, Ie=Ib+Ic. This is not a mosfet that has a conductive state between drain and source. And why a bias has to be established between the emitter base junction for the transistor to operate. Apr 26, 2021 at 15:54
• @Hearth, We are talking about a ratio between current and voltage changes called "differential conductance di/du" ("differential resistance du/di"). In the case of the base-emitter junction, both are applied to the same place - the current is through and the voltage is across the junction. In the case of the whole transistor, the (input) voltage is applied at one place (the base-emitter junction) while the current flows through another place (collector-emitter part); hence the name "transconductance". Apr 26, 2021 at 17:53
• @David Mikeska, I try to imagine what "the base to emitter junction is a diode that shares one of its elements with the collector" means but I can't. Also, how "the emitter to collector is getting forward biased"... and "This is not a mosfet that has a conductive state between drain and source". IMO as a current flows between the collector and emitter, there is a conductive state between them. MOSFET also needs biasing. It is in my interest to know what you mean and I am trying to do so... Apr 26, 2021 at 18:04