Need help understanding a circuit

[Zooty]

Hi all, the below circuit is one used in a window motor. I understand what it is doing but i don't understand why the top transistor (mps8599) is a pnp and not an npn. Could we use an npn here? It looks as though the two transistors are isolating the battery from the coil during the field collapse but i don't get why an upside down pnp is used and not just another npn.

[nilrehob]

the way bipolar transistors work,
npn's are preferably used "under" the load
and pnp's "above" the load
take a look at the pics in Bipolar junction transistor - Wikipedia, the free encyclopedia
it might help

btw, the circuit is a half h-bridge.
H-bridge - Wikipedia, the free encyclopedia
a few implementations here:
Power Smart H-Bridge - BEAM Robotics Wiki
notice the last circuit, the PS-driver, would be interesting to try in a SSG-Bedini-like motor.

/Hob

[Zooty]

What would be the difference in the above circuit using another npn instead of a pnp? Is it something to with the way the base current flows? I can see that if we put another npn above the coil then the coil resistance is added to the base resistance.. is that right?

[nilrehob]

exactly!

npn: base-current into tranststor
pnp: base-current out from tranststor

/Hob

[Zooty]

Thanks for your help

[nilrehob]

Next time someone ask I'll remember to go straight to the bottom line about the base-currencies.



/Hob

[Zooty]

The 3k is the base resistance for the pnp and npn drive transistors. Do you know what the 1.5k and 2.2k resistors do in this circuit?

[citfta]

Hi Zooty, the 1.5 k and the 2.2 k resistors provide the proper bias to the bases of the PNP and NPN transistors so that when the 2n3904 is turned off the PNP and NPN transistors will turn off. When the 2n3904 is turned on the bias is reduced on the PNP transistor and raised on the NPN transistor. This turns both of them on. Hope this helps.

Carroll

[faramog]

The arrow tells you the current flow direction: out = pull down; in = pull up. Remember in both cases the base emitter turn on is in the range 0.3-0.7 volts. Except as an amplifier a transistor is essentially a switch with the collector being dragged to the emitter voltage. If you look at the circuit and the pnp is 'switched on' the collector is dragged up to the power rail. In the npn is dragged down to the neg voltage.

Hope that helps

[Zooty]

Hi Zooty, the 1.5 k and the 2.2 k resistors provide the proper bias to the bases of the PNP and NPN transistors so that when the 2n3904 is turned off the PNP and NPN transistors will turn off. When the 2n3904 is turned on the bias is reduced on the PNP transistor and raised on the NPN transistor. This turns both of them on. Hope this helps.

Carroll

If the 1.5k and 2.2 resistors were not there, when the 2n3904 switched off, would the other transistors stay on?

[faramog]

If the 1.5k and 2.2 resistors were not there, when the 2n3904 switched off, would the other transistors stay on?

The bipolar transisors would not switch on at all as the base and emitter would be tied. The resistor is needed to create a volt drop with respect to the current flow to create the turn on voltage.

[Zooty]

The bipolar transisors would not switch on at all as the base and emitter would be tied. The resistor is needed to create a volt drop with respect to the current flow to create the turn on voltage.

When the 2n3904 switches on, wouldn't the current from the battery run through the emitter of the pnp without the 1.5k resistor/track? Isn't that the lowest resistance path for the current? Isn't that what switches the pnp on in the first place? I'm confused

*EDIT*

I just ran a basic simulation and it works without the two resistors mentioned. It occurred to me that these two resistors are there to switch both transistors on at the same strength because the npn in my simulation above would be getting more juice than the pnp. Is that it?

[faramog]

Ahh .. when you said remove , I was rushing at lunch ... and had in my mind, remove resistor, keep connection...Sure if you connect like that it may work. I thought you were asking about the difference between npn & pnp ..

[gyula]

Hi Zooty,

The 1.5k insures a voltage drop for the pnp's base-emitter and the 2.2k does this for the npn. This happens whenever the 2N3904 is switched on by the trigger coil. When the voltage drops across these 2 resistors reach or exceed the typical 0.65-0.7V base-emitter forward voltage, both the pnp and npn transistors can switch on (more or less in the same time) so current can flow in the drive coil. Then, when the 2N3904 switches off, both the pnp and npn transistors also switch off, i.e. both ends of the drive coil get isolated from the battery and the collapsing magnetic field induces a voltage in the drive coil and this voltage (called counter emf) can go back to the battery via the diode bridge when switch is set to the CEMF position.

The pnp type is preferred in place of the top transistor because whenever it is switched off it "shows" its collector-base and collector-emitter junction towards the upper end of the drive coil, just like the npn does the same at the lower end of the drive coil. The reverse voltage rate for collector-base junctions is at least as high as the collector-emitter max rated voltage (normally 80V for the MPS8599) and if you used a npn in place of the top transistor, it would "show" its emitter-base junction (and its emitter-collector junction) towards the coil upper end and the reverse max rated base-emitter voltage is normally between 5-8V only (5V for the MPS8099), above it it breaks down, i.e. start conducting like a Zener diode. So the induced voltage in the coil may find a path and establish current via the base-emitter junction, even though the transistor is off, you would loose energy from the collapsed flux.

I just noticed your simulation circuit. of course it can work without the 2 resistors you mention but now you lost the possibility of adjusting separately the base-emitter bias for the pnp and the npn: when the 2N3904 is switched on, the same current can flow into both bases, dictated by exclusively the 6.8kOhm resistor only. With the 1.5k and 2.2k in place, you individually can influence the base currents by the respective voltage drops across the resistors, this is a needed feature here to achieve the pnp and npn switch on or off simultaneously.

rgds, Gyula

[Zooty]

Thanks to all for the explanations. I get it 100% now. It is almost the same as the SSG circuit but the pnp is missing. So this circuit is primarily used to send the radiant spike back to the source but unlike the SSG, the battery is fully disconnected from the rest of the circuit when it receives the spike? Is there an advantage with this circuit when charging another battery instead of the source compared to the SSG circuit?